Complete Question
C2B.7
Suppose I drop a 60-kg anvil from rest and from such a height that the anvil reaches a speed of 10 m/s just before hitting the ground. Assume the earth was at rest before I dropped the anvil.
(a) What is the earth's speed just before the anvil hits?
b) How long would it take the earth to travel [tex]1.0 \mu m[/tex] (about a bacterium's width) at this speed?
Answer:
a
[tex]|v_1| = 1.0*10^{-22} \ m/s[/tex]
b
[tex]t = 9.95 *10^{15} \approx 10 *10^{15} \ s[/tex]
Explanation:
From the question we are told that
The mass of the anvil is [tex]m_a = 60\ kg[/tex]
The speed at which it hits the ground is [tex]v = 10 \ m/s[/tex]
Generally the mass of the earth has a value [tex]m_e = 5972*10^{24} \ kg[/tex]
Now according to the principle of momentum conservation
[tex]P_i = P_f[/tex]
Where [tex]P_i[/tex] is the initial momentum which is zero given that both the anvil and the earth are at rest
Now [tex]P_f[/tex] is the final momentum which is mathematically represented as
[tex]P_f = m_a * v + m_e * v_1[/tex]
So
[tex]0 = m_a * v + m_e * v_1[/tex]
substituting values
[tex]0 = 60 * 10 + 5.972 *10^{24} * v_1[/tex]
=> [tex]v_1 = -1.0*10^{-22} \ m/s[/tex]
Here the negative sign show that it is moving in the opposite direction to the anvil
The magnitude of the earths speed is
[tex]|v_1| = 1.0*10^{-22} \ m/s[/tex]
The time it would take the earth is mathematically represented as
[tex]t = \frac{d}{|v_1|}[/tex]
substituting values
[tex]t = \frac{1.0*10^{-6}}{1.0 *10^{-22}}[/tex]
[tex]t = 10 *10^{15} \ s[/tex]
a What CE describes the way energy is stored in a sandwich
A wave travels at a speed of 242 m/s. If the distance between crests is 0.11
m, what is the frequency of the wave? Use |
A. 0.00045 Hz
B. 27 Hz
C. 2200 Hz
D. 190 Hz
Answer:
f = 2200 Hz
Explanation:
It is given that,
Speed of a wave is 242 m/s
The distance between crests is 0.11 m
We need to find the frequency of the wave. The distance between crests is called wavelength of a wave. So,
[tex]v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{242}{0.11}\\\\f=2200\ Hz[/tex]
So, the frequency of the wave is 2200 Hz.
Answer:2200 hz
Explanation:
An astronaut is in an all-metal chamber outside the space station when a solar storm results in the deposit of a large positive charge on the station. Which statement is correct?
a. The astronaut must abandon the chamber immediately to avoid being electrocuted.
b. The astronaut will be safe only if she is wearing a spacesuit made of non-conducting materials.
c. The astronaut does not need to worry: the charge will remain on the outside surface.
d. The astronaut must abandon the chamber if the electric field on the outside surface becomes greater than the breakdown field of air.
d. The astronaut must abandon the chamber immediately because the electric field inside the chamber is non-uniform.
Answer:
c. The astronaut does not need to worry: the charge will remain on the outside surface.
Explanation:
The astronaut need not worry because according to Gauss's law of electrostatic, a hollow charged surface will have a net zero charge on the inside. This is the case of a Gauss surface, and all the charges stay on the surface of the metal chamber. This same principle explains why passengers are safe from electrostatic charges, in an enclosed aircraft, high up in the atmosphere; all the charges stay on the surface of the aircraft.
1. The smallest shift you can reliably measure on the screen is about 0.2 grid units. This shift corresponds to the precision of positions measured with the best Earth-based optical telescopes. If you cannot measure an angle smaller than this, what is the maximum distance at which a star can be located and still have a measurable parallax
Answer:
Explanation:
each grid corresponding 0.1s⁻¹.
0.2grid unit = 0.2×0.1 =0.02s⁻¹
distance of the star from telescope
d = 1/p
d= 1/0.02= 50 par sec
1par sec = 3.26 light year
1 light year = 9.5×10¹²km
3.26ly=3.084×10¹³km
d= 50×3.084×10¹³=1.55×10¹⁵km
What is a possible state for an object in the absence of a net force?
There is only one possible state: constant uniform motion. That means constant speed in a straight line.
(If the constant speed happens to be zero, this description also covers the case where the object isn't moving. That special case is called "at rest".)
Answer:
at restzero accelerationconstant speedHope this helps
A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mass. It has a rotation rate of 8.0 rev/s. Two antennas deploy in the plane of rotation extending from the center of mass of the satellite. Each antenna can be approximated as a rod of mass 150.0 kg and length 6.6 m. What is the new rotation rate of the satellite (in rev/s)
Answer:
Therefore, the new rotation rate of the satellite is 6.3 rev/s.
Explanation:
The expression for conservation of the angular momentum (L) is
[tex]L_{i} = L_{f} I_{i}\times\omega_{i} = I_{f}\times\omega_{f}[/tex]
Where
[tex]I_{i}\ and \ \omega_{i}[/tex] initial moment of inertia and angular velocity
[tex]I_{f}\ and \ \omega_{f}[/tex] is the final moment of inertia and angular velocity
The expression of moment of inertia of the satellite (a solid sphere) is
[tex]I_{i} = \frac{2}{5}m_{s}r^{2}[/tex]
Where [tex]m_{s}[/tex] is the satellite mass
r is the radus of the sphere
Substititute 1900kg for m and 4.6m for r
[tex]I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}[/tex]
The final moment of inertia of the satellite about the centre of mass
[tex]I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}[/tex]
Where [tex]m_{x}[/tex] is the antenna's mass and
I is the length of the antenna
[tex]I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}[/tex]
So, the Final rotation rate of the satellite is:
[tex]I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s[/tex]
Therefore, the new rotation rate of the satellite is 6.3 rev/s.
wo parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude of the electric field strength between them, if the potential 7.05 cm from the zero volt plate (and 2.95 cm from the other) is 393 V?
Answer:
-18896.49 V/m
Explanation:
Distance between the two plates = 10 cm = 10 x [tex]10^{-2}[/tex] m = 0.1 m
Also, one of the plates is taken as zero volt.
a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V
b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V
Potential field strength = -dV/dx
where dV is voltage difference between these points,
dx is the difference in distance between these points
For the first case above,
potential field strength = -393/0.0705 = -5574.46 V/m
For the second case ,
potential field strength = -393/0.0295 = -13322.03 V/m
Magnitude of the field strength across the plates will be
-5574.46 + (-13322.03) = -5574.46 + 13322.03 = -18896.49 V/m
High voltage power is often carried in wire bundles made up of individual strands. In your initial post to the discussion, discuss the forces on the strands of wire due to the current flowing through them. What would happen if the force acted opposite of the known behavior
Answer:
More current will be loss through the metal wire strands if the force on them was repulsive, and more stress will be induced on the wire strands due to internal and external flexing.
Explanation:
A wire bundle is made up of wire strands bunched together to increase flexibility that is not always possible in a single solid metal wire conductor. In the strands of wire carrying a high voltage power, each strand carries a certain amount of current, and the current through the strands all travel in the same direction. It is know that for two conductors or wire, separated by a certain distance, that carries current flowing through them in the same direction, an attractive force is produced on these wires, one on the other. This effect is due to the magnetic induction of a current carrying conductor. The forces between these strands of the high voltage wire bundle, pulls the wire strands closer, creating more bond between these wire strands and reducing internal flex induced stresses.
If the case was the opposite, and the wires opposed themselves, the effect would be that a lot of cost will be expended in holding these wire strands together. Also, stress within the strands due to the repulsion, will couple with external stress from the flexing of the wire, resulting in the weakening of the material.
The biggest problem will be that more current will be lost in the wire due to increased surface area caused by the repulsive forces opening spaces between the strand. This loss is a s a result of the 'skin effect' in wire transmission, in which current tends to flow close to the surface of the metal wire. The skin effect generates power loss as heat through the exposed surface area.
When a charge q is placed at a certain point in an electric field, it experiences a force toward the west of magnitude F. If instead a change 2q were placed at that same point what force would it experience?
Answer:
If instead a charge 2q were placed at that same point the force will be 2F.
Explanation:
The electric force is equal to:
[tex] F = q*E [/tex] (1)
Where:
F: is the electric force
q: is the charge
E: is the electric field
We can see that in equation (1) the electric force (F) is proportional to the charge q, thus, if now the charge it's the double (2q) then the force will be the double too:
Initially:
[tex] F_{1} = q_{1}*E [/tex]
Now,
[tex](2q_{1}})*E = 2(q_{1}*E) = 2F_{1}[/tex]
Therefore, if instead a charge 2q were placed at that same point the force will be 2F.
I hope it helps you!
During a particular time interval, the displacement of an object is equal to zero. Must the distance traveled by this object also equal to zero during this time interval? Group of answer choices
Answer: No, we can have a displacement equal to 0 while the distance traveled is different than zero.
Explanation:
Ok, let's write the definitions:
Displacement: The displacement is equal to the difference between the final position and the initial position.
Distance traveled: Total distance that you moved.
So, for example, if at t = 0s, you are in your house, then you go to the store, and then you return to your house, we have:
The displacement is equal to zero, because the initial position is your house and the final position is also your house, so the displacement is zero.
But the distance traveled is not zero, because you went from you traveled the distance from your house to the store two times.
So no, we can have a displacement equal to zero, but a distance traveled different than zero.
A box on a ramp is connected by a rope to a winch. The winch is turned so that the box moves down the ramp at a constant speed. The box experiences kinetic friction with the ramp. Which forces on the box do zero work as the box moves down the ramp?
a. Weight (gravitational force)
b. Normal force
c. Kinetic friction force
d. Tension force
e. None
Answer:
Option B:
The normal force
Explanation:
The normal force does no work as the box slides down the ramp.
Work can only be done when the force succeeds in moving the object in the direction of the force.
All the other forces involved have a component that is moving the box in their direction.
However, the normal force does not, as it points downwards into the ramp. Since the normal force is pointing into the ramp, and the box is sliding down the ramp, we can say that no work is being done by the normal force because the box is not moving in its direction (which would have been the box moving into the ramp)
At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the left side of the pivot an adult pushes straight down on the teeter-totter with a force of 151 N. Part A In which direction does the teeter-totter rotate if the adult applies the force at a distance of 3.0 m from the pivot?Part B
In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.5 m from the pivot?
(clockwise/counterclockwise)
Part C
In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.0 m from the pivot?
(clockwise/counterclockwise)
Answer:
By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.
here weight of the child =21kgx9.8m/s2 = 205.8N
the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.
torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.
net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.
b) Ta = 2.5x151 = 377.5N-m
Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.
c)Ta = 2x151 = 302N-m
Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.
In 1898, the world land speed record was set by Gaston Chasseloup-Laubat driving a car named Jeantaud. His speed was 39.24 mph (63.15 km/h), much lower than the limit on our interstate highways today. Repeat the calculations of Example 2.7 (assume the car accelerates for 6 miles to get up to speed, is then timed for a one-mile distance, and accelerates for another 6 miles to come to a stop) for the Jeantaud car. (Assume the car moves in the +x direction.)
Find the acceleration for the first 6 miles.
Answer:
the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]
Explanation:
Given that:
the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.
So; the acceleration for the first 6 miles can be calculated by using the formula:
v₂² = v₁² + 2a (Δx)
Making acceleration a the subject of the formula in the above expression ; we have:
v₂² - v₁² = 2a (Δx)
[tex]a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}[/tex]
[tex]a = \dfrac{(63.15 \ km/s)^2 - (0 \ m/s)^2 }{2 (6 \ miles)}[/tex]
[tex]a = \dfrac{(17.54 \ m/s)^2 - (0 \ m/s)^2 }{2 (9.65*10^3 \ m)}[/tex]
[tex]a =0.0159 \ m/s^2[/tex]
Thus;
Assume the car moves in the +x direction;
the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]
The Sun's energy comes from which nuclear reaction?
A. Nuclear fission
B. Gamma decay
C. Positron emission
D. Nuclear fusion
SUBMIT
Sun's energy comes from the nuclear fusion taking place inside. In nuclear fusion two light nuclei fuses together to form a heavy nuclei with the release of greater amount of energy.
What is nuclear fusion :Nuclear fusion is the process of combining two light nuclei to form a heavy nuclei. In this nuclear process, tremendous energy is released. This is the source of heat and light in stars.
On the other hand, nuclear fission is the process of breaking of a heavy nuclei into two lighter nuclei. Fission also produces massive energy. But in comparison, more energy is produced by nuclear fusion.
Nuclear fission is used in nuclear power generators. The light energy and heat energy comes form the nuclear fusion of hydrogens to form helium nuclei. Hence, option D is correct.
Find more on nuclear fusion:
https://brainly.com/question/12701636
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A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. When the flask and mercury are warmed together to a temperature of 52.0°C , a volume of 8.50 cm^3 of mercury overflows the flask.Required:If the coefficient of volume expansion of mercury is βHg = 1.80×10^−4 /K , compute βglass, the coefficient of volume expansion of the glass. Express your answer in inverse kelvins.
Answer:
the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
Explanation:
Given that:
Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³
temperature of the glass flask and mercury= 1.00° C
After heat is applied ; the final temperature = 52.00° C
Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C
Volume of the mercury overflow = 8.50 cm^3 = 8.50 × 10⁻⁶ m³
the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K
The increase in the volume of the mercury = 10⁻³ m³ × 51.00 × 1.80 × 10⁻⁴
The increase in the volume of the mercury = [tex]9.18*10^{-6} \ m^3[/tex]
Increase in volume of the glass = 10⁻³ × 51.00 × [tex]\beta _{glass}[/tex]
Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask
the mercury overflow = [tex](9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3[/tex]
[tex]8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )[/tex]
[tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
Thus; the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and 0.1250 T, respectively. The particle passes out of the electric field, but the magnetic field continues, and the particle makes a semicircle of diameter 25.05 cm.
Part A. What is the particle's charge-to-mass ratio?
Part B. Can you identify the particle?
a. can't identify
b. proton
c. electron
d. neutron
Answer:
Explanation:
Given that
The electric fields of strengths E = 187,500 V/m and
and The magnetic fields of strengths B = 0.1250 T
The diameter d is 25.05 cm which is converted to 0.2505m
The radius is (d/2)
= 0.2505m / 2 = 0.12525m
The given formula to find the magnetic force is [tex]F_{ma}=BqV---(i)[/tex]
The given formula to find the electric force is [tex]F_{el}=qE---(ii)[/tex]
The velocity of electric field and magnetic field is said to be perpendicular
Electric field is equal to magnectic field
Equate equation (i) and equation (ii)
[tex]Bqv=qE\\\\v=\frac{E}{B}[/tex]
[tex]v=\frac{187500}{0.125} \\\\v=15\times10^5m/s[/tex]
It is said that the particles moves in semi circle, so we are going to consider using centripetal force
[tex]F_{ce}=\frac{mv^2}{r}---(iii)[/tex]
magnectic field is equal to centripetal force
Lets equate equation (i) and (iii)
[tex]Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg[/tex]
Therefore, the particle's charge-to-mass ratio is [tex]958.1\times10^5C/kg[/tex]
b)
To identify the particle
Then 1/ 958.1 × 10⁵ C/kg
The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg
Therefore the particle is proton.
A brass ring of diameter 10.00 cm at 19.0°C is heated and slipped over an aluminum rod with a diameter of 10.01 cm at 19.0°C. Assuming the average coefficients of linear expansion are constant. What if the aluminum rod were 10.02 cm in diameter?
Answer:
the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]
Explanation:
The change in length of a bar can be expressed with the relation;
[tex]\Delta L = L_f - L_i[/tex] ---- (1)
Also ; the relative or fractional increase in length is proportional to the change in temperature.
Mathematically;
ΔL/L_i ∝ kΔT
where;
k is replaced with ∝ (the proportionality constant )
[tex]\dfrac{ \Delta L}{L_i}=\alpha \Delta T[/tex] ---- (2)
From (1) ;
[tex]L_f = \Delta L + L_i[/tex] --- (3)
from (2)
[tex]{ \Delta L}=\alpha \Delta T*{L_i}[/tex] ---- (4)
replacing (4) into (3);we have;
[tex]L_f =(\alpha \Delta T*{L_i} ) + L_i[/tex]
On re-arrangement; we have
[tex]L_f = L_i + \alpha L_i (\Delta T )[/tex]
from the given question; we can say that :
[tex](L_f)_{brass}}} = (L_f)_{Al}[/tex]
So;
[tex]L_{brass} + \alpha _{brass} L_{brass}(\Delta T) = L_{Al} + \alpha _{Al} L_{Al}(\Delta T)[/tex]
Making the change in temperature the subject of the formula; we have:
[tex]\Delta T = \dfrac{L_{Al}-L_{brass}}{\alpha _ {brass} L_{brass}-\alpha _{Al}L_{Al}}[/tex]
where;
[tex]L_{Al}[/tex] = 10.02 cm
[tex]L_{brass}[/tex] = 10.00 cm
[tex]\alpha _{brass}[/tex] = 19 × 10⁻⁶ °C ⁻¹
[tex]\alpha_{Al}[/tex] = 24 × 10⁻⁶ °C ⁻¹
[tex]\Delta T = \dfrac{10.02-10.00}{19*10^{-6} \ \ {^0}C^{-1} *10.00 -24*10^{-6} \ \ {^0}C^{-1} *10.02}[/tex]
[tex]\Delta T[/tex] = −396.1965135 ° C
[tex]\Delta T[/tex] ≅ −396.20 °C
Given that the initial temperature [tex]T_i = 19^0 C[/tex]
Then ;
[tex]\Delta T = T_f - T_i[/tex]
[tex]T_f = \Delta T + T_I[/tex]
Thus;
[tex]T_f =(-396.20 + 19.0)^0 C[/tex]
[tex]\mathbf{T_f = -377.2^0 C}[/tex]
Thus; the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]
A light source simultaneously emits light of two wavelengths, 480 nm and 560 nm, respectively. The source is used in a double-slit interference experiment where the slit spacing is a 0.040 mm, and the distance between double slits and the screen is 1.2 m. What is the separation between the second-order bright fringes of the two wavelengths as they appear on the screen
Answer:
0.48 cm
Explanation:
given data
wavelength = 480 nm
wavelength = 560 nm
slit spacing = 0.040 mm
distance between double slits and the screen = 1.2 m
solution
we know that (1 nm= [tex]10^{-9}[/tex] m)
we wil take here equation of equations of interference that is
ym = R × (m λ)/d ..........................1
here m = 2 R i.e distance of screen and slit
so put here value and we get
separation between the second-order bright fringes = 0.48 cm
A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.
Required:
a. If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?
b. What does this tell you about the shape of the nearsighted eye?
1. This distance is greater than for the normal eye.
2. This distance is shorter than for the normal eye.
Answer:
a) The distance from the cornea vertex to the retina is 2.37 cm
b) This distance is shorter than for the normal eye.
Explanation:
a) Let refractive index of air,
n(air) = x = 1
Let refractive index of lens,
n(lens) = y = 1.4
Object distance, s = 36 cm
Radius of curvature, R = 0.65 cm
The distance from the cornea vertex to the retina is the image distance because image is formed in the retina.
Image distance, s' = ?
(x/s) + (y/s') = (y-x)/R
(1/36) + (1.4/s') = (1.4 - 1)/0.65
1.4/s' = 0.62 - 0.028
1.4/s' = 0.592
s' = 1.4/0.592
s' = 2.37 cm
Distance from the cornea vertex to the retina is 2.37 cm
(b) For a normal eye, the distance between the cornea vertex and the retina is 2.60 cm. Since 2.37 < 2.60, this distance is shorter than for normal eye.
A 50-loop circular coil has a radius of 3 cm. It is oriented so that the field lines of a magnetic field are perpendicular to the coil. Suppose that the magnetic field is varied so that B increases from 0.10 T to 0.35 T in 2 ms. Find the induced emf in the coil.
Answer:
-17.8 V
Explanation:
The induced emf in a coil is given as:
[tex]E = \frac{-NdB\pi r^2}{dt}[/tex]
where N = number of loops
dB = change in magnetic field
r = radius of coil
dt = elapsed time
From the question:
N = 50
dB = final magnetic field - initial magnetic field
dB = 0.35 - 0.10 = 0.25 T
r = 3 cm
dt = 2 ms = 0.002 secs
Therefore, the induced emf is:
[tex]E = \frac{-50 * 0.25 * \pi * 0.03^2}{0.002} \\E = -17.8 V[/tex]
Note: The negative sign implies that the EMf acts in an opposite direction to the change in magnetic flux.
A disk-shaped merry-go-round of radius 2.83 m and mass 185 kg rotates freely with an angular speed of 0.701 rev/s . A 63.4 kg person running tangential to the rim of the merry-go-round at 3.51 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. Part A What is the final angular speed of the merry-go-round
Answer:
The final angular speed of the merry-go-round is [tex]3.118\,\frac{rad}{s}[/tex] [tex]\left(0.496\,\frac{rev}{s} \right)[/tex].
Explanation:
Given the absence of external forces, the final angular speed of the merry-go-round can be determined with the resource of the Principle of Angular Momentum Conservation, which is described in this case as:
[tex]I_{g, m} \cdot \omega_{o,m} + I_{g, p}\cdot \omega_{o,p} = (I_{g,m} + I_{g, p})\cdot \omega_{f}[/tex]
Where:
[tex]I_{g,m}[/tex] - Moment of inertia of the merry-go-round with respect to its axis of rotation, measured in [tex]kg\cdot m^{2}[/tex].
[tex]I_{g,p}[/tex] - Moment of inertia of the person with respect to the axis of rotation of the merry-go-round, measured in [tex]kg\cdot m^{2}[/tex].
[tex]\omega_{o, m}[/tex] - Initial angular speed of the merry-go-round with respect to its axis of rotation, measured in radians per second.
[tex]\omega_{o,p}[/tex] - Initial angular speed of the merry-go-round with respect to the axis of rotation of the merry-go-round, measured in radians per second.
[tex]\omega_{f}[/tex] - Final angular speed of the merry-go-round-person system, measured in radians per second.
The final angular speed is cleared:
[tex]\omega_{f} = \frac{I_{g,m}\cdot \omega_{o,m}+I_{g,p}\cdot \omega_{o,p}}{I_{g,m}+I_{g,p}}[/tex]
Merry-go-round is modelled as uniform disk-like rigid body, whereas the person can be modelled as a particle. The expressions for their moments of inertia are, respectively:
Merry-go-round
[tex]I_{g,m} = \frac{1}{2}\cdot M \cdot R^{2}[/tex]
Where:
[tex]M[/tex] - The mass of the merry-go-round, measured in kilograms.
[tex]R[/tex] - Radius of the merry-go-round, measured in meters.
Person
[tex]I_{g,p} = m\cdot r^{2}[/tex]
Where:
[tex]m[/tex] - The mass of the person, measured in kilograms.
[tex]r[/tex] - Distance of the person with respect to the axis of rotation of the merry-go-round, measured in meters.
If [tex]M = 185\,kg[/tex], [tex]m = 63.4\,kg[/tex], [tex]R = r = 2.83\,m[/tex], the moments of inertia are, respectively:
[tex]I_{g,m} = \frac{1}{2}\cdot (185\,kg)\cdot (2.83\,m)^{2}[/tex]
[tex]I_{g,m} = 740.823\,kg\cdot m^{2}[/tex]
[tex]I_{g,p} = (63.4\,kg)\cdot (2.83\,m)^{2}[/tex]
[tex]I_{g,p} = 507.764\,kg\cdot m^{2}[/tex]
The angular speed experimented by the person with respect to the axis of rotation of the merry-go-round is:
[tex]\omega_{o,p} = \frac{v_{p}}{r}[/tex]
[tex]\omega_{o,p} = \frac{3.51\,\frac{m}{s} }{2.83\,m}[/tex]
[tex]\omega_{o,p} = 1.240\,\frac{rad}{s}[/tex]
Given that [tex]I_{g,m} = 740.823\,kg\cdot m^{2}[/tex], [tex]I_{g,p} = 507.764\,kg\cdot m^{2}[/tex], [tex]\omega_{o,m} = 4.405\,\frac{rad}{s}[/tex] and [tex]\omega_{o,p} = 1.240\,\frac{rad}{s}[/tex], the final angular speed of the merry-go-round is:
[tex]\omega_{f} = \frac{(740.823\,kg\cdot m^{2})\cdot \left(4.405\,\frac{rad}{s} \right)+(507.764\,kg\cdot m^{2})\cdot \left(1.240\,\frac{rad}{s} \right)}{740.823\,kg\cdot m^{2}+507.764\,kg\cdot m^{2}}[/tex]
[tex]\omega_{f} = 3.118\,\frac{rad}{s}[/tex]
[tex]\omega_{f} = 0.496\,\frac{rad}{s}[/tex]
The final angular speed of the merry-go-round is [tex]3.118\,\frac{rad}{s}[/tex] [tex]\left(0.496\,\frac{rev}{s} \right)[/tex].
When mapping the equipotentials on the plates with different electrode configurations you may find that some have significant areas with uniform distribution of the equipotential lines. If the distance between such lines is 0.5 cm, what is the electric field there (in units SI)
Answer:
E = V/5 x10⁻³
Explanation:
if the potential difference is V
then electric field E is given by
E = V/d
d = 0.5cm = 5 x 10⁻³m
E = V/5 x10⁻³
An electron of mass 9.11 x 10^-31 kg has an initial speed of 4.00 x 10^5 m/s. It travels in a straight line, and its speed increases to 6.60 x10^5 m/s in a distance of 5.40 cm. Assume its acceleration is constant.
Required:
a. Determine the magnitude of the force exerted on the electron.
b. Compare this force (F) with the weight of the electron (Fg), which we ignored.
Answer:
a. F = 2.32*10^-18 N
b. The force F is 2.59*10^11 times the weight of the electron
Explanation:
a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:
[tex]v^2=v_o^2+2ax[/tex] (1)
v: final speed of the electron = 6.60*10^5 m/s
vo: initial speed of the electron = 4.00*10^5 m/s
a: acceleration of the electron = ?
x: distance traveled by the electron = 5.40cm = 0.054m
you solve the equation (2) for a and replace the values of the parameters:
[tex]a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}[/tex]
Next, you use the second Newton law to calculate the force:
[tex]F=ma[/tex]
m: mass of the electron = 9.11*10^-31kg
[tex]F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N[/tex]
The magnitude of the force exerted on the electron is 2.32*10^-18 N
b. The weight of the electron is given by:
[tex]F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N[/tex]
The quotient between the weight of the electron and the force F is:
[tex]\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}[/tex]
The force F is 2.59*10^11 times the weight of the electron
Underline your answer for each situation: If you advance the movie one frame, the knot at point A would be a) in the same place b) higher c) lower d) to the right e) to the left If the person generates a new pulse like the first but more quickly, the pulse would be a) same size b) wider c) narrower If the person generates another pulse like the first but he moves his hand further, the pulse would be a) same size b) taller c) shorter If the person generates another pulse like the first but the rope is tightened, the pulse will move a) at the same rate b) faster c) slower Now the person moves his hand back and forth several times to produce several waves. You freeze the movie and get this snapshot. Underline your answer for each situation: If you advance the movie one frame, the knot at point A would be a) in the same place b) higher c) lower d) to the right e) to the left If you advance the movie one frame, the pattern of the waves will be _________relative to the hand. a) in the same place b) shifted right c) shifted left d) shifted up e) shifted down If the person starts over and moves his hand more quickly, the peaks of the waves will be a) the same distance apart b) further apart c) closer together If you lower the frequency of a wave on a string you will lower its speed. b) increase its wavelength. c). lower its amplitude. d) shorten its period.
Answer:
a) correct answer is b higher , b) correct answer is b higher , c) correct answer is b faster , d) traveling wave , e)
Explanation:
A traveling wave is described by the expression
y = A sin (kx - wt)
where k is the wave vector and w is the angular velocity
let's examine every situation presented
a) a new faster pulse is generated
A faster pulse should have a higher angular velocity
equal speed is related to the period and frequency
w = 2π f = 2π / T
therefore in this case the period must decrease so that the angular velocity increases
the correct answer is c narrower
b) Generate a pulse, but move your hand more.
Moving the hand increases the amplitude (A) of the pulse
the correct answer is b higher
c) generates a pulse but the force is tightened
Set means that more tension force is applied to the string, so the velicate changes
v = √ (T /μ)
the correct answer is b faster
d) move your hand back and forth
in this case you would see a pulse series whose sum corresponds to a traveling wave
e) Advance a frame the movie
in this case the wave will be displaced a whole period to the right
the correct answer is b
f) move your hand faster
the waves will have a maximum fast, so they are closer
answer C
g) decrease wave frequency
Since the speed of the wave is a constant m ak, decreasing the frequency must increase the wavelength to keep the velocity constant.
the correct answer is b increases its wavelength
Space vehicles traveling through Earth's radiation belts can intercept a significant number of electrons. The resulting charge buildup can damage electronic components and disrupt operations. Suppose a spherical metallic satellite 1.7 m in diameter accumulates 3.1 µC of charge in one orbital revolution. (a) Find the resulting surface charge density. (b) Calculate the magnitude of the electric field just outside the surface of the satellite, due to the surface charge.
Answer:
(a) σ = 3.41*10⁻7C/m^2
(b) E = 38,530.1 N/C
Explanation:
(a) In order to calculate the resulting surface charge density, you use the following formula:
[tex]\sigma=\frac{Q}{S}[/tex] (1)
σ: surface charge density
Q: charge of the satellite = 3.1 µC = 3.1*10^-6C
S: surface area of the satellite
The satellite has a spherical form, then, the area of the surface is given by:
[tex]S=4\pi r^2[/tex] (2)
r: radius of the satellite = d/2 = 1.7m/2 = 0.85m
You replace the equation (2) into the equation (1) and solve for the surface charge density:
[tex]\sigma=\frac{3.1*10^{-6}C}{4\pi (0.85m)^2}=3.41*10^{-7}\frac{C}{m^2}[/tex]
The surface charge density acquired by the satellite on one orbit is 3.41*10⁻7C/m^2
(b) The electric field just outside the surface is calculate d by using the following formula:
[tex]E=k\frac{Q}{R^2}[/tex] (3)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
R: radius of the satellite = 0.85m
[tex]E=(8.98*10^9Nm^2/C^2)\frac{3.1*10^{-6}C}{(0.85m)^2}=38530.1\frac{N}{C}[/tex]
The magnitude of the electric field just outside the sphere is 38,530.1 N/C
The block on this incline weighs 100 kg and is connected by a cable and pulley to a weight of 10 kg. If the coefficient of friction between the block and incline is o.3, the block will:
Answer:
a. 94.54 N
b. 0.356 m/s^2
Explanation:
Given:-
- The mass of the inclined block, M = 100 kg
- The mass of the vertically hanging block, m = 10 kg
- The angle of inclination, θ = 20°
- The coefficient of friction of inclined surface, u = 0.3
Find:-
a) The magnitude of tension in the cable
b) The acceleration of the system
Solution:-
- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.
- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.
- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:
[tex]T - m*g = m*a\\\\[/tex] ... Eq 1
Where,
a: The acceleration of the system
- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.
- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.
- We will employ the static equilibrium of the inclined block in the normal direction and we have:
[tex]N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )[/tex]
- The frictional force ( F ) is proportional to the contact force ( N ) as follows:
[tex]F = u*N\\\\F = u*M*g *cos ( Q )[/tex]
- Now we will apply the Newton's second law of motion parallel to the plane as follows:
[tex]M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q) = M*a\\[/tex] .. Eq2
- Add the two equation, Eq 1 and Eq 2:
[tex]M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}[/tex]
- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).
- Using equation 1, we determine the tension ( T ) in the cable as follows:
[tex]T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N[/tex]
Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above the reference point. If the mass of the car is 1000-kg, and the car traveled a distance of 400 m, estimate the magnitude of the frictional force between the car and the track.
Answer:
The magnitude of the frictional force between the car and the track is 367.763 N.
Explanation:
The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:
[tex]U_{g,1} = U_{g,2} + W_{dis}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]W_{dis}[/tex] - Dissipated work due to friction, measured in joules.
Gravitational potential energy is described by the following formula:
[tex]U = m \cdot g \cdot y[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]y[/tex] - Height with respect to reference point, measured in meters.
In addition, dissipated work due to friction is:
[tex]W_{dis} = f \cdot \Delta s[/tex]
Where:
[tex]f[/tex] - Friction force, measured in newtons.
[tex]\Delta s[/tex] - Travelled distance, measured in meters.
Now, the energy equation is expanded and frictional force is cleared:
[tex]m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s[/tex]
[tex]f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}[/tex]
If [tex]m = 1000\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 40\,m[/tex], [tex]y_{2} = 25\,m[/tex] and [tex]\Delta s = 400\,m[/tex], then:
[tex]f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}[/tex]
[tex]f = 367.763\,N[/tex]
The magnitude of the frictional force between the car and the track is 367.763 N.
In 1949, an automobile manufacturing company introduced a sports car (the "Model A") which could accelerate from 0 to speed v in a time interval of Δt. In order to boost sales, a year later they introduced a more powerful engine (the "Model B") which could accelerate the car from 0 to speed 2.92v in the same time interval. Introducing the new engine did not change the mass of the car. Compare the power of the two cars, if we assume all the energy coming from the engine appears as kinetic energy of the car.
Answer: [tex]\frac{P_B}{P_A}[/tex] = 8.5264
Explanation: Power is the rate of energy transferred per unit of time: P = [tex]\frac{E}{t}[/tex]
The energy from the engine is converted into kinetic energy, which is calculated as: [tex]KE = \frac{1}{2}.m.v^{2}[/tex]
To compare the power of the two cars, first find the Kinetic Energy each one has:
K.E. for Model A
[tex]KE_A = \frac{1}{2}.m.v^{2}[/tex]
K.E. for model B
[tex]KE_B = \frac{1}{2}.m.(2.92v)^{2}[/tex]
[tex]KE_B = \frac{1}{2}.m.8.5264v^{2}[/tex]
Now, determine Power for each model:
Power for model A
[tex]P_{A}[/tex] = [tex]\frac{m.v^{2} }{2.t}[/tex]
Power for model B
[tex]P_B = \frac{m.8.5264.v^{2} }{2.t}[/tex]
Comparing power of model B to power of model A:
[tex]\frac{P_B}{P_A} = \frac{m.8.5264.v^{2} }{2.t}.\frac{2.t}{m.v^{2} }[/tex]
[tex]\frac{P_B}{P_A} =[/tex] 8.5264
Comparing power for each model, power for model B is 8.5264 better than model A.
Imagine that you want to make sure the battery for your string of lights will last as long as possible. A battery will last longer if it powers a circuit with low current. How could you hook up a battery and 2 light bulbs so the least amount of current flows through the battery
Answer:
Hooking up the bulb to the battery in a series arrangement will draw the least amount of current.
Explanation:
In this case now, the bulb will serve as the load on the battery (resistance).
For the current to last longer, the least amount of energy must be drawn.
The least amount of energy will be drawn when the arrangement provides the maximum resistance possible.
Let us take the resistance of each bulb as 'R'
If we arrange the bulbs in series, then, the total resistance will be
Rt = R + R = 2R
at a EMF of V from the battery, current I through the battery will be
I = V/2R
If we arrange the bulbs in parallel, then , the total resistance will be
1/Rt = 1/R + 1/R
1/Rt = 2/R
therefore
Rt = R/2
at an EMF of V from the battery, the current I that will be drawn through the battery will be
I = 2V/R
we see that arranging the bulbs in parallel draws 4 times the current compared to arranging the bulb in series
From the above, we see that arranging the bulbs in series provides the maximum resistance, which means a lesser amount of current is drawn from the battery
A circuit contains two elements, but it is not known if they are L, R, or C. The current in this circuit when connected to a 120-V 60 Hz source is 5.3 A and lags the voltage by 65∘.
Part A. What are the two elements?
Part B. What are their values?
Express your answer using two significant figures
Answer:
the two elements are resistor R and inductor L
answers in two significant figures
R = 9.6Ω
L = 54mH
Explanation: