Answer:
[tex]F = -1.107*10^{-8} N\\|F| = 1.107*10^{-8} N[/tex]
Explanation:
[tex]q_1 = 4.5 * 10^{-9} C[/tex]
[tex]q_2 = -2.8 * 10^{-9} C[/tex]
The distance separating the two charges, r = 3.2 m
According to Coulomb's law of electrostatic attraction, the electrostatic force between the two charges can be given by the formula:
[tex]F = \frac{kq_{1} q_{2} }{r^2}[/tex]
Where [tex]k = 9.0 * 10^9 Nm^2/C^2[/tex]
[tex]F = \frac{9*10^9 * 4.5*10^{-9} * (-2.8*10^{-9}}{3.2^2} \\F = \frac{-113.4*10^{-9}}{10.24}\\F = -11.07 *10^{-9}\\F = -1.107*10^{-8}N[/tex]
Answer:
F = 1.1074 × [tex]10^{-8}[/tex] N
Explanation:
An electrostatic force is either a force or attraction or repulsion between two charges. It can be determined by:
F = [tex]\frac{kq_{1}q_{2} }{r^{2} }[/tex]
where: F is the force, k is a constant, [tex]q_{1}[/tex] is the first charge, [tex]q_{2}[/tex] is the second charge and r the distance between the charges.
Given that: k = 9 × [tex]10^{9}[/tex] N[tex]m^{2}[/tex][tex]C^{-2}[/tex], [tex]q_{1}[/tex] = 4.5 × [tex]10^{-9}[/tex]C, [tex]q_{2}[/tex] = -2.8 × [tex]10^{-9}[/tex]C and r = 3.2 m.
Then,
F = [tex]\frac{9*10^{9}*4.5*10^{-9}*2.8*10^{-9} }{3.2^{2} }[/tex]
= [tex]\frac{1.134*10^{-7} }{10.24}[/tex]
= 1.1074 × [tex]10^{-8}[/tex]
The electrostatic force exerted is 1.1074 × [tex]10^{-8}[/tex] N, and it is a force of attraction.
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.
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
1- A 30 gram bullet travels at 300 m/s. How much kinetic energy does it have?
Answer:
1.35 kJ
Explanation:
KE = ½mv² = ½ × 0.030 kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ
Given:-
Mass (m) of the bullet = 30 gramsVelocity of the bullet (v) = 300 m/sTo Find: Kinetic energy of the bullet.
We know,
Eₖ = ½mv²
where,
Eₖ = Kinetic energy,m = Mass &v = Velocity.thus,
Eₖ = ½(30 g)(300 m/s)²
= (15 g)(90000 m²/s²)
= 1350000 g m²/s²
= 1350 kg m²/s²
= 1350 J
= 1.35 kJ (Ans.)
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.
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.
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
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 flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 30.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 230 complete revolutions.
a) At what rate is the flywheel spinning when the power comes back on?
b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?
Answer:
Explanation:
Hello,
Let's get the data for this question before proceeding to solve the problems.
Mass of flywheel = 40kg
Speed of flywheel = 590rpm
Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.
Time = 30s = 0.5 min
During the power off, the flywheel made 230 complete revolutions.
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + ω₂) / 2] × 0.5
But ∇θ = 230 revolutions
∇θ/t = (530 + ω₂) / 2
230 / 0.5 = (530 + ω₂) / 2
Solve for ω₂
460 = 295 + 0.5ω₂
ω₂ = 330rpm
a)
ω₂ = ω₁ + αt
but α = ?
α = (ω₂ - ω₁) / t
α = (330 - 590) / 0.5
α = -260 / 0.5
α = -520rev/min
b)
ω₂ = ω₁ + αt
0 = 590 +(-520)t
520t = 590
solve for t
t = 590 / 520
t = 1.13min
60 seconds = 1min
X seconds = 1.13min
x = (60 × 1.13) / 1
x = 68seconds
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + 0) / 2] × 1.13
∇θ = 333.35 rev/min
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.
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:
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
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.
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:
a What CE describes the way energy is stored in a sandwich
A 432 kg merry-go-round in the shape of a horizontal disk with a radius of 2.3 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 3.1 rad/s in 2.1 s
Answer:
τ = 1679.68Nm
Explanation:
In order to calculate the required torque you first take into account the following formula:
[tex]\tau=I\alpha[/tex] (1)
τ: torque
I: moment of inertia of the merry-go-round
α: angular acceleration
Next, you use the following formulas for the calculation of the angular acceleration and the moment of inertia:
[tex]\omega=\omega_o+\alpha t[/tex] (2)
[tex]I=\frac{1}{2}MR^2[/tex] (3) (it is considered that the merry-go-round is a disk)
w: final angular speed = 3.1 rad/s
wo: initial angular speed = 0 rad/s
M: mass of the merry-go-round = 432 kg
R: radius of the merry-go-round = 2.3m
You solve the equation (2) for α. Furthermore you calculate the moment of inertia:
[tex]\alpha=\frac{\omega}{t}=\frac{3.1rad/s}{2.1s}=1.47\frac{rad}{s^2}\\\\I=\frac{1}{2}(432kg)(2.3)^2=1142.64kg\frac{m}{s}[/tex]
Finally, you replace the values of the moment of inertia and angular acceleration in the equation (1):
[tex]\tau=(1142.64kgm/s)(1.47rad/s^2)=1679.68Nm[/tex]
The required torque is 1679.68Nm
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
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
3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charge on the paper of the same magnitude, find the weight of the paper in newtons. Remember to convert the distance to meters and show your work here.
Answer:
[tex]\mathbf{1.51\times10^{-15}N}[/tex]
Explanation:
The computation of the weight of the paper in newtons is shown below:
On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.
Therefore the paper charge is
[tex]q_{paper}=-4.1\times10^{-15}C[/tex]
Now the distance from the charge is
[tex]r=1cm=0.01m[/tex]
Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.
[tex]mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}[/tex]
[tex]\Rightarrow W=mg[/tex]
[tex]=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}[/tex]
[tex]=\mathbf{1.51\times10^{-15}N}[/tex]
Dr. Jones performed an experiment to monitor the effects of sunlight exposure on stem density in aquatic plants. In the study, Dr. Jones measured the mass and volume of stems grown in 5 levels of sun exposure. The data is represented below.
Sun exposure Stem mass (g) Stem volume (mL)
30 275 1100
45 415 1215
60 563 1425
75 815 1610
90 954 1742
a. Convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters (mº).
b. Calculate the density of the samples using the equation d = m/v. d = density m = mass (kg) v = volume (m)
c. Convert the density values to scientific notation.
Given that,
Sun exposure = 30%, 45%, 60%, 75%, 90%
Stem mass (g) = 275, 415, 563, 815, 954
Stem volume (ml) = 1100, 1215, 1425, 1610, 1742
(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters
Using conversion of mass
[tex]1\ g=0.001\ kg[/tex]
Conservation of volume
[tex]1\ Lt=0.001\ m^3[/tex]
[tex]1\ mL=1\times10^{-6}\ m^3[/tex]
So, mass in kg
Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954
Volume in m³,
Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742
(b). We need to calculate the density of the samples
Using formula of density
[tex]\rho=\dfrac{m}{V}[/tex]
Where, m = mass
V = volume
If the m = 0.275 kg and V = 0.0011 m³
Put the value into the formula
[tex]\rho=\dfrac{0.275}{0.0011}[/tex]
[tex]\rho=250\ kg/m^3[/tex]
If the m = 0.415 kg and V = 0.001215 m³
Put the value into the formula
[tex]\rho=\dfrac{0.415}{0.001215}[/tex]
[tex]\rho=341.56\ kg/m^3[/tex]
[tex]\rho=342\ kg/m^3[/tex]
If the m = 0.563 kg and V = 0.001425 m³
Put the value into the formula
[tex]\rho=\dfrac{0.563}{0.001425}[/tex]
[tex]\rho=395.08\ kg/m^3[/tex]
If the m = 0.815 kg and V = 0.001610 m³
Put the value into the formula
[tex]\rho=\dfrac{0.815}{0.001610}[/tex]
[tex]\rho=506.21\ kg/m^3[/tex]
If the m = 0.954 kg and V = 0.001742 m³
Put the value into the formula
[tex]\rho=\dfrac{0.954}{0.001742}[/tex]
[tex]\rho=547.6\ kg/m^3[/tex]
[tex]\rho=548\ kg/m^3[/tex]
(c). We need to convert the density values to scientific notation
In scientific notation
The densities are
[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]
Hence, This is required solution.
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)
A piston of small cross-sectional area a is used in a hydraulic press to exert a small force f on the enclosed liquid. A connecting pipe leads to a larger piston of cross sectional area A. a) What force F will the larger piston sustain
Answer:
force on larger piston = [tex]\frac{fA}{a}[/tex]
Explanation:
we label the pistons as piston A and piston B
small piston A:
area = a
force = f
large piston B:
area = A
force = ?
Pascal's law of pressure state that the pressure delivered to a liquid is transmitted undiminished to every portion of the fluid.
we know that pressure = force ÷ area
pressure of piston A = [tex]\frac{f}{a}[/tex]
pressure of piston B = [tex]\frac{(force on piston B)}{A}[/tex]
obeying Pascal's law, the system pressures must be equal. Therefore
[tex]\frac{f}{a} = \frac{(force on piston B)}{A}[/tex]
force on large piston (B) = F = [tex]\frac{fA}{a}[/tex]
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]
A skydiver of mass 87 kg falls straight down. If air resistance exerts 355 N of force on him as he falls, what is the net force on the skydiver?
Answer:
497.6 N
Explanation:
From the question,
The net force on the skydiver = weight of the skydiver- the resistive force of air
F' = W-F...................... Equation 1
Where W = weight of the skydiver, F = resistive force of air.
But,
W = mg................ Equation 2
Where m = mass of the skydiver, g = acceleration due to gravity.
Substitute equation 2 into equation 1
F' = mg-F............ Equation 3
Given: m = 87 kg, F = 355 N, g = 9.8 m/s²
Substitute these values into equation 3
F' = 87(9.8)-355
F' = 852.6-355
F' = 487.6 N
Answer:
498 N down
Explanation:
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].
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 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
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
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.
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.
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