Answer:
a) v = 11.24 m / s , θ = 17.76º b) Kf / K₀ = 0.4380
Explanation:
a) This is an exercise in collisions, therefore the conservation of the moment must be used
Let's define the system as formed by the two cars, therefore the forces during the crash are internal and the moment is conserved
Recall that moment is a vector quantity so it must be kept on each axis
X axis
initial moment. Before the crash
p₀ₓ = m₁ v₁
where v₁ = -25.00 me / s
the negative sign is because it is moving west and m₁ = 900 kg
final moment. After the crash
[tex]p_{x f}[/tex]= (m₁ + m₂) vx
p₀ₓ = p_{x f}
m₁ v₁ = (m₁ + m₂) vₓ
vₓ = m1 / (m₁ + m₂) v₁
let's calculate
vₓ = - 900 / (900 + 1200) 25
vₓ = - 10.7 m / s
Axis y
initial moment
[tex]p_{oy}[/tex]= m₂ v₂
where v₂ = - 6.00 m / s
the sign indicates that it is moving to the South
final moment
p_{fy}= (m₁ + m₂) [tex]v_{y}[/tex]
p_{oy} = p_{fy}
m₂ v₂ = (m₁ + m₂) v_{y}
v_{y} = m₂ / (m₁ + m₂) v₂
we calculate
[tex]v_{y}[/tex] = 1200 / (900+ 1200) 6
[tex]v_{y}[/tex] = - 3,428 m / s
for the velocity module we use the Pythagorean theorem
v = √ (vₓ² + v_{y}²)
v = RA (10.7²2 + 3,428²2)
v = 11.24 m / s
now let's use trigonometry to encode the angle measured in the west clockwise (negative of the x axis)
tan θ = [tex]v_{y}[/tex] / Vₓ
θ = tan-1 v_{y} / vₓ)
θ = tan -1 (3,428 / 10.7)
θ = 17.76º
This angle is from the west to the south, that is, in the third quadrant.
b) To search for loss of the kinetic flow, calculate the kinetic enegy and then look for its relationship
Kf = 1/2 (m1 + m2) v2
K₀ = ½ m₁ v₁² + ½ m₂ v₂²
Kf = ½ (900 + 1200) 11.24 2
Kf = 1.3265 105 J
K₀ = ½ 900 25² + ½ 1200 6²
K₀ = 2,8125 10⁵ + 2,16 10₅4
K₀ = 3.0285 105J
the wasted energy is
Kf / K₀ = 1.3265 105 / 3.0285 105
Kf / K₀ = 0.4380
this is the fraction of kinetic energy that is conserved, transforming heat and transforming potential energy
A 300 g bird flying along at 6.2 m/s sees a 10 g insect heading straight toward it with a speed of 35 m/s (as measured by an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch.
Required:
What is the bird's speed immediately after swallowing?
Answer:
The velocity of the bird is [tex]v_f = 4.87 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the bird is [tex]m_1 = 300 \ g = 0.3 \ kg[/tex]
The initial speed of the bird is [tex]u_1 = 6.2 \ m/s[/tex]
The mass of the insect is [tex]m_2 = 10 \ g = 0.01 \ kg[/tex]
The speed of the insect is [tex]u_ 2 =-35 \ m/s[/tex]
The negative sign is because it is moving in opposite direction to the bird
According to the principle of linear momentum conservation
[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v_f[/tex]
substituting values
[tex](0.3 * 6.2 ) + (0.01 * (-35)) = (0.3 + 0.01 )v_f[/tex]
[tex]1.51 = 0.31 v_f[/tex]
[tex]v_f = 4.87 \ m/s[/tex]
The Final velocity of Bird = 4.87 m/s
Mass of the bird = 300 g = 0.3 kg
Velocity of bird = 6.2 m/s
Momentum of Bird = Mass of bird [tex]\times[/tex] Velocity of Bird = 0.3 [tex]\times[/tex] 6.2 = 1.86 kgm/s
Mass of the insect = 10 g = 0.01 kg
Velocity of insect = - 35 m/s
Momentum of the Insect = Mass of Insect [tex]\times[/tex] Velocity of Insect = - 0.35 kgm/s
According to the law of conservation of momentum We can write that
In the absence of external forces on the system , the momentum of system remains conserved in that particular direction.
The bird opens the mouth and enjoys the free lunch hence
Let the final velocity of bird is [tex]v_f[/tex]
Initial momentum of the system = Final momentum of the system
1.86 -0.35 = [tex]v_f[/tex] ( 0.01 + 0.3 )
1.51 = [tex]v_f[/tex] 0.31
[tex]v_f[/tex] = 4.87 m/s
The Final velocity of Bird = 4.87 m/s
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https://brainly.com/question/18066930
A ray in glass (n = 1.51) reaches a boundary with air at 49.2 deg. Does it reflect internally or refract into the air? Enter 0 for reflect, and 1 for refract.
Answer:
0 - Then, the ray is totally reflected
Explanation:
The ray reaches the boundary between the two mediums at 49.2°.
If the ray is totally reflected it is necessary that the crictical angle is lower that the incidet angle.
You use the following to calculate the critical angle:
[tex]\theta_c=sin^{-1}(\frac{n_2}{n_1})[/tex] (1)
n2: index of refraction of the second medium (air) = 1.00
n1: index of refraction of the first medium (glass) = 1.51
You replace the values of the parameters in the equation (1):
[tex]\theta_c=sin^{-1}(\frac{1.00}{1.51})=41.47\°[/tex]
The critical angle is 41.47°, which is lower than the incident angle 49.2°.
Then, the ray is totally reflected.
0
American eels (Anguilla rostrata) are freshwater fish with long, slender bodies that we can treat as uniform cylinders 1.0 m long and 10 cm in diameter. An eel compensates for its small jaw and teeth by holding onto prey with its mouth and then rapidly spinning its body around its long axis to tear off a piece of flesh. Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way. Although this feeding method is costly in terms of energy, it allows the eel to feed on larger prey than it otherwise could. The eel has a certain amount of rotational kinetic energy when spinning at 14 spins per second. If it swam in a straight line instead, about how fast would the eel have to swim to have the same amount of kinetic energy as when it is spinning? (a) 0.5 m/s; (b) 0.7 m/s; (c) 3 m/s; (d) 5 m/s.
Answer:
(c) 3 m/s;
Explanation:
Moment of inertia of the fish eels about its long body as axis
= 1/2 m R ² where m is mass of its body and R is radius of transverse cross section of body .
= 1/2 x m x (5 x 10⁻² )²
I = 12.5 m x 10⁻⁴ kg m²
angular velocity of the eel
ω = 2 π n where n is revolution per second
=2 π n
= 2 π x 14
= 28π
Rotational kinetic energy
= 1/2 I ω²
= .5 x 12.5 m x 10⁻⁴ x(28π)²
= 4.8312m J
To match this kinetic energy let eel requires to have linear velocity of V
1 / 2 m V² = 4.8312m
V = 3.10
or 3 m /s .
What is the on ohooke benden
er ord power
What is the main difference between work, power and energy
Answer:Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.
A butcher grinds 5 and 3/4 lb of meat then sells it for 2 and 2/3 pounds to the customer what is the maximum amount me that the butcher can sell to the next customer
Answer:
The maximum amount of meat that the butcher can sell is [tex]3\frac{1}{12}\:lb[/tex]
Explanation:
The maximum amount can be found by taking the difference of mixed numbers.
[tex]5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\[/tex]
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Calculate the amount of kinetic energy the car stores if it has a mass of 1200 kg and speed of 15 m/s
Answer:
KE = 135,000 j or 135 KJ
Explanation:
KE=0.5mv^2
KE=0.5*1200*15^2
KE = 135,000 joules or 135 Kilo Joules
what is the most likely elevation of point x?
A. 150 ft
B. 200 ft
C. 125 ft
D. 250 ft
Difference between regular and irregular object.
Hope this helps....
Good luck on your assignment.....
The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s) of the orbit and shaded segments are the same?
Answer: not sure
Explanation:
Location C is 0.021 m from a small sphere that has a charge of 5 nC uniformly distributed on its surface. Location D is 0.055 m from the sphere. What is the change in potential along a path from C to D?
Answer:
ΔV = -1321.73V
Explanation:
The change in potential along the path from C to D is given by the following expression:
[tex]\Delta V=-\int_a^bE dr[/tex] (1)
E: electric field produced by a charge at a distance of r
a: distance to the sphere at position C = 0.021m
b: distance to the sphere at position D = 0.055m
The electric field is given by:
[tex]E=k\frac{Q}{r^2}[/tex] (2)
Q: charge of the sphere = 5nC = 5*10^-9C
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
You replace the expression (2) into the equation (1) and solve the integral:
[tex]\Delta V=-kQ\int_a^b \frac{dr}{r^2}=-kQ[-\frac{1}{r}]_a^b[/tex] (3)
You replace the values of a and b:
[tex]\Delta V=(8.98*10^9Nm^2/C^2)(5*10^{-9}C)[\frac{1}{0.055m}-\frac{1}{0.021m}]\\\\\Delta V=-1321.73V[/tex]
The change in the potential along the path C-D is -1321.73V
A stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two fragments, each 1.0 kg, move in directions that make 60o angle above and below the positive x-axis and their speeds are 60 m/s each. What is the velocity of the 4.0-kg fragment
Answer:
-15 m/s
Explanation:
The computation of the velocity of the 4.0 kg fragment is shown below:
For this question, we use the correlation of the momentum along with horizontal x axis
Given that
Weight of stationary shell = 6 kg
Other two fragments each = 1.0 kg
Angle = 60
Speed = 60 m/s
Based on the above information, the velocity = v is
[tex]1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0[/tex]
[tex]\frac{60}{2} + \frac{60}{2} - 4\ v = 0[/tex]
[tex]v = \frac{60}{4}[/tex]
= -15 m/s
A wire with mass 90.0g is stretched so that its ends are tied down at points 88.0cm apart. The wire vibrates in its fundamental mode with frequency 80.0Hz and with an amplitude of 0.600cm at the antinodes.a) What is the speed of propagation of transverse waves in the wire?b) Compute the tension in the wire.
Answer:
a) V = 140.8 m/s
b) T = 2027.52 N = 2.03 KN
Explanation:
a)
The formula for the speed of the wave is given as follows:
f₁ = V/2L
V = 2f₁L
where,
V = Speed of Wave = ?
f₁ = Fundamental Frequency = 80 Hz
L = Length of Wire = 88 cm = 0.88 m
Therefore,
V = (2)(80 Hz)(0.88 m)
V = 140.8 m/s
b)
Another formula for the speed of wave is:
V = √T/μ
V² = T/μ
T = V²μ
where,
T = Tension in String = ?
μ = Linear Mass Density of Wire = Mass of Wire/L = 0.09 kg/0.88 m
μ = 0.1 kg/m
Therefore,
T = (140.8 m/s)²(0.1 kg/m)
T = 2027.52 N = 2.03 KN
A wire carries a current of 4 A travelling to the left (-x direction). It is placed in a constant magnetic field of magnitude 0.05 T, pointing upward ( z direction). a. If 25 cm of the wire is in the magnetic field, what is the force on the current
Answer:
0.05 N
Explanation:
Data provided in the question
The Wire carries a current of 4A to the left direction
The constant magnetic field of magnitude = 0.05 T
Pointing upward i.e Z direction
The wire is in the magnetic field = 25 cm
Based on the above information, the force on the current is
[tex]= Current \times constant\ magnetic\ field\ of\ magnitude \times magnetic\ field[/tex]
[tex]= 4 \times 0.05 \times 0.25[/tex]
= 0.05 N
The direction will be the negative Y direction
The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ????0 . Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 2.47×1014 s−1. g
Answer:
E = 0.965eV
Explanation:
In order to calculate the minimum energy needed to eject the electrons you use the following formula:
[tex]E=h \nu[/tex] (1)
h: Planck' constant = 6.626*10^{-34}J.s
v: threshold frequency = 2.47*10^14 s^-1
You replace the values of v and h in the equation (1):
[tex]E=(6.262*10^{-34}J.s)(2.47*10^{14}s^{-1})=1.54*10^{-19}J[/tex]
In electron volts you obtain:
[tex]1.54*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=0.965eV[/tex]
The minimum energy needed is 0.965eV
When using a mercury barometer , the vapor pressure of mercury is usually assumed to be zero. At room temperature mercury's vapor pressure is about 0.0015 mm-Hg. At sea level, the height hhh of mercury in a barometer is about 760 mm.Required:a. If the vapor pressure of mercury is neglected, is the true atmospheric pressure greater or less than the value read from the barometer? b. What is the percent error? c. What is the percent error if you use a water barometer and ignore water's saturated vapor pressure at STP?
Answer:
Explanation:
(a)
The true atmospheric pressure will has more value than the reading in the barometer. If Parm is the atmospheric
pressure in the tube then the resulting vapour pressure is
Patm - pgh = Prapor
The final reading ion the barometer is
pgh = Palm - Proper
Hence, the true atmospheric pressure is greater.
you can find the answer in this book
physics principles with Applications, Global Edition Problem 67P: Chapter: CH 13 Problem:67p
The force a spring exerts on a body is a conservative force because:
a. a spring always exerts a force parallel to the displacement of the body.
b. the work a spring does on a body is equal for compressions and extensions of equal magnitude.
c. the net work a spring does on a body is zero when the body returns to its initial position.
d. the work a spring does on a body is equal and opposite for compressions and extensions of equal magnitude.
e. a spring always exerts a force opposite to the displacement of the body.
Answer:
c. the net work a spring does on a body is zero when the body returns to its initial position
Explanation:
A force is conservative when the net work done over any path that returns to the initial position is zero. Choice C matches that definition.
An ideal spring of the kind used in physics problems has the characteristic that it applies the same force at the same distance always. So any work required to extend or compress the spring is reversed when the reverse motion takes place.
A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Answer:
If the particle is an electron [tex]E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton, [tex]E_y = 6.08 * 10^6 N/C[/tex]
Explanation:
Initial speed at the origin, [tex]u = 3 * 10^6 m/s[/tex]
[tex]\theta = 38^0[/tex] to +ve x-axis
The particle crosses the x-axis at , x = 1.5 cm = 0.015 m
The particle can either be an electron or a proton:
Mass of an electron, [tex]m_e = 9.1 * 10^{-31} kg[/tex]
Mass of a proton, [tex]m_p = 1.67 * 10^{-27} kg[/tex]
The electric field intensity along the positive y axis [tex]E_y[/tex], can be given by the formula:
[tex]E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\[/tex]
If the particle is an electron:
[tex]E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton:
[tex]E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 6.08 * 10^6 N/C[/tex]
From a height of 40.0 m, a 1.00 kg bird dives (from rest) into a small fish tank containing 50.5 kg of water. Part A What is the maximum rise in temperature of the water if the bird gives it all of its mechanical energy
Answer:
0.00185 °C
Explanation:
From the question,
The potential energy of the bird = heat gained by the water in the fish tank.
mgh = cm'(Δt)................... Equation 1
Where m = mass of the bird, g = acceleration due to gravity, h = height, c = specific heat capacity of water, m' = mass of water, Δt = rise in temperature of water.
make Δt the subject of the equation
Δt = mgh/cm'............... Equation 2
Given: m = 1 kg, h = 40 m, m' = 50.5 kg
constant: g = 9.8 m/s², c = 4200 J/kg.K
Substitute into equation 2
Δt = 1(40)(9.8)/(50.5×4200)
Δt = 392/212100
Δt = 0.00185 °C
A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.24 m from the slits. The first bright fringe is located 3.30 mm from the center of the interference pattern. What is the wavelength of the laser light?
Answer:
λ = 509 nm
Explanation:
In order to calculate the wavelength of the light you use the following formula:
[tex]y=m\frac{\lambda D}{d}[/tex] (1)
where
y: distance of the mth fringe to the central peak = 3.30 mm = 3.30*10^-3 m
m: order of the bright fringe = 1
D: distance from the slits to the screen = 3.24 m
d: distance between slits = 0.500 mm = 0.500*10^-3 m
You first solve the equation (1) for λ, and then you replace the values of the other parameters:
[tex]\lambda=\frac{dy}{mD}\\\\\lambda=\frac{(0.500*10^{-3}m)(3.30*10^{-3}m)}{(1)(3.24m)}=5.09*10^7m\\\\\lambda=509*10^{-9}m=509nm[/tex]
The wavelength of the light is 509 nm
At least how many Calories does a mountain climber need in order to climb from sea level to the top of a 5.42 km tall peak assuming the muscles of the climber can convert chemical energy to mechanical energy with an efficiency of 16.0 percent. The total mass of the climber and the equipment is 78.4 kg. (Enter your answer as a number without units.)
Answer:
Ec = 6220.56 kcal
Explanation:
In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.
You use the following formula:
[tex]W_c=Mgh[/tex] (1)
Wc: work done by the climber
g: gravitational constant = 9.8 m/s^2
M: mass of the climber = 78.4 kg
h: height reached by the climber = 5.42km = 5420 m
You replace in the equation (1):
[tex]W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J[/tex] (2)
Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:
[tex]0.16(E_c)=4,164,294.4J[/tex]
Ec: Calories
You solve for Ec and convert the result to Cal:
[tex]E_c=\frac{4,164,294.4}{016}=26,026,840J*\frac{1kcal}{4184J}\\\\E_c=6220.56\ kcal[/tex]
The amount of Calories needed by the climber was 6220.56 kcal
A depiction of a famous scientific experiment is given. Consider how the beam changes when the magnet is off compared to when the magnet is on. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned off. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned off, the beam path bends toward the positively charged plate and ends at the lower half of the wide end of the tube. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned n. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned on, the beam path travels in a straight path to the center of the wide end of the tube. What type of beam was used in this experiment?
Answer:
The beam used is a negatively charged electron beam with a velocity of
v = E / B
Explanation:
After reading this long statement we can extract the data to work on the problem.
* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.
* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced
[tex]F_{e} = F_{m}[/tex]
q E = qv B
v = E / B
this configuration is called speed selector
They ask us what type of beam was used.
The beam used is a negatively charged electron beam with a velocity of v = E / B
You are trying to overhear a juicy conversation, but from your distance of 25.0 m, it sounds like only an average whisper of 25.0 dB. So you decide to move closer to give the conversation a sound level of 80.0 dB instead. How close should you come?
Answer:
r₂ = 1,586 m
Explanation:
For this problem we are going to solve it by parts, let's start by finding the sound intensity when we are 25 m
β = 10 log (I / I₀)
where Io is the sensitivity threshold 10⁻¹² W / m²
I₁ / I₀ = [tex]e^{\beta/10}[/tex]
I₁ = I₀ e^{\beta/10}
let's calculate
I₁ = 10⁻¹² e^{25/10}
I₁ = 1.20 10⁻¹¹ W / m²
the other intensity in exercise is
I₂ = 10⁻¹² e^{80/10}
I₂ = 2.98 10⁻⁹ W / m²
now we use the definition of sound intensity
I = P / A
where P is the emitted power that is a constant and A the area of the sphere where the sound is distributed
P = I A
the area a sphere is
A = 4π r²
we can write this equation for two points of the found intensities
I₁ A₁ = I₂ A₂
where index 1 corresponds to 25m and index 2 to the other distance
I₁ 4π r₁² = I₂ 4π r₂²
I₁ r₁² = I₂ r₂²
r₂ = √ (I₁ / I₂) r₁
let's calculate
r₂ = √ (1.20 10⁻¹¹ / 2.98 10⁻⁹) 25
r₂ = √ (0.40268 10⁻²) 25
r₂ = 1,586 m
A disk between vertebrae in the spine is subjected to a shearing force of 640 N. Find its shear deformation taking it to have the shear modulus of 1.00 109 N/m2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.30 cm in diameter.
Answer:
3.08*10^-6 m
Explanation:
Given that
Total shearing force, F = 640 N
Shear modulus, S = 1*10^9 N/m²
Height of the cylinder, L = 0.7 cm
Diameter of the cylinder, d = 4.3 cm
The solution is attached below.
We have our shear deformation to be 3.08*10^-6 m
Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12.0-V battery, the current from the battery is 1.78 A. When the resistors are connected in parallel to the battery, the total current from the battery is 11.3 A. Determine the two resistances.
Answer:
R(smaller) = 1.3 Ω and R(larger) = 5.4 Ω
Explanation:
Ohm's Law states that:
V = IR
R = V/I
where,
R = Resistance
V = Potential Difference
I = Current
Therefore, for series connection:
Rs = Vs/Is
where,
Rs = Resistance when connected in series = R(smaller) + R(larger)
Vs = Potential Difference when connected in series = 12 V
Is = Current when connected in series = 1.78 A
Therefore,
R(smaller) + R(larger) = 12 V/1.78 A
R(smaller) + R(larger) = 6.74 Ω --------------- equation 1
R(smaller) = 6.74 Ω - R(larger) --------------- equation 2
Therefore, for series connection:
Rp = Vp/Ip
where,
Rp = Resistance when connected in parallel = [1/R(smaller) + 1/R(larger)]⁻¹
Rp = [{R(smaller) + R(larger)}/{R(smaller).R(larger)]⁻¹
Rp = R(smaller).R(larger)/[R(smaller) + R(larger)]
Vp = Potential Difference when connected in parallel = 12 V
Ip = Current when connected in parallel = 11.3 A
Therefore,
R(smaller).R(larger)/[R(smaller) + R(larger)] = 12 V/11.3 A
using equation 1 and equation 2, we get:
[6.74 Ω - R(larger)].R(larger)/6.74 Ω = 1.06 Ω
6.74 R(larger) - R(larger)² = (6.74)(1.06)
R(larger)² - 6.74 R(larger) + 7.16 = 0
solving this quadratic equation we get:
R(larger) = 5.4 Ω (OR) R(larger) = 1.3 Ω
using these values in equation 2, we get:
R(smaller) = 1.3 Ω (OR) R(smaller) = 5.4 Ω
Since, it is given in the question that R(smaller)<R(larger).
Therefore, the correct answers will be:
R(smaller) = 1.3 Ω and R(larger) = 5.4 Ω
C2B.7Suppose 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
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 2-kg block is released from rest at the top of a 20-mlong frictionless ramp that is 4 m high. At the same time, an identical block is released next to the ramp so that it drops straight down the same 4 m. What are the values for each of the following for the blocks just before they reach ground level.
Required:
a. Gravitational potential energy Block a_____ J Block b _____ J
b. Kinetic energy Block a _____ J Block b _____
c. Speed Block a _____ J Block b _____ J
d. Momentum Block a _____ J Block b _____ J
Answer:
A.) 78.4 J for both
B.) 78.4 J for both
C.) 8.85 m/s for both
D.) 17.7 kgm/s
Explanation:
Given information:
Mass m = 2 kg
Distance d = 20 m
High h = 4 m
A.) Gravitational potential energy can be calculated by using the formula
P.E = mgh
P.E = 2 × 9.8 × 4
P.E = 78.4 J
Since the two objects are identical, the gravitational potential energy of the block for both a and b will be 78.4 J
B.) According to conservative energy,
Maximum P.E = Maximum K.E.
Therefore, the kinetic energy of the two blocks will be 78.4 J
C.) Since K.E = 1/2mv^2 = mgh
V = √(2gh)
Solve for velocity V by substituting g and h into the formula
V = √(2 × 9.8 × 4)
V = √78.4
V = 8.85 m/s
The velocities of both block will be 8.85 m/s
D.) Momentum is the product of mass and velocity. That is,
Momentum = MV
Substitute for m and V into the formula
Momentum = 2 × 8.85 = 17.7 kgm/s
Both block will have the same value since the ramp Is frictionless.
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands
Complete question is;
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.
Answer:
Tangential speed of foot just before it lands is; v = 5.37m/s
Explanation:
Let U (potential energy) be zero on the ground.
So, initially, U = mgh
where, h = 0.98/2 = 0.49m (midpoint of the leg)
Now just before the leg hits the floor it would have kinetic energy as;
K = ½Iω²
where ω = v/r and I = ⅓mr²
So, K = ½(⅓mr²)(v/r)²
K = (1/6) × (mr²)/(v²/r²)
K = (1/6) × mv²
From principle of conservation of energy, we have;
Potential energy = Kinetic energy
Thus;
mgh = (1/6) × mv²
m will cancel out to give;
gh = (1/6)v²
Making v the subject, we have;
v = √6gh
v = √(6 × 9.81 × 0.49)
v = √28.8414
v = 5.37m/s
Giving quadrilateral a(2,-1 ) b ( 1,3) c(6,5) d(7,1) you want to prove that it is a parallelogram by showing opposite sides are congruent . what formula would you use ? show that sb is congruent to cd
Answer:
AB = CD = √17
Explanation:
The distance formula is used to find the length of a line segment between two points. Here, we want to show the distance AB is the same as the distance CD.
d = √((x1 -x1)² +(y2 -y1)²)
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AB: d = √((1 -2)² +(3 -(-1))²) = √((-1)² +4²) = √17
CD: d = √((7-6)² +(1-5)²) = √(1² +(-4)²) = √17 . . . . same as AB
Segment AB is congruent to segment CD.
A bicycle rider has a speed of 20.0 m/s at a height of 60 m above sea level when he begins coasting down hill. Sea level is the zero level for measuring gravitational potential energy. Ignoring friction and air resistance, what is the rider's speed when he coasts to a height of 18 m above sea level?
Answer:
The rider's speed will be approximately 35 m/s
Explanation:
Initially the rider has kinetic and potential energy, and after going down the hill, some of the potencial energy turns into kinetic energy. So using the conservation of energy, we have that:
[tex]kinetic_1 + potencial_1 = kinetic_2 + potencial_2[/tex]
The kinetic and potencial energy are given by:
[tex]kinetic = mass * speed^2 / 2[/tex]
[tex]potencial = mass * gravity * height[/tex]
So we have that:
[tex]m*v^2/2 + mgh = m*v'^2/2 + mgh'[/tex]
[tex]20^2/2 + 9.81*60 = v'^2/2 + 9.81*18[/tex]
[tex]v'^2/2 + 176.58 = 788.6[/tex]
[tex]v'^2/2 = 612.02[/tex]
[tex]v'^2 = 1224.04[/tex]
[tex]v' = 34.99\ m/s[/tex]
So the rider's speed will be approximately 35 m/s
a cannon is fired with an initial horizontal velocity of 20m/s and an initial velocity of 25m/s. After 3s in the air, the cannon hits its target. How far away(in meters) was the cannon from its target
Answer:
60 m
Explanation:
After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.
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The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.