Answer:
(a) 108
(b) 110.500 kW
(c) 920.84 A
Solution:
As per the question:
Voltage at primary, [tex]V_{p} = 120\ V[/tex] (rms voltage)
Voltage at secondary, [tex]V_{s} = 13000\ V[/tex] (rms voltage)
Current in the secondary, [tex]I_{s} = 8.50\ mA[/tex]
Now,
(a) The ratio of secondary to primary turns is given by the relation:
[tex]\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}[/tex]
where
[tex]N_{p}[/tex] = No. of turns in primary
[tex]N_{s}[/tex] = No. of turns in secondary
[tex]\frac{N_{s}}{N_{p}} = \frac{13000}{120}[/tex] ≈ 108
(b) The power supplied to the line is given by:
Power, P = [tex]V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW[/tex]
(c) The current rating that the fuse should have is given by:
[tex]\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}[/tex]
[tex]\frac{13000}{120} = \frac{I_{p}}{8.50}[/tex]
[tex]I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A[/tex]
A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheric pressure is Pa = 1.01 × 105 Pa. The volume of the container is V0 = 4.4 × 10-4 m3. The maximum difference between the pressure inside and outside that this particular container can withstand before bursting or imploding is ΔPmax = 2.26 × 105 Pa. For this problem, assume that the density of air maintains a constant value of rhoa = 1.20 kg / m3 and that the density of seawater maintains a constant value of rhos = 1025 kg / m3.What is the maximum height h in meters above the ground that the container can be lifted before bursting?
To determine the maximum height h that the container can be lifted before bursting, we can use the formula ΔP = ρgh, where ΔP is the pressure difference, ρ is the density of the fluid inside the container, g is the acceleration due to gravity, and h is the height difference.
Explanation:To determine the maximum height h that the container can be lifted before bursting, we need to consider the difference in pressure inside and outside the container.
The maximum pressure difference that the container can withstand before bursting is given as ΔPmax = 2.26 × 105 Pa.
To calculate the maximum height h, we can use the formula: ΔP = ρgh, where ΔP is the pressure difference, ρ is the density of the fluid inside the container (assuming it is the same as seawater density, rhos = 1025 kg/m3), g is the acceleration due to gravity (approximately 9.81 m/s2), and h is the height difference.
Plugging in the values, we get ΔPmax = 1025 * 9.81 * h. Solving for h, we find h ≈ 231 meters.
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The wheel of a car has a radius of 0.350 m. The engine of the car applies a torque of 295 N · m to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a counter torque. Moreover, the car has a constant velocity, so this counter torque balances the applied torque. What is the magnitude of the static frictional force?
A police officer in hot pursuit of a criminal drives her car through an unbanked circular (horizontal) turn of radius 300 m at a constant speed of 22.2 m/s. Her mass is 55.0 kg. To the nearest degree, what is the angle (relative to vertical) of the net force of the car seat on the officer?
Answer:
The angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.
Explanation:
Given:
Mass of the driver is, [tex]m=55\ kg[/tex]
Radius of circular turn is, [tex]R=300\ m[/tex]
Linear speed of the car is, [tex]v=22.2\ m/s[/tex]
Since, the car makes a circular turn, the driver experiences a centripetal force radially inward towards the center of the circular turn. Also, the driver experiences a downward force due to her weight. Therefore, two forces act on the driver which are at right angles to each other.
The forces are:
1. Weight = [tex]mg=55\times 9.8=539\ N[/tex]
2. Centripetal force, 'F', which is given as:
[tex]F=\frac{mv^2}{R}\\F=\frac{55\times (22.2)^2}{300}\\\\F=\frac{55\times 492.84}{300}\\\\F=\frac{27106.2}{300}=90.354\ N[/tex]
Now, the angle of the net force acting on the driver with respect to the vertical is given by the tan ratio of the centripetal force (Horizontal force) and the weight (Vertical force) and is shown in the triangle below. Thus,
[tex]\tan \theta=\frac{90.354}{539}\\\tan \theta=0.1676\\\theta=\tan^{-1}(0.1676)=9.52\approx 10[/tex]°
Therefore, the angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.
The net force angle relative to the vertical is found by calculating the centripetal and vertical forces. The angle is approximately 10°. This involves understanding forces in circular motion.
To find the angle of the net force of the car seat on the police officer, we need to consider the forces involved and the direction of the net force.
The car is executing horizontal circular motion, which means there is a centripetal force acting horizontally towards the center of the circle. This force is provided by the friction between the car tires and the road. The centripetal force Fc can be calculated by:
Fc = m * ac
where m is the mass of the officer and ac is the centripetal acceleration, which is given by:
ac = v2 / r
Given:
Radius of the turn, r = 300 m
Speed of the car, v = 22.2 m/s
Mass of the officer, m = 55.0 kg
First, calculate the centripetal acceleration:
ac = (22.2 m/s)2 / 300 m = 1.6444 m/s2
Now, calculate the centripetal force:
Fc = 55.0 kg * 1.6444 m/s2 = 90.442 N
The vertical force acting on the officer is her weight:
Fw = m * g = 55.0 kg * 9.8 m/s2 = 539 N
Now, to find the angle θ relative to the vertical, use the tangent function, as the net force's components are the horizontal centripetal force and the vertical weight:
θ = tan-1(Fc / Fw) = tan-1(90.442 / 539)
θ ≈ 9.5°
Thus, the angle of the net force relative to the vertical is approximately 10° (to the nearest degree).
You are adding vectors of length 20 and 40 units. Which of the following choices is a possible resultant magnitude?
a. 64
b. 18
c. 0
d. 100
e. 37
Answer: e. 37
Explanation:
When two vectors are added, the maximum and minimum posible values, happen when both vectors are aligned each other.
If both vectors aim in the same direction, the maximum value is just the arithmetic sum of their lengths, in this case, 60.
If they aim in opposite direction, the resultant is the substraction of their magnitudes, which yields 20.
Any other posible value (depending on the angle between vectors, which can span from 0º to 180º) must be between those values.
So, the only choice that fits within this interval, is 37.
A satellite explodes in outer space, far from any other body, sending thousands of pieces in all directions. Is the linear momentum of the satellite before the explosion less than, equal to, or greater than the total linear momentum of all the pieces after the explosion?
The answer to this problem can be given through energy conservation as well as Newton's first law.
Newton's first law states that as long as there is no force exerted on a body, its movement will be constant or at rest. In this way the amount of linear movement of the satellite before the explosion will be equal to the sum of the movement generated in the pieces after the explosion. In the absence of external forces but the preservation of the internal force as the only one to act, these will not change the total momentum of the system.
A historian claims that a cannonball fired at a castle wall would melt on impact with the wall. Let's examine this claim. Assume that the kinetic energy of the cannonball is completely transformed into the internal energy of the cannonball on impact with no energy transferred to the wall. The cannonball is made of iron, which has a specific heat capacity of 450 J/kg-K and a melting point of 1811 K.
1) If the cannonball has an initial temperature of 298 K, how fast would the cannonball need to travel in order to reach its melting point on impact?
_________m/s
Answer:
v = 1166.9 m / s
Explanation:
The equation for caloric energy is
Q = m [tex]c_{e}[/tex] ΔT = m ce ([tex]T_{f}[/tex]-T₀)
Where m is the mass, ce is the specific heat and DT is the temperature variation
In this case the cannonball has kinetic energy
Em = K = ½ m v²
They indicate that mechanical energy is completely transformed into heat
Q = Em
m [tex]c_{e}[/tex] ([tex]T_{f}[/tex] - To) = ½ m v²
v = √ 2 [tex]c_{e}[/tex] ([tex]T_{f}[/tex]-To)
Let's calculate
v = √ (2 450 (1811-298))
v = 1166.9 m / s
A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is 6.68 × 10-3 Wb. What is the flux that passes through the circular loop?
Answer:
0.0085 Wb
Explanation:
a = Side of square
r = Radius of circle
[tex]\phi_s[/tex] = Magnetic flux through square loop = [tex]6.68\times 10^{-3}\ Wb[/tex]
Magnetic flux is given by
[tex]\phi=BA[/tex]
For square
[tex]\phi_{s}=Ba^2[/tex]
The length of the square will be equal to the circumference of the circle
[tex]4a=2\pi r\\\Rightarrow r=\frac{2a}{\pi}[/tex]
For circle
[tex]\phi_{c}=B\pi r^2\\\Rightarrow \phi_{c}=B\pi \left(\frac{2a}{\pi}\right)^2\\\Rightarrow \phi_c=Ba^2\frac{4}{\pi}\\\Rightarrow \phi_c=\phi_s\frac{4}{\pi}\\\Rightarrow \phi_c=6.68\times 10^{-3}\frac{4}{\pi}\\\Rightarrow \phi_c=0.0085\ Wb[/tex]
The flux that passes through the circular loop is 0.0085 Wb
The magnetic flux that passes through the circular loop is 0.00848 Wb.
What is the magnetic flux?The magnetic flux is defined as the measurement of the total magnetic field which passes through a given area.
Given that the magnetic flux that passes through the square loop is 6.68 × 10-3 Wb.
Let's consider that r is the radius of the circle and a is the side of the square.
The magnetic flux through the square loop is given below.
[tex]\phi_s = Ba^2[/tex]
Where B is the magnetic field.
The length of the square will be equal to the circumference of the circle
[tex]4a = 2\pi r[/tex]
[tex]r = \dfrac {2a}{\pi}[/tex]
The magnetic flux through the circular loop is given below.
[tex]\phi = BA[/tex]
Where A is is the cross-sectional area of the circular loop.
[tex]\phi = B \pi r^2[/tex]
[tex]\phi = B \pi (\dfrac {2a}{\pi})^2[/tex]
[tex]\phi = B\pi \times \dfrac {4a^2}{\pi^2}[/tex]
[tex]\phi = Ba^2 \dfrac {4}{\pi}[/tex]
[tex]\phi = \phi_s \dfrac {4}{3.14}[/tex]
[tex]\phi = 6.68 \times 10^{-3} \times 1.27[/tex]
[tex]\phi = 0.00848 \;\rm Wb[/tex]
Hence we can conclude that the magnetic flux that passes through the circular loop is 0.00848 Wb.
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If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced a distance 0.125m from its equilibrium position and released with zero initial speed. Then after a time 0.800s its displacement is found to be a distance 0.125m on the opposite side, and it has passed the equilibrium position once during this interval. Find : Amplitude, period, and frequency.
Explanation:
An object is attached to the spring and then released. It begins to oscillate. If it is displaced a distance 0.125 m from its equilibrium position and released with zero initial speed. The amplitude of a wave is the maximum displacement of the particle. So, its amplitude is 0.125 m.
After 0.800 seconds, its displacement is found to be a distance 0.125 m on the opposite side. The time period will be, [tex]t=2\times 0.8=1.6\ s[/tex]
We know that the relation between the time period and the time period is given by :
[tex]f=\dfrac{1}{t}[/tex]
[tex]f=\dfrac{1}{1.6}[/tex]
f = 0.625 Hz
So, the frequency of the object is 0.625 Hz. Hence, this is the required solution.
Answer:
The amplitude is 0.125 m, period is 1.64 sec and frequency is 0.61 Hz.
Explanation:
Given that,
Distance = 0.125 m
Time = 0.800 s
Since the object is released form rest, its initial displacement = maximum displacement .
(a). We need to calculate the amplitude
The amplitude is maximum displacement.
[tex]A=0.125 m[/tex]
(b). We need to calculate the period
The object will return to its original position after another 0.820 s,
So the time period will be
[tex]T=2\times t[/tex]
Put the value into the formula
[tex]T=2\times0.820[/tex]
[tex]T=1.64\ sec[/tex]
(c). We need to calculate the frequency
Using formula of frequency
[tex]f=\dfrac{1}{T}[/tex]
Put the value into the formula
[tex]f=\dfrac{1}{1.64}[/tex]
[tex]f=0.61\ Hz[/tex]
Hence, The amplitude is 0.125 m, period is 1.64 sec and frequency is 0.61 Hz.
A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.
1) Calculate the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball. (Express your answer to three significant figures.)
Answer:
[tex]\dfrac{K_t}{K_r}=\dfrac{5}{2}[/tex]
Explanation:
Given that,
Mass of the bowling ball, m = 5 kg
Radius of the ball, r = 11 cm = 0.11 m
Angular velocity with which the ball rolls, [tex]\omega=2.8\ rad/s[/tex]
To find,
The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.
Solution,
The translational kinetic energy of the ball is :
[tex]K_t=\dfrac{1}{2}mv^2[/tex]
[tex]K_t=\dfrac{1}{2}m(r\omega)^2[/tex]
[tex]K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2[/tex]
The rotational kinetic energy of the ball is :
[tex]K_r=\dfrac{1}{2}I \omega^2[/tex]
[tex]K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2[/tex]
[tex]K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2[/tex]
Ratio of translational to the rotational kinetic energy as :
[tex]\dfrac{K_t}{K_r}=\dfrac{5}{2}[/tex]
So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2
The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is [tex]\frac{5}{2}[/tex] or 2.50.
Given the data in question;
Mass of the bowling ball; [tex]m = 5.0kg[/tex]Radius; [tex]r = 11.0cm = 0.11m[/tex]Angular velocity; [tex]w = 2.80rad/s[/tex]Ratio of [tex]E_{translational}[/tex] to the [tex]E_{rotational }[/tex]; [tex]R_{\frac{E_{translational}}{E_{rotational}}}= \frac{E_{translational}}{E_{rotational}} = \ ?[/tex]
Translational and Rotational kinetic EnergyRotational energy or angular kinetic energy is simply the kinetic energy due to the rotation of a rigid body while translational kinetic energy is the same as one-half the product of mass and the square of the velocity of the body.
They are expressed as;
[tex]E_{rotational} = \frac{1}{2}Iw^2[/tex]
Where;
[tex]w[/tex] is the angular speed or velocity. [tex]I[/tex] is the moment of inertia of the bowling ball ( [tex]I = \frac{2}{5}mR^2[/tex] ).Hence, [tex]E_{rotational} = \frac{1}{2}(\frac{2}{5}mR^2 )w^2[/tex]
[tex]E_{translational} = \frac{1}{2}mv^2[/tex]
Where;
m is mass of the objectv is the linear velocity ( [tex]v = R*w[/tex] )Hence, [tex]E_{translational} = \frac{1}{2}m(R*w)^2[/tex]
To determine the translational kinetic energy of the bowling ball, we substitute our given values into the equation above.
[tex]E_{translational} = \frac{1}{2}m( R*w)^2\\ \\E_{translational} = \frac{1}{2} * 5.0kg * 0.0121m^2 * 7.84rad/s^2[/tex]
To determine the rotational kinetic energy of the bowling ball, we substitute our given values into the equation above.
[tex]E_{rotational} = \frac{1}{2}( \frac{2}{5}mR^2)w^2\\ \\E_{rotational} = \frac{1}{2}( \frac{2}{5} * 5.0kg* (0.11m)^2)(2.80rad/s)^2\\ \\E_{rotational} = \frac{2}{10} * 5.0kg * 0.0121m^2 * 7.84rad/s^2[/tex]
So, Ratio of [tex]E_{translational}[/tex] to the [tex]E_{rotational }[/tex]; [tex]R_{\frac{E_{translational}}{E_{rotational}}}= \frac{E_{translational}}{E_{rotational}}[/tex]
[tex]R_{\frac{E_{translational}}{E_{rotational}}}= \frac{E_{translational}}{E_{rotational}}\\\\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{\frac{1}{2} * 5.0kg * 0.0121m^2 * 7.84rad/s^2}{\frac{2}{10} * 5.0kg * 0.0121m^2 * 7.84rad/s^2}\\\\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{\frac{1}{2} }{\frac{2}{10} }\\ \\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{\frac{1*10}{2} }{2}\\ \\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{5}{2} = 2.50[/tex]
Therefore, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is [tex]\frac{5}{2}[/tex] or 2.50.
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A beam of monochromatic light is aimed at a slit of width w and forms a diffraction pattern. In which case is the width of the central band greater?
a. When the incident light is blue
b. When the incident light is yellow
c. Same in both cases, blue and yellow
The escape velocity at the surface of Earth is approximately 11 km/s. What is the mass, in units of ME (the mass of the Earth), of a planet with three times the radius of Earth for which the escape speed is three times that for Earth?
Answer:11 km/s
Explanation:
Given
Escape velocity at the surface of earth is 11 km/s
Escape velocity is given by
[tex]V_e=\sqrt{\frac{2GM}{R}}[/tex]
Escape velocity at the surface of earth
[tex]11=\sqrt{\frac{2GM}{R}}[/tex]--------------------1
If Escape velocity is three times and the radius is also the three times
[tex]V_e_2=\sqrt{\frac{2G(3M)}{3R}}[/tex]
[tex]V_e_2=\sqrt{\frac{2GM}{R}}=V_e[/tex]
i.e. [tex]V_e_2=11 km/s[/tex]
You have two springs. One has a greater spring constant than theother. You also have two objects, one with a greater mass than theother. Which object should be attached to which spring, so that theresulting spring-object system has the greatest possible period ofoscillation?
The object with the smaller mass should be attached to the springwith the greater spring constant.
The object with the smaller mass should be attached to the springwith the smaller spring constant.
The object with the greater mass should be attached to the springwith the greater spring constant.
The object with the greater mass should be attached to the springwith the smaller spring constant.
The object with the greater mass should be attached to the spring with the smaller spring constant, so that the resulting spring-object system has the greatest possible period of oscillation.
Answer: Option D
Explanation:
According to the simple harmonic motions, from physics, it gives a relation between deformation force and the deflection. The more deflection results in more time period of oscillation.
F = - k x
where ‘k’ is the spring constant, and ‘F’ is the deformation force.
So, deflection is directly proportionate to forces, and inversely proportionate to its spring constant. Hence, we can derive that the force must be maximum, and hence weight must be maximum, with the spring constant lesser. Then, the deflection will be high. So, time period increases.
The coil in a generator has 100 windings and a cross sectional area of 0.01 m^2. If the coil rotates with constant speed in the magnetic field of the earth (B = 0.5x10^-4 T), how many revolutions per second must it complete to generate a maximum e.m.f. of one volt?
Using this result, is it practical to create a generator that uses only the earths magnetic field?
To develop the problem it is necessary to apply the concepts related to the induced voltage and its definition according to the magnetic field and angular velocity.
By definition the induced voltage or electromotive force is given by
[tex]\epsilon = BAN\omega[/tex]
Where,
B = Magnetic Field
A = Cross-sectional area
N = Number of turns
[tex]\omega[/tex]= Angular velocity
For the given case in the problem we will look for the angular velocity,
[tex]\omega = \frac{\epsilon}{BAN}\\\omega = \frac{1}{(0.5*10^{-4})(0.01)(100)}\\\omega = 2*10^4rad/s \\\omega = 2*10^4rad/s (\frac{1rev}{2\pi rad})\\\omega = 3183.098rev/s[/tex]
Therefore the number of revolutions is 3183.1rev/s.
Although the number of revolutions the magnetic field of the earth is discarded for use as a generator because the magnetic field of the earth compared to other current devices is considered weak. Very little power would be obtained from the generator
A 350 kg mass, constrained to move only vertically, is supported by two springs, each having a spring constant of 250 kN/m. A periodic force with a maximum value of 100 N is applied to the mass with a frequency of 2.5 rad/s. Given a damping factor of 0.125, the amplitude of the vibration is
Answer:
Explanation:
Expression for amplitude of forced damped oscillation is as follows
A = [tex]\frac{F_0}{\sqrt{m(\omega^2-\omega_0^2)^2+b^2\omega^2} }[/tex]
where
ω₀ =[tex]\sqrt{\frac{k}{m} }[/tex]
ω₀ = [tex]\sqrt{\frac{500000}{350} }[/tex]
=37.8
b = .125 ,
ω = 2.5
m = 350
A = [tex]\frac{100}{\sqrt{350(37.8^2-2.5^2)^2+.125^2\times2.5^2} }[/tex]
A = 3.75 mm . Ans
The aorta carries blood away from the heart at a speed of about40 cm/s and has a radius of approximately 1.1 cm.
The aortabranches eventually into a large number of tiny capillaries thatdistribute the blood to the various body organs.
In a capillary,the blood speed is approximately 0.007 cm/s, and the radius isabout 6.0 10-4 cm.
Treat the blood as an incompressiblefluid, and use these data to determine the approximate number ofcapillaries in the human body.
Answer:
The no. of capillaries are [tex]1.92\times 10^{10}[/tex]
Solution:
As per the question:
Speed of the blood carried by the aorta, [tex]v_{a} = 40\ cm/s[/tex]
Radius of the aorta, [tex]R_{a} = 1.1 cm[/tex]
Speed of the blood in the capillaries, [tex]v_{c} = 0.007\ cm/s[/tex]
Radius of the capillaries, [tex]R_{c} = 6.0\times 10^{- 4} cm[/tex]
Now,
To determine the no. of capillaries:
Cross sectional Area of the Aorta, [tex]A_{a} = \pi R_{a}^{2} = \pi \times (1.1)^{2} = 1.21\pi \ m^{2}[/tex]
Cross sectional Area of the Capillary, [tex]A_{c} = \pi R_{c}^{2} = \pi \times (6.0\times 10^{- 4})^{2} = (3.6\times 10^{- 7})\pi \ m^{2}[/tex]
Let the no. of capillaries be 'n'
Also, the volume rate of flow in the aorta equals the sum total flow in the 'n' capillaries:
[tex]A_{a}v_{a} = nA_{c}v_{c}[/tex]
[tex]1.21\pi\times 40 = n\times 3.6\times 10^{- 7}\pi\times 0.007[\tex]
[tex]n = 1.92\times 10^{10}[/tex]
Using the principle of continuity, which states the constant volume flow rate of an incompressible fluid, we calculate that the human body has approximately ten billion capillaries.
Explanation:To determine the approximate number of capillaries in the human body based on the given data, we can use the principle of continuity, which states that the product of the cross-sectional area of a tube and the fluid speed through the tube is constant.
This implies that the blood flow (volume rate of flow) is the same in the aorta as in the capillaries, i.e., Aorta's cross-sectional area × speed of blood flow in Aorta = Capillary's cross-sectional area × speed of blood flow in capillaries × number of capillaries. From this, we can solve for the number of capillaries:
(π(1.1 cm)^2 × 40 cm/s) = (π(6.0 × 10^-4 cm)^2 × 0.007 cm/s) × number of capillaries.
When we do the calculations, we find that there are approximately ten billion capillaries in the human body, a vast network to ensure blood is delivered to every part of the body.
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A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2 m/s in the opposite direction on a horizontal frictionless surface. If the block is brought to rest by the collision, what is the kinetic energy of the bullet as it emerges from the block
Answer:
[tex]K=512J[/tex]
Explanation:
Since the surface is frictionless, momentum will be conserved. If the bullet of mass [tex]m_1[/tex] has an initial velocity [tex]v_{1i}[/tex] and a final velocity [tex]v_{1f}[/tex] and the block of mass [tex]m_2[/tex] has an initial velocity [tex]v_{2i}[/tex] and a final velocity [tex]v_{2f}[/tex] then the initial and final momentum of the system will be:
[tex]p_i=m_1v_{1i}+m_2v_{2i}[/tex]
[tex]p_f=m_1v_{1f}+m_2v_{2f}[/tex]
Since momentum is conserved, [tex]p_i=p_f[/tex], which means:
[tex]m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}[/tex]
We know that the block is brought to rest by the collision, which means [tex]v_{2f}=0m/s[/tex] and leaves us with:
[tex]m_1v_{1i}+m_2v_{2i}=m_1v_{1f}[/tex]
which is the same as:
[tex]v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}[/tex]
Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:
[tex]v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s[/tex]
So kinetic energy of the bullet as it emerges from the block will be:
[tex]K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J[/tex]
A He–Ne laser illuminates a narrow, single slit that is 1850 nm wide. The first dark fringe is found at an angle of 20.0∘ from the central peak. 1) Determine the wavelength of the light from the laser. (Express your answer to three significant figures.)
Answer:
Wavelength of light will be 632.73 nm
Explanation:
We have given width of the single slit [tex]d=1850nm-1850\times 10^{-9}m[/tex]
Angle is given as [tex]\Theta =20^{\circ}[/tex]
We have to find the wavelength of light for first dark fringe
We know that for dark fringe it is given that
[tex]dsin\Theta =n\lambda[/tex]
For first fringe n = 1
So [tex]1850\times 10^{-9}\times sin20^{\circ} =1\times \lambda[/tex]
[tex]\lambda =632.73nm[/tex]
Final answer:
The wavelength of the light from the He-Ne laser is 618 nm.
Explanation:
To determine the wavelength of the light from the He-Ne laser, we can use the equation for the first dark fringe in a single-slit diffraction pattern:
sin(theta) = m * (lambda) / w
Where theta is the angle of the first dark fringe, m is the order of the fringe (which is 1 for the first dark fringe), lambda is the wavelength of the light, and w is the width of the single slit.
Plugging in the given values:
sin(20.0∘) = 1 * (lambda) / 1850 nm
Simplifying and solving for lambda:
(lambda) = 1850 nm * sin(20.0∘)
(lambda) = 618 nm
Therefore, the wavelength of the light from the He-Ne laser is 618 nm.
A heat engine receives an amount of energy Qh= 790 kJ by heat transfer from a high temperature thermal reservoir at Th=950 K. Energy is rejected by heat transfer to a lower temperature thermal reservoir at T1=590 K. If waste heat in the amount of Q1=160 kJ is rejected to the low temperature thermal reservoir during each cycle.
a) Solve for the maximum theoretical efficiency that an engine in this situation could operate with. ANSWER: 0.379
b) Solve for actual efficiency that the engine is operating with.
c) Which of the following best describes the manner in which the cycle is operating...
-Reversibly or Impossibly?
Answer:
(a) [tex]\eta_{max}=37.895\%[/tex]
(b) [tex]\eta=79.75\%[/tex]
(c) Impossibly
Explanation:
Given:
temperature of source reservoir, [tex]T_H=950\ K[/tex]temperature of sink reservoir, [tex]T_L=590\ K[/tex]heat absorbed by the engine, [tex]Q_H=790\ kJ[/tex]heat rejected by the engine, [tex]Q_L=160\ kJ[/tex](a)
Now the maximum theoretical efficiency of the engine:
[tex]\eta_{max}=\frac{T_H-T_L}{T_H}\times 100\% [/tex]
[tex]\eta_{max}=\frac{950-590}{950}\times 100\% [/tex]
[tex]\eta_{max}=37.895\%[/tex]
(b)
Actual efficiency of the heat engine:
[tex]\eta=\frac{Q_H-Q_L}{Q_H}\times 100\% [/tex]
[tex]\eta=\frac{790-160}{790}\times 100\% [/tex]
[tex]\eta=79.75\%[/tex]
(c)
This is impossible because the actual efficiency can never be greater than the ideal (Carnot) efficiency of a heat engine.
The answers are:
(a) the maximum theoretical efficiency is 37.9%
(b) the actual efficiency is 79.7%
(c) The cycle is operating impossibly
The following data is provided to us:
The temperature of the source, [tex]T_1=950K[/tex]
The temperature of the sink, [tex]T_2=590K[/tex]
Heat absorbed by the engine, [tex]Q_1=790kJ[/tex]
Heat rejected by the engine, [tex]Q_2=160kJ[/tex]
(a) Now, the maximum efficiency of the engine is:
[tex]\eta _{max}=1-\frac{T_2}{T1}\\ \\\eta_{max}=1-\frac{590}{950}\\ \\\eta_{max}=0.379[/tex]
maximum efficiency ( [tex]\eta_{max}[/tex] ) = 37.9%
(b) actual efficiency:
[tex]\eta=1-\frac{Q_2}{Q_1}\\ \\\eta=1-\frac{160}{790}\\ \\\eta=0.797[/tex]
actual efficiency ([tex]\eta[/tex]) = 79.7%
(c) it is not possible for a Carnot engine to have more efficiency than the maximum possible efficiency.
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A sled plus passenger with total mass m = 52.3 kg is pulled a distance d = 21.8 m across a horizontal, snow-packed surface for which the coefficient of kinetic friction with the sled is μk = 0.13. The pulling force is constant and makes an angle of φ = 36.7 degrees above horizontal. The sled moves at constant velocity.a) Write an expression for the work done by the pulling force in terms of m, g (acceleration due to gravity), φ, μk, and d.b) What is the work done by the pulling force, in joules?c) Write an expression for the work done on the sled by friction in terms of m, g (acceleration due to gravity), φ, μk, and d.d) What is the work done on the sled by friction, in joules?
Answer:
Explanation:
Since the sled plus passenger moves with constant velocity , force applied will be equal to frictional force. Let the force applied be F
a ) Frictional force = μ R = F cosφ
R = mg - F sinφ
μ(mg - F sinφ) = F cosφ
μmg = F (μsinφ+cosφ)
F = μmg / (μsinφ+cosφ)
Work done
= F cosφ x d
= μmg x cosφ x d / (μsinφ+cosφ)
b )Work done
= 0.13 x 52.3 x 9.8 cos36.7 x 21.8 / ( 0.13 sin36.7 +cos36.7)
= 1164.61 / .87946
1324.23 J
c ) work done on the sled by friction
= - (work done by force)
= - μmg x cosφ x d / (μsinφ+cosφ)
d ) work done on the sled by friction
= - 1324.23 J
The work done by the pulling force is given by the formula W = F * d * cos(φ), and in this case, the force of friction can be calculated as Ff = μk * m * g * cos(φ).
Explanation:The work done by the pulling force is given by the formula W = F * d * cos(φ), where F is the magnitude of the pulling force, d is the distance, and φ is the angle above the horizontal. In this case, the force of friction can be calculated as Ff = μk * m * g * cos(φ). Therefore, the work done by the pulling force can be written as W = Ff * d.
Using the given values, we can substitute in the values to calculate the work done by the pulling force, which is W = μk * m * g * cos(φ) * d.
The work done on the sled by friction can be calculated by multiplying the force of friction by the distance, since the force is acting in the opposite direction. Therefore, the work done on the sled by friction is given by Wfr = -Ff * d. Substituting the given values, we can calculate the work done by friction, which is Wfr = -μk * m * g * cos(φ) * d.
Inserting given values in this formula, we can calculate the work done by friction as Wf = - (0.13) * (52.3 kg) * (9.8 m/s²) * (21.8 m) = -1475.24 Joules.
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A man pushes a block up an incline at a constant speed. As the block moves up the incline,
a. its kinetic energy and potential energy both increase.
b. its potential energy increases and its kinetic energy remains the same.
c. its potential energy increases and its kinetic energy decreases by the same amount.
d. its kinetic energy increases and its potential energy remains the same.
When the block started to move at constant speed it starts to gain a kinetic energy but the gravitational potential energy increases . Thus, option c is correct.
What is kinetic energy?Kinetic energy of an object is the energy generated by virtue of its motion whereas potential energy is generated by virtue of its position in a gravitational filed.
Both kind of energy depends on the mass of the object.
Kinetic energy KE = 1/2 mv² where v is the velocity. Kinetic energy reaches its maximum at maximum speed. Potential energy p = Mgh, where g is acceleration due to gravity and h is the height.
Thus when the block moves in the incline with constant velocity and its mass is also constant, no change will be there for kinetic energy but its potential energy will increase since gravitational pull increases. Thus option c is correct.
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An airplane is flying at Mach 1.48 at an altitude of 7,800.00 meters, where the speed of sound is v = 311.83 m/s. How far away from a stationary observer will the plane be when the observer hears the sonic boom? (Enter the total distance in m.)
Answer:
The plane will be 11545.46 m far when the observer hears the sonic boom
Explanation:
Step 1: Data given
Altitude of the plane = 7800 meters
speed of sound = 311.83 m/s
Step 2:
The mach number M = vs/v
This means v/vs = 1/M
Half- angle = ∅
sin∅= v/vs
∅ = sin^-1 (v/vs)
∅ = sin^-1 (1/M)
∅ = sin^-1(1/1.48)
∅= 42.5 °
tan ∅ = h/x
⇒ with h= the altitude of the plane = 7800 meter
⇒ with x = the horizontal distance moved by the plane
x = h/tan ∅
x = 7800 / tan 42.5
x = 8512.2 meters
d = the distance between the observer and the plane when the observer hears the sonic boom is:
d = √(8512.2² + 7800²)
d = 11545.46 m
The plane will be 11545.46 m far when the observer hears the sonic boom
A power plant uses 365 GJ/hr of energy from a boiler and extracts 18 GJ/hr as work from a turbine and rejects the remaining energy to a large cold resevoir of air (the atmosphere). What is the rate of heat rejection to the atmosphere for this plant? Give your answer in GJ/hr
Answer:
The rate of heat rejection will be 347 GJ/hr
Explanation:
We have given that power plant uses energy of 365 GJ/hr
So energy uses [tex]Q_B=365GJ/hr[/tex]
Work done from turbine [tex]W_T=18GJ/hr[/tex]
We have to find the rate of heat rejection to the atmosphere [tex]Q_C[/tex]
We know that [tex]Q_B=W_T+Q_C[/tex]
[tex]365=18+Q_C[/tex]
[tex]Q_C=347GJ/hr[/tex]
So the rate of heat rejection will be 347 GJ/hr
A 1200 kg car carrying four 80 kg people travels over a rough "washboard" dirt road with corrugations 4.0 m apart which causes the car to bounce on its spring suspension. The car bounces with maximum amplitude when its speed is 15 km/h. The car now stops, and the four people get out. By how much does the car body rise on its suspension owing to this decrease in weight?
Answer:
ΔX = 0.0483 m
Explanation:
Let's analyze the problem, the car oscillates in the direction y and advances with constant speed in the direction x
The car can be described with a spring mass system that is represented by the expression
y = A cos (wt + φ)
The speed can be found by derivatives
[tex]v_{y}[/tex] = dy / dt
[tex]v_{y}[/tex] = - A w sin (wt + φ
So that the amplitude is maximum without (wt + fi) = + -1
[tex]v_{y}[/tex] = A w
X axis
Let's reduce to the SI system
vₓ = 15 km / h (1000 m / 1 km) (1h / 3600s) = 4.17 m / s
As the car speed is constant
vₓ = d / t
t = d / v ₓ
t = 4 / 4.17
t = 0.96 s
This is the time between running two maximums, which is equivalent to a full period
w = 2π f = 2π / T
w = 2π / 0.96
w = 6.545 rad / s
We have the angular velocity we can find the spring constant
w² = k / m
m = 1200 + 4 80
m = 1520 m
k = w² m
k = 6.545² 1520
k = 65112 N / m
Let's use Newton's second law
F - W = 0
F = W
k x = W
x = mg / k
Case 1 when loaded with people
x₁ = 1520 9.8 / 65112
x₁ = 0.22878 m
Case 2 when empty
x₂ = 1200 9.8 / 65112
x₂ = 0.18061 m
The height variation is
ΔX = x₁ -x₂
ΔX = 0.22878 - 0.18061
ΔX = 0.0483 m
Final answer:
Calculating the rise in the car's suspension requires determining the spring constant from the initial displacement due to an 80 kg person's weight and then using it to find the new displacement when that weight is removed.
Explanation:
The problem relates to simple harmonic motion and the effects of weight on a car's suspension system. Upon the removal of the combined mass of the four 80 kg people (total of 320 kg) from the 1200 kg car, we need to find out how the car body rises due to this change in load. First, let's calculate the spring constant (k) using Hooks law (F = kx), where F is the force and x is the displacement.
When an 80 kg person gets in, the displacement (x) is 1.20 cm or 0.012 m, and the force (F) due to their weight is their mass times gravity (F = mg = 80 kg \\times 9.8 m/s²). Next, to find the new displacement caused by the removal of 320 kg, we'll set up the equation kx = mg for both situations and solve for the new displacement (x).
The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay. For this collision,
a. neither momentum nor kinetic energy is conserved.
b. only momentum is conserved.
c. both momentum and kinetic energy are conserved.
d. only kinetic energy is conserved.
Answer:
b
Explanation:
The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay.
This form of collision is called inelastic collision. And inelastic collision momentum is conserved but the kinetic energy is not conserved. Hence the correct option is b. only momentum is conserved.
Answer:
b
Explanation:
This form of collision is called inelastic collision.
g An Atwood machine has two hanging masses, m1 and m2, attached with a massless string over a pulley. If the pulley spins, rather than allowing the string to change direction without spinning, has mass M3, radius R, and moment of inertia equal to that of a disk, what is the tension force down on each side of the pulley?
Answer:
[tex]m_1( g -a) = T_1[/tex]
[tex]T_2 = m_2 (a +g)[/tex]
Explanation:
Given data;
Two hanging mass is given as m1 and m2
Mass of pulley is given as m3
radius of pulley is r
Assuming mass m1 is greater than m2
Take downward direction for mass m1
and upward direction for mass m2
and clockwise direction of pulley is positive
from newton second law on each mass
for Mass m1
[tex]\sum F_y = m_1 a_y[/tex]
[tex]m_1 g -T_1 = m_1 a[/tex]
[tex]m_1( g -a) = T_1[/tex]
for Mass m2
[tex]\sum F_y = m_2 a_y [/tex]
[tex]T_2 - m_2 g = m_2 a[/tex]
[tex]T_2 = m_2 (a +g)[/tex]
fro pulley
[tex]\sum \tau = I\alpha[/tex]
[tex]rT_1 -rT_2 =1/2 m_p r^2 \alpha[/tex]
A satellite is put into an elliptical orbit around the Earth. When the satellite is at its perigee, its nearest point to the Earth, its height above the ground is hp=227.0 km, and it is moving with a speed of vp=8.950 km/s. The gravitational constant G equals 6.67×10−11 m3·kg−1·s−2 and the mass of Earth equals 5.972×1024 kg.When the satellite reaches its apogee, at its farthest point from the Earth, what is its height ha above the ground?
Answer:
6633549.52903 m
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
m = Mass of the Earth = 5.972 × 10²⁴ kg
[tex]h_p[/tex] = Height above ground = 227 km
[tex]v_p[/tex] = Velocity at perigee = 8.95 km/s
Perigee distance is
[tex]R_p=6371+227=6598\ km[/tex]
The apogee distance is given by
[tex]R_a=\dfrac{R_p}{\dfrac{2Gm}{R_pv_p^2}-1}\\\Rightarrow R_a=\dfrac{6598\times 10^3}{\dfrac{2\times 6.67\times 10^{-11}\times 5.972\times 10^{24}}{6598\times 10^3\times (8.950\times 10^3)^2}-1}\\\Rightarrow R_a=13004549.52903\ m[/tex]
The height above the ground would be
[tex]h_a=13004549.52903-6371000=6633549.52903\ m[/tex]
The height above the ground is 6633549.52903 m
A cylindrical barrel is completely full of water and sealed at the top except for a narrow tube extending vertically through the lid. The barrel has a diameter of 80.0 cm, while the tube has a diameter of 1.10 cm. You can cause the lid to pop off by pouring a relatively small amount of water into the tube. To what height do you need to add water to the tube to get the lid to pop off the barrel? The lid pops off when the vector sum of the force of the atmosphere pushing down on the top of the lid and the force of the water pushing up on the bottom of the lid is 390 N up. Also, use g = 9.8 m/s2.
What is the height of water in the tube in cm?
To solve this problem it is necessary to apply the concepts related to pressure as a unit that measures the force applied in a specific area as well as pressure as a measurement of the density of the liquid to which it is subjected, its depth and the respective gravity.
The two definitions of pressure can be enclosed under the following equations
[tex]P = \frac{F}{A}[/tex]
Where
F= Force
A = Area
[tex]P = \rho gh[/tex]
Where,
[tex]\rho =[/tex] Density
g = Gravity
h = Height
Our values are given as,
[tex]d = 0.8m \rightarrow r = 0.4m[/tex]
[tex]A = \pi r^2 = \pi * 0.4^2 = 0.503m^2[/tex]
If we make a comparison between the lid and the tube, the diameter of the tube becomes negligible.
Matching the two previous expressions we have to
[tex]\frac{F}{A} = \rho g h[/tex]
Re-arrange to find h
[tex]h = \frac{F}{A\rho g}[/tex]
[tex]h = \frac{390}{(0.503)(1000)(9.8)}[/tex]
[tex]h = 0.079m[/tex]
[tex]h = 7.9cm[/tex]
Therefore the height of water in the tube is 7.9cm
A long, thin solenoid has 700 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 36.0 A/s . a. What is the magnitude of the induced electric field at a point near the center of the solenoid? b. What is the magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid?
To solve this problem it is necessary to apply the concepts related to the magnetic field in a solenoid.
By definition we know that the magnetic field is,
[tex]B = \mu_0 n I[/tex]
[tex]\frac{dB}{dt} = \frac{d}{dt} \mu_0 NI[/tex]
[tex]\frac{dB}{dt} = \mu_0 n\frac{dI}{dt}[/tex]
At the same tome we know that the induced voltage is defined as
[tex]\epsilon = \frac{\Phi}{dt}[/tex]
[tex]\epsilon = A \frac{dB}{dt}[/tex]
Replacing
[tex]\epsilon = A \mu_0 n\frac{dI}{dt}[/tex]
[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}[/tex]
PART A) Substituting with our values we have that
[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0)(36)}{2}\\\epsilon = 0V/m\\[/tex]
Therefore there is not induced electric field at the center of solenoid.
PART B) Replacing the radius for 0.5cm
[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0.5*10^{-2})(36)}{2}\\\epsilon = 7.9168*10^{-5}V/m[/tex]
Therefore the magnitude of the induced electric field at a point 0.5cm is [tex]7.9168*10^{-5}V/m[/tex]
The magnitude of the induced emf near the center of the solenoid is 0 V/m.
The magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid is 8 x 10⁻⁵ V/m.
The given parameters;
number of turns of the solenoid, N = 700 turns/mradius of the wire, r = 2.5 cmcurrent in the solenoid, I = 36 A/sThe magnitude of the induced emf near the center of the solenoid is calculated as follows;
[tex]B = \mu_0 nI\\\\\frac{dB}{dt} = \mu_0 n \frac{dI}{dt} \\\\\frac{dB}{dt} = (4\pi \times 10^{-7} \times 700 \times 36) = 0.032 \ T/s[/tex]
[tex]E = \frac{r}{2} (\frac{dB}{dt} )\\\\\ E = \frac{0}{2} (0.032)\\\\ E = 0 \ V/m[/tex]
The magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid is calculated as follows;
[tex]E = \frac{r}{2} (\frac{dB}{dr} )\\\\ E= \frac{0.5\times 10^{-2} }{2}( 0.032)\\\\E = 8\times 10^{-5} \ V/m[/tex]
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Two bumper cars in an amusement park ride collide elastically as one approaches the other directly fromthe rear. Car A has a mass of 435 kg and car B a mass of 495 kg, owing to differences in passenger mass. Ifcar A approaches at 4.50 m/s and car B is moving at 3.70 m/s calculatea) their velocities after the collisionb) the change in momentum of each.
Answer:
a) The velocity of car B after the collision is 4.45 m/s.
The velocity of car A after the collision is 3.65 m/s.
b) The change of momentum of car A is - 370.45 kg · m/s
The change of momentum of car B is 370.45 kg · m/s
Explanation:
Hi there!
Since the cars collide elastically, the momentum and kinetic energy of the system do not change after the collision.
The momentum of the system is calculated adding the momenta of each car:
initial momentum = final momentum
mA · vA + mB · vB = mA · vA´ + mB · vB´
Where:
mA = mass of car A
vA = initial velocity of car A
mB = mass of car B
vB = initial velocity of car B
vA´= final velocity of car A
vB´ = final velocity of car B
Let´s replace with the data we have and solve the equation for vA´:
mA · vA + mB · vB = mA · vA´ + mB · vB´
435 kg · 4.50 m/s + 495 kg · 3.70 m/s = 435 kg · vA´ + 495 kg · vB´
3789 kg · m/s = 435 kg · vA´ + 495 kg · vB´
3789 kg · m/s - 495 kg · vB´ = 435 kg · vA´
(3789 kg · m/s - 495 kg · vB´)/435 kg = vA´
Let´s write this expression without units for a bit more clarity:
vA´= (3789 - 495 vB´)/435
The kinetic energy of the system is also conserved, then, the initial kinetic energy is equal to the final kinetic energy:
initial kinetic energy of the system = final kinetic energy of the system
1/2 · mA · vA² + 1/2 · mB · vB² = 1/2 · mA · (vA´)² + 1/2 · mB · (vB´)²
Replacing with the data:
initial kinetic energy = 1/2 · 435 kg · (4.50 m/s)² + 1/2 · 495 kg · (3.70)²
initial kinetic energy = 7792.65 kg · m²/s²
7792.65 kg · m²/s² = 1/2 · 435 kg · (vA´)² + 1/2 · 495 kg · (vB´)²
multiply by 2 both sides of the equation:
15585.3 kg · m²/s² = 435 kg · (vA´)² + 495 kg · (vB´)²
Let´s replace vA´ = (3789 - 495 vB´)/435
I will omit units for clarity in the calculation:
15585.3 = 435 · (vA´)² + 495 · (vB´)²
15585.3 = 435 · (3789 - 495 vB´)²/ 435² + 495 (vB´)²
15585.3 = (3789² - 3751110 vB´ + 245025 vB²) / 435 + 495 (vB´)²
multiply both sides of the equation by 435:
6779605.5 = 3789² - 3751110 vB´ + 245025 vB² + 215325 vB´²
0 = -6779605.5 + 3789² - 3751110 vB´ + 460350 vB´²
0 = 7576915.5 - 3751110 vB´ + 460350 vB´²
Solving the quadratic equation:
vB´ = 4.45 m/s
vB´ = 3.70 m/s (the initial velocity)
a) The velocity of car B after the collision is 4.45 m/s
The velocity of car A will be teh following:
vA´= (3789 - 495 vB´)/435
vA´= (3789 - 495 (4.45 m/s))/435
vA´ = 3.65 m/s
The velocity of car A after the collision is 3.65 m/s
b) The change of momentum of each car is calculated as the difference between its final momentum and its initial momentum:
ΔpA = final momentum of car A - initial momentum of car A
ΔpA = mA · vA´ - mA · vA
ΔpA = mA (vA´ - vA)
ΔpA = 435 kg (3.648387097 m/s - 4.50 m/s) (I have used the value of vA´ without rounding).
ΔpA = - 370.45 kg · m/s
The change of momentum of car A is - 370.45 kg · m/s
ΔpB = mB (vB´ - vB)
ΔpB = 495 kg (4.448387097 m/s - 3.70 m/s) (I have used the value of vB´ without rounding).
ΔpB = 370.45 kg · m/s
The change of momentum of car B is 370.45 kg · m/s
I have used the values of the final velocities without rounding so we can notice that the change of momentum of both cars is equal but of opposite sign.
In Physics, when discussing collisions, the total momentum of a closed and isolated system is conserved. The change in momentum during a collision can be calculated using the principles of conservation of momentum and energy, with specifics depending on whether the collision is elastic or inelastic.
Explanation:The subject of this question is Physics, specifically dealing with the concepts of elastic and inelastic collisions, momentum, and the conservation of momentum principle. When two objects collide, momentum is exchanged between them, but the total momentum of the system remains constant if the system is closed and isolated. The change in momentum depends on whether the collision is elastic, where kinetic energy is conserved, or inelastic, where the objects stick together and kinetic energy is not conserved.
The example given with cars of equal mass indicates that after an elastic collision, the cars will trade velocities, and after a completely inelastic collision, they will move together with a combined momentum equal to their total initial momentum.
However, for the specific question asked by the student, which involves two bumper cars with different masses and velocities, the velocities after the collision and the change in momentum for each car would be determined by applying the elastic collision equations. These include conservation of momentum and conservation of kinetic energy. The precise calculations were not provided as part of the task.
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A hoop, a disk, and a solid sphere each have mass 1.4 kg and diameter 16 cm. Starting from rest, all three objects roll down a 7° slope. If the slope is 3 m long and all bodies roll without slipping, find the speed of each at the bottom.
I know I have to use rotational kinetic energy and translational kinetic energy to get the answer but im not sure how.
The answers are Hoop=1.89 m/s disk=2.18 m/s and sphere=2.26 m/s
Answer:
The velocities of hoop , disk and sphere are 1.89 m/s , 2.18 m/s , 2.26 m/s.
Explanation:
Lets find the speed of any general body of mass 'm' , moment of inertia 'I' , radius 'r'.
Let 'v' be the speed and 'ω' be the angular velocity of the body at bottom of the slope.
Since there is no external force acting on the system (Eventhough friction is acting at the point of contact of the body and slope , it does no work as the point of contact is always at rest and not moving) , we can conserve energy for this system.
Initially the body is at rest and at a vertical height 'h' from the ground.
Here , h=3sin(7°)
Initial energy = mgh.
Finally on reaching bottom h=0 but the body has both rotational and translational kinetic energy.
∴ Final energy = [tex]\dfrac{1 }{2}[/tex]I[tex]ω^{2}[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex].
Since the body is rolling without slipping.
v=rω
and
Initial Energy = Final Energy
mgh = [tex]\dfrac{1 }{2}[/tex]I[tex]ω^{2}[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex]
∴ mgh = [tex]\dfrac{1 }{2}[/tex]I[tex]\dfrac{v^{2} }{r^{2} }[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex]
∴ v = [tex]\sqrt{\frac{2mgh}{\frac{I}{r^{2} }+m } }[/tex]
For a hoop ,
I = m[tex]r^{2}[/tex]
Substituting above value of I in the expression of v.
We get,
v = [tex]\sqrt{gh}[/tex] = [tex]\sqrt{9.81×3sin(7°) }[/tex] = 1.89 m/s
Similarly for disk,
I = [tex]\dfrac{1}{2}[/tex]m[tex]r^{2}[/tex]
We get,
v = [tex]\sqrt{\frac{4gh}{3} }[/tex] = 2.18 m/s
For solid sphere ,
I = [tex]\dfrac{2}{5}[/tex]m[tex]r^{2}[/tex]
v = [tex]\sqrt{\frac{10gh}{7} }[/tex] = 2.26 m/s.