Answer:
same
Explanation:
Acc. to Einstien's postulate of special theory of
Relativity ,
Velocity of the light beam is same in all frames of references
(a) If the freight car is at rest
The frame we can assumed as Non - inertial frame of reference
s
In the inertial frame of reference , velocity of the light beam has its own value as : 3 x 10^8 m/s
(b) If the freight car is moving , the frame we can assumed as Non -inertial frame of reference
In thus case also , The velocity of the light beam will also have the same value as ; 3 x 108 m/s
That depends on what quantity is graphed.
It also depends on what kind of acceleration is taking place ...
continuous change of speed or continuous change of direction.
-- If the graph shows speed vs time, and the acceleration is a change
in speed, then the graph is a connected series of straight-line pieces.
Each straight piece slopes up if speed is increasing, or down if speed
is decreasing.
-- If the graph shows speed vs time, and the acceleration is a change in
direction only, then the graph is a straight horizontal line, since speed is
constant.
-- If the graph shows direction vs time, and the acceleration is a change
in speed only, then the graph is a straight horizontal line, since direction
is constant.
-- If the graph shows direction vs time, and the acceleration is a change
in direction, then the graph is a connected series of pieces of line.
Each piece may be straight if the direction is changing at a constant rate,
or curved if the direction is changing at a rate which grows or shrinks.
Each piece may slope up if the angle that defines the direction is growing,
or may slope down if the angle that defines the direction is decreasing.
-- If the graph shows distance vs time, and the acceleration is a
change in speed, then the graph is a connected series of pieces
of curves. Each piece curves up if speed is increasing, or down if
speed is decreasing.
-- If the graph shows distance vs time, and the acceleration is a change
in direction only, then the graph is a straight line sloping up, since speed
is constant.
That latest value for the Angle is in Grads, not in Kilograms.
Apply law of conservation of momentum along vertical direction.
![m_1v_1sin\theta_1-m_2v_2sin\theta_2=0](https://tex.z-dn.net/?f=m_1v_1sin%5Ctheta_1-m_2v_2sin%5Ctheta_2%3D0)
![v_2=\frac{v_1sin\theta_1}{sin\theta_2}=\frac{sin54}{sin36}v_1 = 1.376v_1](https://tex.z-dn.net/?f=v_2%3D%5Cfrac%7Bv_1sin%5Ctheta_1%7D%7Bsin%5Ctheta_2%7D%3D%5Cfrac%7Bsin54%7D%7Bsin36%7Dv_1%20%3D%201.376v_1)
Apply law of conservation of momentum along the horizontal direction
![m_1u_1=m_1v_1cos\theta_1+m_2v_2cos\theta_2](https://tex.z-dn.net/?f=m_1u_1%3Dm_1v_1cos%5Ctheta_1%2Bm_2v_2cos%5Ctheta_2)
![u_1=v_1(cos\theta_1+1.376cos\theta_2)](https://tex.z-dn.net/?f=u_1%3Dv_1%28cos%5Ctheta_1%2B1.376cos%5Ctheta_2%29)
![u_1=v_1(cos(54)+1.376cos(36))](https://tex.z-dn.net/?f=u_1%3Dv_1%28cos%2854%29%2B1.376cos%2836%29%29)
![u_1=(1.7) v_1](https://tex.z-dn.net/?f=u_1%3D%281.7%29%20v_1)
![v_1= \frac{3.14}{1.7}=1.84m/s](https://tex.z-dn.net/?f=v_1%3D%20%5Cfrac%7B3.14%7D%7B1.7%7D%3D1.84m%2Fs)
The second ball velocity is ![v_2 = (1.376)(1.84)=2.531m/s](https://tex.z-dn.net/?f=v_2%20%3D%20%281.376%29%281.84%29%3D2.531m%2Fs)
The magnitud of final total momentum is
![m(v_1+v_2)=(1.37)(2.531+1.84)=5.98kgm/s](https://tex.z-dn.net/?f=m%28v_1%2Bv_2%29%3D%281.37%29%282.531%2B1.84%29%3D5.98kgm%2Fs)
The magnitude of final energy is
![\frac{1}{2}m(v^2_1+v^2_2)=\frac{1}{2}(1.37)(2.531^2+1.84^2)=6.07J](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dm%28v%5E2_1%2Bv%5E2_2%29%3D%5Cfrac%7B1%7D%7B2%7D%281.37%29%282.531%5E2%2B1.84%5E2%29%3D6.07J)
Answer Expert Verified
Hello here is your answer and also please mark me as brainlest
Since one revolution around a circle is 2 pi radian; hence 1 rpm equals 2 pi radians per minute. And because a minute has 60 seconds, 1 rpm equals 1/60 revolution per second. Therefore, we have 2 pie/60 * 15.3 = 0.2513 rps. The linear velocity v = wr where w is the angular velocity in rad/s and r is the distance. So we have 0.2513 * 10.0 = 2.513 rad/s The centripetal acceleration is given by a = w^2 r = (2.513)^2 * 10 = 63.15 rad/s2 The centripetal force F = mass * centripetal acc = 75 * 63.15 = 4736.25 N The torque = centripetal force * distance = 4736 * 63.15 = 299078.4 Nm Two forces acts on the astronaut. The normal force and acceleration due to gravity.
I hope this help you
brainlest
Answer:
The average power delivered by the engine is 90 hp.