Answer:
The acceleration of the car will be 
Explanation:
We have given that distance from stop sign s = 200 m
Time t = 0.2 sec
We have to find the constant acceleration
Now from second equation of motion 


So the acceleration of the car will be 
The angular speed of the device is 1.03 rad/s.
<h3>What is the conservation of angular momentum?</h3>
A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.
Using the conservation of angular momentum

Here, = the system's angular momentum before the collision
= 0 + mv
= (0.005)(450)(0.752)
= 1.692 kgm²/s
The moment of inertia of the system is given by
I = 2(M₁R₁² + M₂R₂²)+ mR₁²
= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²
= 1.6292 kgm²
Here, = Iω
So,
1.692 = 1.6292(ω)
ω = 1.03 rad/s
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Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
Power P is the amount of energy E transferred or converted per unit time t, and is expression is:
P = E/t
The necessary energy to lift the box its the work of against the Gravity Force, and is given by:
E = mgh
Then the power is:
P = E/t
P = mgh/t
P = (45 kg)(9.8 m/s²)(1.7 m )/ 30
P = 24.99 W
<h2>P ≈ 25 W</h2>
Answer:
mass X velocity
Explanation:
The momentum of a body is the product of its mass and velocity