Answer:
We know from the basic speed distance relation that
Since the car started from rest and it covered the distance between the 2 officer's in 19 minutes we have speed of the car
Which clearly exceeds the limit of 
Answer:
The centripetal acceleration for the first radius; 2.0 m = 50 m/s²
The centripetal acceleration for the second radius; 4.0 m = 25 m/s²
The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²
The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²
The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²
Explanation:
Given;
mass of the object, m = 1 kg
velocity of the object, v = 10 m/s
different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m
The centripetal acceleration for the first radius; 2.0 m
The centripetal acceleration for the second radius; 4.0 m
The centripetal acceleration for the third radius; 6.0 m
The centripetal acceleration for the fourth radius; 8.0 m
The centripetal acceleration for the fifth radius; 10.0 m
<h3>Hello There!!</h3>
<h3><u>Given</u>,</h3>
Force(F) = 150N
Mass(m) = 90kg
<h3><u>To </u><u>Find,</u></h3>
Acceleration(a) = ?
<h3><u>We know,</u></h3>
F= m×a
<h3>Hope this helps</h3>
A 100 g cart is moving at 0.5 m/s that collides elastically from a stationary 180 g cart. Final velocity is calculated to be 0.25m/s.
Collision in which there is no net loss in kinetic energy in the system as a result of the collision is known as elastic collision . Momentum and kinetic energy both are conserved quantities in elastic collisions.
Collision in which part of the kinetic energy is changed to some other form of energy is inelastic collision.
For an elastic collision, we use the formula,
m₁V₁i+ m₂V₂i = m₁V1f + m₂V₂f
For a perfectly elastic collision, the final velocity of the 100g cart will each be 1/2 the velocity of the initial velocity of the moving cart.
Final velocity = 0.5/2
=0.25 m/s.
To know more about elastic collision, refer
brainly.com/question/7694106
#SPJ4
Answer:
a) During the reaction time, the car travels 21 m
b) After applying the brake, the car travels 48 m before coming to stop
Explanation:
The equation for the position of a straight movement with variable speed is as follows:
x = x0 + v0 t + 1/2 a t²
where
x: position at time t
v0: initial speed
a: acceleration
t: time
When the speed is constant (as before applying the brake), the equation would be:
x = x0 + v t
a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:
x = 0m + 26 m/s * 0.80 s = <u>21 m </u>
b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):
v = v0 + a* t
0 = 26 m/s + (-7.0 m/s²) * t
-26 m/s / - 7.0 m/s² = t
t = 3.7 s
With this time, we can calculate how far the car traveled during the deacceleration.
x = x0 +v0 t + 1/2 a t²
x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>
