Two different objects are 3 meters from the reference point on a graph and share the same y-coordinate.

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the 14) car a deceleration of 3.50

sergij07 [2.7K]

To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving.

First, we manipulate the one of the kinematic equations

v^2 = v0^2 + 2 (a) (x) where v = 0 since the car stopped

Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,

we get v0 = sqrt (2(a)(x))

Substituting the known values,

v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s

Therefore, before stopping the car the original speed of the car would be 14.49 m/s

7 0

3 years ago

A 1.53-kg bucket hangs on a rope wrapped around a pulley of mass 7.07 kg and radius 66 cm. This pulley is frictionless in its ax

levacccp [35]

Answer:

$\alpha = 6.431\,\frac{rad}{s^{2}}$

Explanation:

The pulley is modelled by the Newton's Laws, whose equation of equilibrium is:

$\Sigma M = T \cdot R = \frac{1}{2}\cdot M \cdot R^{2}\cdot \alpha$

Given that tension is equal to the weight of the bucket, the angular acceleration experimented by the pulley is:

$T = \frac{1}{2}\cdot M \cdot R \cdot \alpha$

$m_{b}\cdot g = \frac{1}{2}\cdot M \cdot R \cdot \alpha$

$\alpha = \frac{2\cdot m_{b}\cdot g}{M\cdot R}$

$\alpha = \frac{2\cdot (1.53\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{(7.07\,kg)\cdot (0.66\,m)}$

$\alpha = 6.431\,\frac{rad}{s^{2}}$

7 0

2 years ago

A truck travels at a constant speed of 45 kilometers per hour. How far does the truck travel in 20 minutes?

Mekhanik [1.2K]

The answer is 15 kilometers in 20 minutes.

6 0

3 years ago

A car goes from rest to 40m/s in 5 seconds. What is its acceleration?

attashe74 [19]

Answer:

acceleration is 8 m/s^2

Explanation:

a= final velocity -initial velocity/time taken

5 0

3 years ago

A mechanic jacks up the front of a car to an angle of 8.0° with the horizontal in order to change the front tires. the car is 3.

SOVA2 [1]

First draw a free-body diagram. Torque T = Force F x Distance d where force is the component of gravitational force g and d is the lever arm distance to the pivot point. Since the pivot point is at the back tire we subtract that from the length of the car resulting in d = 1.12 - 0.40 = 0.72 meters = d. We are interested in the perpendicular component of the force exerted on the car jack so use sin 8 degrees then T=1130 kg x 9.81 m/s^2 x sin(8 degrees) x0.72 m = 1,110.80 Newton-meters

7 0

3 years ago