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marissa [1.9K]
4 years ago
10

NEED HELP!! WILL GIVE BRAINIEST!! When a driver presses the brake pedal, his car can stop with an acceleration of -5.4 meters pe

r second squared. how far will the car travel while coming to a complete stop if its initial speed was 25 meters per second?
Physics
2 answers:
nexus9112 [7]4 years ago
7 0

Answer:

d = 57.87 m

Explanation:

It is given that,

Acceleration of car due to applied brake is -5.4\ m/s^2

Its initial speed was 25 m/s

Its final speed was 0 as its comes to rest.

Leat d is the distance covered by the car. It can be calculated using third equation of kinematics as :

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-5.4)}\\\\d=57.87\ m

So, the distance covered by the car is 57.87 m.

8_murik_8 [283]4 years ago
3 0

Answer:

d = 57.87 m

Explanation:  have a nice day :)

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Calculate the time it takes for the car to reach the beginning of the hill given that acceleration as already found in part a is 2.89 m/s²

Answer:

5.88 s

Explanation:

Using kinematic equation, v=u+at where v and u are the final and initial velocities respectively, a is acceleration and t is time.

Considering the first part, acceleration is already found as 2.89 m/s and the final velocity is given as 17 m/s while the initial velocity is zero since it is at rest.

Making t the subject of formula then

t=(v-u)/a

Substituting the given figures then

t=(17-0)/2.89=5.8823529411764s

Rounded off, t=5.88 s

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3 years ago
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3 years ago
A car moving at 95 km/h passes a 1.00-km-long train traveling in the same direction on a track that isparallel to the road. If t
victus00 [196]

Answer:

Time to pass the train=0.05 h

How far the car traveled in this time=4.75 Km

Explanation:

We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:

\Delta V=V_{car}-V_{train}=95km/h-75km/h=20km/h

We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:

V=\dfrac{x}{t}

Where x is the distance, t is the time and v is the speed. using the data that we have:

V=\dfrac{x}{t}=\dfrac{1km}{20km/h}=0.05h

This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:

V=\dfrac{x}{t}\\x=v\cdot t\\x=95km/h\cdot 0.05h\\x=4.75km

7 0
3 years ago
A child's top is held in place upright on a frictionless surface. The axle has a radius of ????=3.21 mm . Two strings are wrappe
tekilochka [14]

Answer:

Angular momentum, L=6.47\times 10^{-3}\ m

Explanation:

It is given that,

Radius of the axle, r=3.21\ mm=3.21\times 10^{-3}\ m

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Time taken by the string to unwind, t = 0.32 s

We know that the rate of change of angular momentum is equal to the torque acting on the torque. The relation is given by :

\tau=\dfrac{dL}{dt}

Torque acting on the top is given by :

\tau=F\times r

Here, F is the tension acting on it. Torque acting on the top is given by :

\tau=2F\times r

2T\times r=\dfrac{L}{t}

L=2T\times r \times t

L=2\times 3.15\times 3.21\times 10^{-3}\times 0.32

L=6.47\times 10^{-3}\ m

So, the angular momentum acquired by the top is 6.47\times 10^{-3}\ m. Hence, this is the required solution.

7 0
4 years ago
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