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2 years ago

A car accelerates uniformly from rest at a speed of 1.67 ft s^2 over a distance of 5 yards.What is the acceleration of a car?

Physics

2 answers:

stepladder [879]2 years ago

7 0

Answer:

a= 0.256 ft/s^2

Explanation:

Solution:

Since the car has started from rest the initial velocity will be zero

i.e. Vi=0

It gets a speed of 1.6 fts^2,

i.e. Vf= 1.6 fts^2

The distance is 5 yards

i.e d= 5 yards

According to the equation of motion

2as= VF^2- Vi^2

Thus a= (VF^2- Vi^2)/2s

a= (1.6^2 - 0)/ 2*5

a= 0.256 ft/s^2

Nina [5.8K]2 years ago

5 0

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

velocity= distance/time

23= 30/t

t= 30/23

t= 1.30 seconds

hence

acceleration= 23/1.30

accelaration= 17.69 m/s^2

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The block is made of A) Tin, as its specific heat capacity is $0.225 J/(g^{\circ}C)$

Explanation:

When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by $\Delta T$ , according to the following equation :

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In this problem, we have:

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2 years ago

An green hoop with mass mh = 2.8 kg and radius rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass

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The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
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the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

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