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2 years ago

A car came to a stop from a speed of 35 m/s in a time of 8.1 seconds. What was the acceleration of the car?

2 answers:

5 0

Simply subtract the two velocities and divide by 8.1,

$\frac{0 - 35}{8.1} = - 4.32$

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey

Ne4ueva [31]2 years ago

4 0

Answer:

The acceleration of the car is -4.32 m/s^2.

Explanation:

Given that,

Speed of car u= 35 m/s

Time t = 8.1 s

We need to calculate the acceleration

using equation of motion

$v = u+at$

Where, v = final velocity

u = initial velocity

t = time

a = acceleration

$0=35+a\times8.1$

$a = -\dfrac{35}{8.1}$

$a = -4.32\ m/s^2$

Hence, The acceleration of the car is -4.32 m/s^2.

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The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi

ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done = mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = $= \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})$

kinetic energy = $= \frac{1}{2}*4*(1.5^{2}-11^{2})$ = -237.5 kg m/s2

-237.5 = 4*F

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2 years ago

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is

Harman [31]

Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
$T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}$
where
$k= \frac{M_{a}}{M_{b}}$

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2 years ago

Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approache

FromTheMoon [43]

Answer:

y = 54.9 m

Explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

W = -fr d

The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis

N - W cos θ = 0

The equation for the force of friction is

fr = μ N

fr = μ mg cos θ

We replace at work

W = - μ m g cos θ d

Mechanical energy in the lower part of the embankment

Em₀ = K = ½ m v²

The mechanical energy in the highest part, where it stopped

$Em_{f}$ = U = m g y

W = ΔEm = $Em_{f}$ - Em₀

- μ m g d cos θ = m g y - ½ m v²

Distance d and height (y) are related by trigonometry

sin θ = y / d

y = d sin θ

- μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

d (g syn θ + μ g cos θ) = ½ v²

d = v²/2 g (sintea + myy cos tee)

d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

d = 1555.85 /7.8145

d = 199.1 m

Let's use trigonometry to find the height

sin 16 = y / d

y = d sin 16

y = 199.1 sin 16

y = 54.9 m

8 0

2 years ago

a 10.0 kg block on a smooth horizontal surface is acted upon by two forces: a horizontal force of 30 N acting to the right and a

Eddi Din [679]

Answer:

$a=-3\ m/s^2$

Explanation:

<u>Second Newton's Law</u>

It allows to compute the acceleration of an object of mass m subject to a net force Fn. The relation is given by

$F_n=m.a$

The net force is the sum of all vector forces applied to the object. The block has two horizontal forces applied (in absence of friction): The 30 N force acting to the right and the 60 N force to the left. The positive horizontal direction is assumed to the right, so the net force is

$F_n=30\ N-60\ N=-30\ N$