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3 years ago

6

A car traveling at 38 m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is its acceleration?

1 answer:

Katena32 [7]3 years ago

8 0

a = 38 m/s / 7s

a = 5.43 ms⁻²

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Please help me quickly!!!

vichka [17]

Hey! So referring to the data the thing we can clearly see is that in a vacuum, everything, regardless of its mass, falls at the same speed.

Acceleration is often confused with speed, or velocity, but the difference is, acceleration by definition is the rate of which an object falls with respect to its mass and time.

Every single thing in the world falls at the same acceleration, this is because of gravity. The difference is the speed of which it falls. In space, there is not any gravity, and so, the objects are able to fall at the same speed regardless of their mass.

6 0

3 years ago

The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x

Diano4ka-milaya [45]

Answer:

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Explanation:

Given that

At X=0 V=Vo

At X=X1 V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

$V^2=U^2-2aS$

$0=V_o^2-2a X_1$

$a=\dfrac{V_o^2}{2X_1}$

So the friction force on the box

Ff= m x a

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Where m is the mass of the box.

4 0

3 years ago

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

4 0

2 years ago

calculate the work done in kilo joules in lifting a mass of 20kg at steady velocity through a vertical height of 20m

bulgar [2K]

Answer:

$\huge\boxed{\sf Work\ done = 4 kJ}$

Explanation:

Since work done is in the form of potential energy, we will use the formula of potential energy here.

We know that,

<h3>P.E. = mgh </h3>

Where,

m = mass = 20 kg

g = acceleration due to gravity = 10 m/s²

h = vertical height = 20 m

So,

<h3>Work done = mgh</h3>

Work done = (20)(10)(20)

Work done = 4000 joules

Work done = 4 kJ

$\rule[225]{225}{2}$

5 0

1 year ago

The space shuttle is accelerated off its launch pad to a velocity of 525 m/s in 18.0 seconds.

Eva8 [605]

Answer: 29.17m/s^2

Explanation:

Given the following :

Velocity = 525 m/s

Time = 18 seconds

Acceleration = change in Velocity with time

Using the motion equation:

v = u + at

Where v = final Velocity

u = Initial Velocity and t = time

Plugging our values

525 = 0 + a × 18

525 = 18(a)

a = 525 / 18

a = 29.166666

a = 29.17 m/s^2

8 0

3 years ago