All categories

4 years ago

A car is traveling at 10 m/s starts to decelerate steadily. It comes to a complete stop in 20 seconds what is it’s acceleration

Physics

2 answers:

zhannawk [14.2K]4 years ago

6 0

Hello!

A car traveling at 10 m/s starts to decelerate steadily. It comes to a complete stop in 20 seconds. What is it's acceleration ?

We have the following data:

Vi (initial velocity) = 10 m/s (starts)

Vf (final velocity) = 0 m/s (stop)

t (time) = 20 s

a (acceleration) = ? (in m/s²)

We apply the data to the formula of the hourly function of the velocity, let us see:

$V_f = V_i + a*t$

$0 = 10 + a*20$

$- 10 = 20\:a$

$20\:a = - 10$

$a = \dfrac{-1\diagup\!\!\!\!0}{2\diagup\!\!\!\!0}$

$\boxed{\boxed{a = - 0.5\:m/s^2}}\Longrightarrow(the\:car\:slows\:down)\:\:\:\:\:\:\bf\green{\checkmark}$

Answer:

The acceleration is -0.5 m/s² (decelerate)

________________________________

$\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}$

Ludmilka [50]4 years ago

3 0

Answer:

-0.5 m/s

Explanation:

given:

u (initial velocity) = 10 m/s

v (Final velocity) = 0 m/s (because it is given that it comes to a complete stop)

t (time taken) = 20 s

therefore,

acceleration = v - u / t

a = 0 - 10 / 20

a = -10 / 20

a = -1/2

a = -0.5

There is a acceleration of -0.5 m/s

You might be interested in

Two point charges 3q and −8q (with q > 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so

riadik2000 [5.3K]

Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

2√2 x = √3 (L+x)

2√2 x - √3 x = √3 L

x(2√2 - √3 ) = √3 L

x = √3 L / (2√2 - √3 )

Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q 8q / L²

Q = 8q (x² / L²)

so charge required = - 8q (x² / L²)

and its distance from x on negative x side = √3 L / (2√2 - √3 )

3 0

3 years ago

A client experiences difficulty in performing the prone iso-abs exercise. Which of the following should be suggested as an regre

butalik [34]

Answer:

a. Quadruped arm and opposite leg raise

Explanation:

Quadruped arm and opposite leg lift

- Kneel on the floor, lean forward and place your hands down.

- Keep your knees in line with your hips and hands directly under your shoulders.

- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.

- Go back to the starting position.

This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.

It is also used together with other exercises for the treatment of hyperlordosis.

8 0

3 years ago

4. What is the magnitude and direction of the gravitational force that acts on a man

Cloud [144]

The gravitational force acting on the man is 800 N towards the Earth's centre

Explanation:

The weight of an object on the Earth is exactly the gravitational force exerted by the Earth on the object.

The gravitational force exerted by the Earth on an object located at the Earth's surface is given by:

$F=G\frac{Mm}{R^2}$

where

G is the gravitational constant

M is the Earth's mass

m is the mass of the object

R is the radius of the Earth

And the direction of the force is towards the Earth's centre.

Since G, M and R are constant, they are grouped into a single constant called g, acceleration of gravity:

$g=\frac{GM}{R^2}$

therefore the gravitational force can be rewritten as

$F=mg$

And this is the usual equation that we use to calculate the weight of an object.

Therefore, weight and gravitational force acting on an object on Earth are the same thing: so, the gravitational force acting on the man is equal to his weight, 800 N, and it acts towards the Earth's center.

Learn more about weight and gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$