Ask question

Ask question

All categories

3 years ago

5

An object experiences an acceleration of -6.8 m/s2. As a result, it accelerates from 54 m/s to a complete stop. How much dista

nce did it travel during that acceleration?

1 answer:

inessss [21]3 years ago

3 0

Answer:

The distance traveled during its acceleration, d = 214.38 m

Explanation:

Given,

The object's acceleration, a = -6.8 m/s²

The initial speed of the object, u = 54 m/s

The final speed of the object, v = 0

The acceleration of the object is given by the formula,

a = (v - u) / t m/s²

∴ t = (v - u) / a

= (0 - 54) / (-6.8)

= 7.94 s

The average velocity of the object,

V = (54 + 0)/2

= 27 m/s

The displacement of the object,

d = V x t meter

= 27 x 7.94

= 214.38 m

Hence, the distance the object traveled during that acceleration is, a = 214.38 m

You might be interested in

The EMF does not depend on which of the following?

zaharov [31]

Answer:

Explanation:

When a wire of length L is moved with velocity v , perpendicular to magnetic field B , EMF will be produced at its two ends . EMF can be calculated with the help of following expression

EMF = BLv

As per this formula , EMF produced will not depend upon number of electrons in the magnetic field.

So B is the right choice.

6 0

3 years ago

The sun radiates about 3.6 x 1026 joules of energy each second. How much mass does the lose each second? Show all of your work a

AlekseyPX

E=mc^2

hence m= E/c^2
= (3.6x10^26)/(9×10^16)
= 4x 10^9 kg

8 0

3 years ago

Read 2 more answers

A car is travelling at 20m/s, accelerates at 4m/s2 . Calculate its velocity after 5s?

IgorLugansk [536]

Answer:

<u>40 m/s</u>

Explanation:

The final velocity of an object that experiences a period of acceleration is given by:

v - vi = A*t

Initial velocity, vi, is = 20m/s

Acceleration, A, is 4m/2^2

Time, t, is 5 seconds

v - vi = A*t

v - 20m/s = (4m/2^2)/(5 sec)

v = (20 m/s) + (20 m/s)

v = 40 m/s

5 0

2 years ago

Review Conceptual Example 7 before starting this problem. A uniform plank of length 5.0 m and weight 225 N rests horizontally on

Andreas93 [3]

Answer:

0.82 m

Explanation:

We are given that

Weight, $w_1$ =225 N

Length,l=5 m

d=1.1 m

We have to find the distance x for which person weighs 385 N walk on the overhanging part of the plank before it just begins to tip.

Half length= $\frac{l}{2}=\frac{5}{2}=2.5 m$

$r=\frac{l}{2}-d=2.5-1.1=1.4 m$

$w_2=385 N$

When system in equilibrium then

$w_1r=w_2 x$

$225\times 1.4=385\times x$

$x=\frac{225\times 1.4}{385}$

$x=0.82 m$

6 0

4 years ago

Two balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L

olga_2 [115]

Answer:

Explanation:

Given

mass of balls $m= 5 kg$

$N=45.6 rev/s$

angular velocity $\omega =2\pi N=286.55 rad/s$

Length of Rod $2L=1.1 m$

Tension in the Second half of rod

$T_2=m\omega ^2(2L)=2m\omega ^2L$

$T_2=5\times (286.55)^2\times 1.1$

$T_2=451.609 kN$

For First Part

$T_1-T_2=m\omega ^2L$

$T_1=T_2+m\omega ^2L$

$T_1=3 m\omega ^2L$

$T_1=3\times 5\times (286.55)^2\times 0.55$

$T_1=677.41 kN$

7 0

3 years ago