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3 years ago

2. What is the kinetic energy of a 7.26 kg bowling ball that is rolling at a speed of 2 m/s?

2 answers:

6 0

Kinetic energy is the energy possessed by an object when that object is moving in space. The higher the mass of an object or higher the speed of an object the higher the kinetic energy will be.

So to calculate the Kinetic Energy we can use the following formula

K.E=(1/2)*m*v^2

Inserting the values in formula gives:

K.E=1/2*7.26*2^2

14.52J

This is the final answer which gives the kinetic energy of the ball.

Dimas [21]3 years ago

3 0

Answer:

The kinetic energy of bowling ball is 14.52J

Explanation:

Given that the formula of kinetic energy is (1/2)mv². Then substitute the following value into the formula:

m = 7.26kg

v = 2m/s

k.e = (1/2)×7.26×2²

= 14.52J

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If an object is moving eastward and slowing down, then the direction of its acceleration is

Alchen [17]

C. Eastward. Acceleration is the change in speed so it can be a positive (speeding up) or negative (slowing down) acceleration

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2 years ago

A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

ElenaW [278]

Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

Weight of the beam, ( $W_{b}$ ) = 332 N

Weight of the Suki, ( $W_{s}$ ) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$

$-W_{s} \times x + W_{b} \times 1.5$ = 0

x = $\frac{W_{b} \times 1.5}{W_{s}}$

= $\frac{332 N \times 1.5}{505 N}$

= 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

8 0

4 years ago

) you carry a 7.0 kg bag of groceries 1.2 m above the ground at constant velocity across a 2.7 m room. how much work do you do o

lapo4ka [179]

M = 7.0 kg, the mass of the groceries
h = 1.2 m, the elevation of the bag of groceries

The bag of groceries moves a constant velocity over the 2.7-m room.
At constant velocity, there is no applied force, and the kinetic energy remains constant.

At an elevation of 1.2 m, there is an increase in PE (potential energy) given by
V = m*g*h
= (7.0 kg)*(9.8 m/s²)*(1.2 m)
= 82.32 J

The change in PE is equal to the work done.

Answer: 82.3 J

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4 years ago

Two ropes are attached to a 40-kg object. The first rope applies a force of 25 N and the second, 40 N. If the two ropes are perp

kifflom [539]

Answer:

1.2 m/s^2

Explanation:

I will ignore gravity and friction....and will choose 25 N as UP and 40 N to Left (see diagram)

Resultant force: R^2 = 25^2 + 40^2 R = sqrt(2225)=47 N

F = ma

F/m = a

47 N / 40 kg = a = 1.2 m/s^2