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2 years ago

If a bowling ball of mass 7.0 kg has a potential energy of 1100 Joules, how high above the ground is the bowling ball?

Physics

1 answer:

kirza4 [7]2 years ago

6 0

Answer:

16.03m(2dp)

Explanation:

Ep=m x g x h

1100=7.0x 9.8( gravitational field strength) x h

Height= 1100/7.0 x 9.8

=16.03498542

= 16.03m (2dp)

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A long electric cable is suspended above the earth and carries a current of 345 A parallel to the surface of the earth. The eart

jok3333 [9.3K]

Answer:

0.906 N

Explanation:

Formula for magnetic force acting on current carrying cable:

$F = IBLsin(\theta)$

Where I = 345A is the current in the wire, B = $5.6*10^{-5} T$ is the magnetic magnitude generated by Earth. L = 46.9 m is the cable length. $\theta = 88.2^o$ is the angle between vector B and cable direction.

$F = 345*5.6*10^{-5}*46.9*sin(88.2^o)$

$F = 0.906 N$

7 0

2 years ago

I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t

dezoksy [38]

Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }$
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
$f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz$

The wavelength of the standing wave is instead twice the length of the string:
$\lambda=2 L= 2 \cdot 24 m=48 m$

So the speed of the wave is
$v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s$

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
$t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s$

7 0

3 years ago

A baseball with a mass of 0.80 kg is given an acceleration of 20.00 m/s. How much net force was applied to the ball

dybincka [34]

a x m = f

.80 x 20 = 16

4 0

3 years ago

Driving along a boring stretch of interstate in Illinois, you start experimenting using the average speed equation you learned i

astra-53 [7]

The average speed would be 33.29m/s.
The average speed equation is:

$Average speed = \frac{total distance}{total time}$

First you will need to solve for the distance you traveled in each scenario. So we can solve this by getting the product of speed and the time traveled.

Scenario 1:
Speed = 29m/s
Time = 120s
Distance = ?

Distance = (29m/s)(120s)
= 3,480m

Scenario 2
Speed = 35m/s
Time = 300s
Distance = ?

Distance = (35m/s)(300s)
= 10,500m

Now that you have the distance of both, you can solve for your average speed.

$Average speed = \frac{total distance}{total time}$
$= \frac{3,480m+10,500m}{120s+300s}$
$= \frac{13,980m}{420s}$
$= 33.29m/s$

5 0

2 years ago

A person walks first at a constant speed of 5.50 m/s along a straight line from point A to point B and then back along the line

Sav [38]

Answer:

4.25 m/s

Explanation:

They walked the first distance at 5.50 m/s, then the same distance at 3 m/s.

Since the distances are equal, the average speed is simply the average of 5.50 and 3.

(5.50 + 3) / 2 = 4.25

Her average speed over the entire trip is 4.25 m/s.

8 0

3 years ago