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3 years ago

What type of issue does the effects manufacturer need

Physics

1 answer:

SOVA2 [1]3 years ago

4 0

Answer:

A) Availability

Explanation:

Right on Edge 2021

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A kid on a trampoline has 1,000 J of potential energy when they are at the top of a jump. How much kinetic energy will the kid h

emmasim [6.3K]

C: 500 J
Hope this helped!

7 0

3 years ago

Two air track carts move along an air track towards each other. Cart A has a mass of 450 g and moves toward the right with a spe

ra1l [238]

Answer:

0.465 kgm/s

Explanation:

Given that

Mass of the cart A, m1 = 450 g

Speed of the cart A, v1 = 0.85 m/s

Mass of the cart B, m2 = 300 g

Speed of the cart B, v2 = 1.12 m/s

Now, using the law of conservation of momentum.

It is worthy of note that our cart B is moving in opposite directions to A

m1v1 + m2v2 =

(450 * 0.85) - (300 * 1.12) =

382.5 - 336 =

46.5 gm/s

If we convert to kg, we have

46.5 / 100 = 0.465 kgm/s

Thus, the total momentum of the system is 0.465 kgm/s

6 0

4 years ago

A box of mass 1 kg is hung from a spring scale. The reading on the spring scale is approximately 10N. What would be the reading

8090 [49]

The answer will be 50N.
This is because the spring reads weight and weight is mass times acceleration due to gravity.5kg*10m/s2=50N

8 0

3 years ago

Read 2 more answers

A 174 pound Jimmy Cheek is riding on a 54 ft diameter Ferris Wheel. The normal force on Jimmy Cheek is 146 pounds when Jimmy is

Veronika [31]

To solve this problem we will apply the concepts related to the balance of Forces, the centripetal Force and Newton's second law.

I will also attach a free body diagram that allows a better understanding of the problem.

For there to be a balance between weight and normal strength, these two must be equivalent to the centripetal Force, therefore

$F_c = W-N$

$m\omega^2r = W-N$

Here,

m = Net mass

$\omega$ = Angular velocity

r = Radius

W = Weight

N = Normal Force

$m\omega^2r = 174-146$

The net mass is equivalent to

$F = mg \rightarrow m = \frac{F}{g}$

Then,

$m = \frac{174lb}{32.17ft/s^2}$

Replacing we have then,

$(\frac{174lb}{32.17ft/s^2})\omega^2 (54ft) =174lb-146lb$

Solving to find the angular velocity we have,

$\omega = 0.309rad/s$

Therefore the angular velocity is 0.309rad/s

6 0

3 years ago

in a young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in the distance th

blagie [28]

Answer:

Zero

Explanation:

Because using

Deta X= dsinစ x n(lambda)

But we know that for central maxima

n is zero

So after substituting

Deta x = 0

3 0

3 years ago