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expeople1 [14]
3 years ago
7

Wow free poin ts !!!!

Physics
2 answers:
zubka84 [21]3 years ago
8 0
Thank you very much, very much appreciated.
Eddi Din [679]3 years ago
5 0
Ended Wooooooooovjdnxnx
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which object has the most gravitational potential energy? A. an 8 kg book at a height of 3m B. an 5 kg book at a height of 3 m C
astra-53 [7]
I think A is the correct answer because its high is more higher compared to the others, and the mass really does not matter, to know the gravitational potential energy, we need to know how high the object is located because gravity does not show any favor to an object that has more mass or an object that doesnt
6 0
4 years ago
Read 2 more answers
Yuri built a model of the Solar System on his desk. But, it is not exactly the same as the real Solar System. What limits his mo
agasfer [191]

Answer:

It would be hard to make a model that shows the real sizes and distances between planets ( B )

Explanation:

Yuri building a model of the solar system will face the difficult of replicating the correct distances between the planets and the real sizes of the planets, because in model building the key factors of the Model must be represented properly.

The size of the planets and the distance between the planets are key factors when trying to model the solar system. but the distance between the planets depends on the position of the planets on their orbits which means the distances are not constant ( fixed ) hence that  would be the limitation of his model.

5 0
3 years ago
A baton twirler has a baton of length 0.36 m with masses of 0.48 kg at each end. Assume the rod itself is massless. The rod is f
AleksAgata [21]

Answer:

Part a)

When rotated about the mid point

\tau = 0.021Nm

Part b)

When rotated about its one end

\tau = 0.042 Nm

Explanation:

As we know that the angular acceleration of the rod is rate of change in angular speed

so we will have

\alpha = \frac{\Delta \omega}{\Delta t}

\alpha = \frac{2.4 - 0}{3.6}

\alpha = 0.67 rad/s^2

Part a)

When rotated about the mid point

I = 2mr^2

I = 2(0.48)(0.18)^2

I = 0.0311 kg m^2

now torque is given as

\tau = 0.0311 (0.67)

\tau = 0.021Nm

Part b)

When rotated about its one end

I = m(2r)^2

I = (0.48)(0.36)^2

I = 0.0622 kg m^2

now torque is given as

\tau = (0.0622)(0.67)

\tau = 0.042 Nm

3 0
3 years ago
Can I please have help
NeTakaya

Answer:

it will usually increase

Explanation:

potato

6 0
3 years ago
Two parallel slits are illuminated by light composed of two wavelengths. Wavelength A is 564 nm and the other is wavelength B an
anygoal [31]

Answer:

423nm

Explanation:

To find the unknown wavelength you take into account the distance y to the maximum central fringe, for light fringes and dark fringes.

- for light fringes:

dsin\theta=m\lambda\\\\sin\theta\approx\theta=\frac{y}{D}\\\\y=\frac{m\lambda_1D}{d}

- for dark fringes:

y=\frac{m\lambda_2/2 D}{d}

The third-order bright fringe (m= 3) of wavelength A coincides with the fourth dark fringe (m=4) of the wavelength B. Hence you have that:

\frac{(3)\lambda_1D}{d}=\frac{(4)\lambda_2D}{d}\\\\\lambda_2=\frac{3}{4}\lambda_1=\frac{3}{4}(564nm)=423nm

hence, the wavelength B is 423nm

7 0
4 years ago
Read 2 more answers
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