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lina2011 [118]
3 years ago
8

A velocity-time graph can give you

Physics
1 answer:
N76 [4]3 years ago
3 0

Answer:

We can use them to determine whether or not the object is moving at any point in time. We can also use them to see what speed the object is travelling at that point in time. Using data from the graph, we can calculate any acceleration , the change in speed and the change in time.

Explanation:

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You are 12 miles north of your base camp when you begin walking north at a speed of 2 miles per hour. What is your location, rel
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If you walk at a pace of 2 miles per hour for 5 hours, you should have walked 10 miles. You would be 2 miles away from your base camp. 
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Mandy is testing an unknown solution to determine whether it is an acid or a base. She places a piece of red litmus paper into t
Marina CMI [18]
The solution is a base
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Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.
Harrizon [31]

1.549×10-19lJ is the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.

The equation E= hcE =hc, where h is Planck's constant and c is the speed of light, describes the inverse relationship between a photon's energy (E) and the wavelength of light ().

The Rydberg formula is used to determine the energy change.

Rydberg's original formula used wavelengths, but we may rewrite it using units of energy instead. The result is the following.

aaΔE=R(1n2f−1n2i) aa

were

2.17810-18lJ is the Rydberg constant.

The initial and ultimate energy levels are ni and nf.

As a change of pace from

n=5 to n=3 gives us

ΔE

=2.178×10-18lJ (132−152)

=2.178×10-18lJ (19−125)

=2.178×10-18lJ×25 - 9/25×9

=2.178×10-18lJ×16/225

=1.549×10-19lJ

Learn more about Rydberg formula here-

brainly.com/question/13185515

#SPJ4

8 0
2 years ago
2. A test reveals that 150 J of work is required to lift an object 3 m at a
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Answer:

50N

Explanation:

W=Fd

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7 0
2 years ago
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sammy [17]

Answer:

-0.25 rad/s^2

Explanation:

The equivalent of Newton's second law for rotational motions is:

\tau = I \alpha

where

\tau is the net torque applied to the object

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\alpha is the angular acceleration

In this problem we have:

\tau = -12.5 Nm (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

I=50.0 kg m^2 is the moment of inertia

Solving for \alpha, we find the angular acceleration:

\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2

3 0
3 years ago
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