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kvv77 [185]
3 years ago
9

Help me please

Physics
1 answer:
insens350 [35]3 years ago
6 0

Answer: a or b can u pls give me brainlest

Explanation:A straight line is a curve with constant slope. Since slope is acceleration on a velocity-time graph, each of the objects represented on this graph is moving with a constant acceleration.

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olga2289 [7]

There are two particular cases, the first is when Object A is attracted to the neutral wall. This would indicate that the object is not neutral, as there is an attraction.

At the same time we know that Object A is attracted to an object B. And therefore, the load of A must be opposite to that of B. Remember that opposite charges attract each other. If the charge of object B is positive, then the charge of object A will be negative.

Option B is correct: It has a negative charge.

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Why the specific heat capacity of the sun remain constant<br><br>​
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The heat remains constant because there’s nothing to cool it down
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Deadpool is doing a superhero landing from a 21 meter tall building what would his velocity be right before he hits the ground?
Morgarella [4.7K]

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4 0
2 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
What is the direction of the force on a positive charge when passing through a magnetic field as indicated in this diagram? Expl
maria [59]

Answer:

The direction of the magnetic force on a moving charge is perpendicular to the plane formed by v and B and follows right hand rule–1 (RHR-1)

Explanation:

hope this helps

3 0
3 years ago
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