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1 year ago

Please help me with this,

Physics

1 answer:

Naddik [55]1 year ago

6 0

Answer:

HOLA AMIGOS

Explanation:

THE FORCE FLIED IS SO PROTECTED THAT WITH ITS POWER IT CAN GENAERATE A WHOLE MACHINE

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Which acceleration-time graph corresponds to the motion of the car if it moves toward the right, while slowing down at a steady

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We define acceleration as the rate of change of the velocity

Thus, if you have positive velocity and positive acceleration, your <u>speed increases.</u>

If you have positive velocity and negative acceleration, your speed decreases.

Now you get the idea, we will see that the correct option is graph 1.

We know that the car moves towards the right (let's define this as "the car has positive velocity") and we also know that te car is slowing down constantly (thus the acceleration needs to be negative and constant).

By looking at the graphs, the only one with these properties is graph 1.

If you want to learn more, you can read:

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2 years ago

A force that needs to touch something to affect it

bija089 [108]

I would say friction because it requires to surfaces in order for the force to take place

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3 years ago

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Newton’s third law of motion explains the two forces namely ‘action’ and ‘reaction’ coming into action when the two bodies are i

SIZIF [17.4K]

(b) Always act on the different bodies in opposite directions

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3 years ago

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Place your hands on the side of your head, with your palms facing forward. This should look like you have given yourself large e

nirvana33 [79]

Answer:

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2 years ago

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$

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2 years ago