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3 years ago

A cheetah can go from a state of resting to running at 20 m/s in just two seconds what is the cheetah's average acceleration?

Physics

2 answers:

Sunny_sXe [5.5K]3 years ago

8 0

It's C i believe. To solve it we just take 20 and divide it by 2. Which gives us the average of 10 m/s

-Steel jelly

Oxana [17]3 years ago

4 0

The answer would be C. 10 m/s

Hope this helps

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Provide a general expression to clearly describe how to use yourknowlegde of an atom's composition to determine the sign and mag

nikdorinn [45]

The net charge on an atom is equal to the overall difference between the number of protons in the nucleus versus the number of electrons around the nucleus, where a negative sign represents less protons and a positive sign represents more protons (than electrons).

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3 years ago

A 6.5 x 104 W engine exerts a constant force on of 5.5 x 103 N on a car, the resulting velocity is?

jek_recluse [69]

Answer:

12m/s

Explanation:

Given parameters:

Power = 6.5 x 10⁴W

Force = 5.5 x 10³N

Unknown:

The resulting velocity = ?

Solution:

The velocity of a body is related to force and power using the expression below;

Power = Force x velocity

Insert the parameters and solve for velocity

6.5 x 10⁴ = 5.5 x 10³ x velocity

velocity = $\frac{6.5 x 10^{4} }{5.5 x 10^{3} }$ = 12m/s

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3 years ago

Determine whether each of the statements below is true or false, and place it in the appropriate bin. Objects with equal speeds

lisov135 [29]

Objects with equal speeds definitely have equal velocities. -- FALSE. For equal velocities, they also have to be going in the same direction.

If you are given an object's velocity, you can definitely determine its speed. -- TRUE. If you know the velocity, then you know both the object's speed and its direction.

If you know the distance an object travels, and the time it takes to do so, you can determine the object's velocity. -- FALSE. Knowing the distance and time, you can figure out the object's speed. But if you don't also know the direction it's moving, then you can't say what its velocity is.

If an object moves at constant speed, it must also be moving at constant velocity. -- FALSE. Besides constant speed, it also needs to move in a straight line to have constant velocity. If it turns, its velocity changes, even if its speed doesn't.

If an object moves at constant velocity, it must also be moving at constant speed. -- TRUE. Constant velocity means its speed AND its direction are not changing.

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After laboring through this one, I'm wondering if there can possibly be any more ways to say the same thing.

7 0

3 years ago

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

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3 years ago

Quasars are thought to be the nuclei of active galaxies in the early stages of their formation. A typical quasar radiates energy

uranmaximum [27]

Answer:

1.7531 smu/yn

Explanation:

check the attached file below for answer and explanation.

6 0

3 years ago