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Paul [167]
3 years ago
5

Examine the distance-time graph.

Physics
1 answer:
kakasveta [241]3 years ago
6 0

Answer:

A

Explanation:

because it is accelerating and as the vehicle accelerates the speed increas

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A friend says that the reason one's hair stands out while touching a charged Van de Graaff generator is simply that the hair str
Llana [10]

Answer:

Explanation:

Yes I agree with the statement .

When a person who is perfectly insulated from the earth , touches a Van de Graaff , his body acquires charge . when the hair acquires it, it stands out due to mutual repulsion . It is to be noted here that at pointed areas on a surface , there is larger accumulation of charge. Accumulation of charge is greater at hair tops .

It is also a general observation that when a bird sits on high tension wire , his feather stands out due to the same reason.

5 0
3 years ago
A ray at which angle would produce the most glare? A. 25 degrees B. 59 degrees C. 37 degrees D. 70 degrees
GREYUIT [131]

A ray at which angle would produce the most glare is at 70 degrees. The answer is letter D. the greater the incident of light is, the greater is its index of refraction and thus having greater angle to produce a light ray.

3 0
4 years ago
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(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
3 years ago
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Please help with this question!!!!
Maslowich
The answer is B hope this helps
4 0
3 years ago
A 200 kg roller coaster is located at the top of a hill where the height is 30 m. What is the velocity when it reches the bottom
maw [93]

Assuming it starts at rest, the roller coaster only has potential energy at the top of the hill, which is

E_P=(200\,\mathrm{kg})(30\,\mathrm m)g

When it reaches the bottom, its potential energy will have converted to kinetic energy,

E_K=\dfrac12(200\,\mathrm{kg})v^2

where v is its velocity at that point. By the law of conservation of energy, assuming no loss of energy due to other sources (e.g. sound, heat), we have

E_P=E_K\iff(6000\,\mathrm{kg}\cdot\mathrm m)g=(100\,\mathrm kg})v^2

\implies v=\sqrt{60g}\,\dfrac{\rm m}{\rm s}\approx24.2\,\dfrac{\rm m}{\rm s}

8 0
3 years ago
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