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dusya [7]
3 years ago
15

1 point

Physics
1 answer:
Oduvanchick [21]3 years ago
5 0
The distance it traveled is 147
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If
nadya68 [22]

Answer:

hhjjkkkksksksjskkskakakkskskksksksoao

Explanation:

hiiii look forward but I don't know how to do it

8 0
4 years ago
How much potential energy does a 50-N box have when lifted at a height of 1.5M?
nikitadnepr [17]

The correct answer is: Option (A) 75 J

Explanation:

First, be careful with the units here. As you can see it is mentioned that there is a 50N box. It means that the weight (<em>mg</em>) of the box is given as the unit is <em>Newton</em>, not its mass (which is in kg).

As,

Potential-energy = mass * acceleration-due-to-gravity * height

PE = m*g*h --- (A)


In equation (A), mg is actually the weight of the box, which is given.

mg = 50N

h = height = 1.5m

Plug the values in equation (A):

PE = 50 * 1.5  = <em>75 J (Option A)</em>

3 0
4 years ago
I NEED HELP I AM SO CONFUSED, WILL GIVE BRAIN.
MakcuM [25]

Answer:

210

Explanation:

A ball rolls horizontally off the cliff at a speed of 30 m/s. It takes 7 seconds for the ball to hit the ground. What is the height of the cliff and the horizontal distance traveled by the ball?

S = (1/2)*9.8 m/s^2 * 7^2 = 240.1 m if the ball is very dense so air resistance, and therefore terminal velocity, can be ignored.

S = v * t = 30 m/s * 7 s = 210 m for the horizontal distance, again assuming negligible air resistance.

5 0
3 years ago
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
storchak [24]

Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

The formula for the force exerted between two charges is

F=k \dfrac{ q_1q_2}{r^2}

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

F=k\dfrac{q^{2}}{r^2}

\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

7 0
3 years ago
He graph below shows the velocity f(t) of a runner during a certain time interval:
Annette [7]
<span>The following that describes the intercepts on the graph is "The initial velocity of the runner was 4 m/s, and the runner stopped after 8 seconds." It is because the starting point of the line is at 4 and then the ending point is at 8.

</span>
5 0
3 years ago
Read 2 more answers
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