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umka21 [38]
3 years ago
7

On the moon what would be the force of gravity acting on an object that has a mass of 7 kg?

Physics
1 answer:
Rudiy273 years ago
4 0

Answer:

0

Explanation:

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A person hits a tennis ball with a mass of 0.058 kg against a wall.
horrorfan [7]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

F=\dfrac{m(v-u)}{t}

F=\dfrac{0.058\times (11-(-11))}{2.1}

F = 0.607 N

(b) Area of wall, A=3\ m^2

Let P is the average pressure on that area. It is given by :

P=\dfrac{F}{A}

P=\dfrac{0.607\ N}{3\ m^2}

P = 0.202 Pa

Hence, this is the required solution.

8 0
3 years ago
What is a target ceiling ?
zloy xaker [14]

Answer:

Target ceiling. the upper limit of your physical activity. Target fitness zone. Above the threshold of training and below the target ceiling.

Hope this helps. Can u give me brainliest

Explanation:

5 0
3 years ago
What three variable factors determine the force of gravity between any two objects?
brilliants [131]
The force of gravity between two objects is:
F = G*m1*m2/r^2

So, it is dependent of the two masses and the distance between their centers of mass.  
5 0
3 years ago
The <br> of a wave is the density of the medium’s particles at the compression of the wave.
andrew11 [14]
<span>Amplitude is the correct answer. I hope this helps.</span>
6 0
3 years ago
Read 2 more answers
What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed
RoseWind [281]
<h3><u>Minimum uncertainty in the vertical component of the momentum of each photon:</u></h3>

According to Heisenberg's Uncertainty principle, both the “position and velocity of the particle” cannot be measured exactly at the same time. The momentum of the particle equals the product of its mass and velocity. And it can be inferred that the “product of the uncertainties” in the “momentum and the position” of a particle equals \frac{h}{4 \pi}.  

Immediately after the photon has passed through the slit, given particle has a momentum uncertainty of \Delta P_{x} and its position uncertainty is \Delta x, then the minimum uncertainty in its momentum will be

\Delta P_{x} . \Delta x \geq \frac{h}{4 \pi}

8 0
3 years ago
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