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Sophie [7]
3 years ago
7

Is there a sound loud enough to kill a person ?

Physics
2 answers:
Svetllana [295]3 years ago
7 0

Yes, In space I believe go to the NASA website and search your question there.

shepuryov [24]3 years ago
6 0

In space there is. Items such as a supernova will instantly end your life.

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7. Delia Gonzalez, Mother of Felix, said Felix wanted to play Football but
Gelneren [198K]
Determinatio/Goal setting
4 0
2 years ago
A communication satellite is in a circular path orbit around Earth. If the speed of the satellite is constant, the net force act
allsm [11]

Answer:

is changing in direction, but constant in magnitude

Explanation:

This question is a bit tricky since the velocity of the satellite is changing, but the speed is constant.

Speed is simply a measure of how fast you are going. It doesn't matter where you're going, just how quickly.

Velocity, on the other hand, does care about which direction you're going. For example, it could be then when you travel right, your velocity is positive, and when you travel left, your velocity is negative. This is the similar for a 2D shape like a circular orbit

Since we know velocity is changing, there must be acceleration which changes that velocity (since acceleration <em>is</em><em> </em>the change in velocity: going from 0 to 60 mph, for example)

Thus, with a non-zero net acceleration, we know that there must be a force that is changing in direction, but constant in magnitude (since the orbit is a circle, and always attracted to the center of the Earth at equal distance).

4 0
3 years ago
A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.
Nataliya [291]

Answer:

Approximately 1.2 \times 10^{4}\; {\rm J} (assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: g = 9.81\; {\rm N \cdot kg^{-1}}.

The change in the gravitational potential energy ({\rm GPE}) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass m increased by \Delta h, the {\rm GPE} of that object would have increased by m\, g\, \Delta h.

In this question, the height of this object increased by \Delta h = 6.0\; {\rm m}. The mass of this object is m = 200\; {\rm kg}. Thus, the {\rm GPE} of this object would have increased by:

\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}.

(Note that 1\; {\rm N \cdot m} = 1\; {\rm J}.)

3 0
2 years ago
Which of following is typical stored energy used to power todays vehicles?
Gala2k [10]
Electrical energy.................
3 0
4 years ago
A 0.140-kg baseball is dropped from rest from a height of 1.8 m above the ground. It rebounds to a height of 1.4 m. What change
aliina [53]

Answer:

\Delta p=-1.56\ kg-m/s

Explanation:

It is given that,

Mass of the baseball, m = 0.14 kg

It is dropped form a height of 1.8 m above the ground. Let u is the velocity when it hits the ground. Using the conservation of energy as :

u=\sqrt{2gh}

h = 1.8 m  

u=\sqrt{2\times 9.8\times 1.8}

u = 5.93 m/s

Let v is the speed of the ball when it rebounds. Again using the conservation of energy to find it :

v=\sqrt{2gh'}

h' = 1.4 m  

v=-\sqrt{2\times 9.8\times 1.4}

v = -5.23 m/s

The change in the momentum of the ball is given by :

\Delta p=m(v-u)

\Delta p=0.14(-5.23-5.93)

\Delta p=-1.56\ kg-m/s

So, the change in the ball's momentum occurs when the ball hits the ground is 1.56 kg-m/s. Hence, this is the required solution.

4 0
4 years ago
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