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Andre45 [30]
2 years ago
9

A lound sound has large

Physics
1 answer:
Simora [160]2 years ago
7 0

Answer:

amplitude

Explanation:

A loud sound has large very high <em>amplitude</em>

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A(n) 10.1 g bullet is fired into a(n) 2.41 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9.8
fomenos

Answer:

v = 186.90\,\frac{m}{s}

Explanation:

The motion of ballistic pendulum is modelled by the appropriate use of the Principle of Energy Conservation:

\frac{1}{2}\cdot (m_{p}+m_{b})\cdot v^{2} = (m_{p}+m_{b})\cdot g \cdot h

The final velocity of the system formed by the ballistic pendulum and the bullet is:

v = \sqrt{2\cdot g\cdot h}

v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.031\,m)}

v\approx 0.78\,\frac{m}{s}

Initial velocity of the bullet can be calculated from the expression derived of the Principle of Momentum:

(0.0101\,kg)\cdot v = (2.41\,kg + 0.0101\,kg)\cdot (0.78\,\frac{m}{s} )

v = 186.90\,\frac{m}{s}

5 0
3 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
3 years ago
For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mas
inn [45]

Answer:

Yes is large enough

Explanation:

We need to apply the second Newton's Law to find the solution.

We know that,

F= ma

And we know as well that

a= \frac{v}{t}

Replacing the aceleration value in the equation force we have,

F= \frac{mv}{t}

Substituting our values we have,

F= \frac{(0.060)(55)}{4*10^{-3}}

F=825N

The weight of the person is then,

W = mg

W = (60)(9.8) = 558N

<em>We can conclude that force on the ball is large to lift the ball</em>

6 0
2 years ago
Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of
FrozenT [24]
Gravity obeys the inverse square law.  At 6400 km above the center of the Earth (Earth's surface) you weigh x.  Twice that reduces your weight to 1/4th.  Four times that height reduces your weight to 1/16th.  4 times 6400 km is 25,600 km.  But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer:  19,200 km.

Incidentally, it doesn't exactly work the opposite way.  At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.
6 0
3 years ago
According to Newton’s Second Law of Motion, an object will accelerate if which kind of force is applied?
Kobotan [32]

Answer:

unbalanced force

Explanation:

this is a guess so just look it up

7 0
2 years ago
Read 2 more answers
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