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3 years ago

How can you decrease the momentum of an object?

2 answers:

7 0

You will have to "put force against the object to slow it down." Momentum is the force that is keeping a object moving in a certain direction so if you would want to slow down the object you will have to put another force against the object to slow it down or stop it. For example: a person kicks a ball, the ball moving is the momentum. So, if you would want to stop the ball you would have t put something in its path to slow it down which is the decreasing of it's momentum. Therefore you would put a bump in the wall and when the ball hits the bump it slows down.

Hope this helps!

boyakko [2]3 years ago

6 0

The momentum of an object is the product of (its mass)
and (its velocity). So, if you want to decrease the object's
momentum, then you're going to have to decrease one of
those quantities.

You have exactly two choices:

1). Reduce its mass. (Put the object on a diet, or cut a piece off.)
or
2). Reduce its speed.

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A 2.0-mm-long, 1.0-mmmm-diameter wire has a variable resistivity given by rho(x)=(2.5×10−6)[1+(x1.0m)2]Ωmrho(x)=(2.5×10−6)[1+(x1

Tresset [83]

Answer:

1.144 A

Explanation:

given that;

the length of the wire = 2.0 mm

the diameter of the wire = 1.0 mm

the variable resistivity R = $\rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]$

Voltage of the battery = 17.0 v

Now; the resistivity of the variable (dR) can be expressed as = $\frac{\rho dx}{A}$

$dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}$

Taking the integral of both sides;we have:

$\int\limits^R_0 dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx$

$R = 3.185 [x + \frac {x^3}{3}}]^2__0$

$R = 3.185 [2 + \frac {2^3}{3}}]$

R = 14.863 Ω

Since V = IR

$I = \frac{V}{R}$

$I = \frac{17}{14.863}$

I = 1.144 A

∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A

8 0

3 years ago

Could a mixture be made only element and no compounds

Veseljchak [2.6K]

Take the tiny bit of carbon dioxide and the tiny bit of water vapor out of the air,
and the rest of what you're breathing right now is a mixture of elements.

4 0

3 years ago

Humans can typically perceive frequencies that range from 20 Hz–20 kHz. Dogs can hear sounds up to 45 kHz. Dog whistles produce

Aleksandr [31]

Pitch is directly related to the frequency of the sound. In this item, we are given that the frequency of the sound is higher compared to those which are audible to the human being's ears. The pitch therefore of the dog's whistle is high.

On the other hand, the frequency and the wavelength of a certain wave are inversely proportional. This means that the high frequency wave will have a short wavelength.

Hence, the answer to this item would have to be "high pitch with a short wavelength"

The answer to this item is the second option.

6 0

3 years ago

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A child is swinging back and forth with a constant period and amplitude. Somewhere in front of the child, a stationary horn is e

Amanda [17]

Answer:

Explanation:

We shall apply concept of Doppler's effect of apparent frequency to this problem . Here observer is moving sometimes towards and sometimes away from the source . When observer moves towards the source , apparent frequency is more than real frequency and when the observer moves away from the source , apparent frequency is less than real frequency . The apparent frequency depends upon velocity of observer . The formula for apparent frequency when observer is going away is as follows .

f = f₀ ( V - v₀ ) / V , f is apparent , f₀ is real frequency , V is velocity of sound and v is velocity of observer .

f will be lowest when v₀ is highest .

velocity of observer is highest when he is at the equilibrium position or at middle point .

So apparent frequency is lowest when observer is at the middle point and going away from the source while swinging to and from before the source of sound .

3 0

3 years ago

Compare the signs of ƒ for lenses and mirrors.

STALIN [3.7K]

Answer:

simple

Explanation:

<h3>CONCAVE MIRRORS AND LENSES</h3>

<h3>f= negative</h3>

<h3>CONVEX MIRRORS AND LENSES</h3><h3 /><h3>f= positive</h3>

<h3>PLEASE FOLLOW ME AND MARK IT BRAINLIEST</h3>

3 0

3 years ago

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