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2 years ago

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If the velocity of an object doubles and it’s mass triples, the objects momentum increases by how much? A)1.5 times B)2 times C)

3 times D)6 times

1 answer:

shtirl [24]2 years ago

6 0

Answer:

(D)6 times

Explanation:

Hello !

Because when we see the equation of momentum P=mv, where p=momentum

m=mass of the object&

v=velocity of the object

And, p is directly propositional to both mass and velocity.

Thus, p=mv

p=3m.2v....giving on the values
p=6(mv)

Therefore, we can conclude by when the mass is tripled and the velocity is doubled the momentum is increased by 6 units.

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3 years ago

A man tries to push a 200 kg Car that moves at a acceleration 0.50 m/s2. The man is able to displace the car 10 m. How much work

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Given the data in the question;

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Distance covered by the car; $d = 10m$

Work done; $W = \ ?$

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Next we substitute our values into the expression of work done above.

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2 years ago

A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?

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3 years ago

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3 years ago

Una tubería con secciones circulares se emplea para el traslado de agua entre el tanque central y un tanque auxiliar. El tanque

charle [14.2K]

A) 20 L/s

B) 2.55 m/s

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Explanation:

a)

In this problem, we know that the volume of the tank:

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b)

The volume flow rate of water through the pipe can be also written as

$Q=Av$

where

A is the cross-sectional area of the pipe

v is the speed of the water

In section B, we have:

r = 0.050 m is the radius of section B

so, the cross-sectional area of section B is:

$A=\pi r^2 = \pi (0.050)^2=0.00785 m^2$

The volume flow rate in SI units is

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Therefore, the speed of the water in section B is:

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c)

As in part B), we know that the volume flow rate must remain constant through the entire pipe.

So, the volume flow rate in section A of the pipe is still

$Q=0.02 m^3/s$

The radius of the pipe in section A is

$r=0.025 m$

Therefore, the cross-sectional area in section A of the pipe is

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we can find the speed of water in section A:

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d)

Here we want to find the gauge pressure in section B.

We know that:

$p_A = 2.0 atm = 105,000 Pa$ is the pressure in section A

$h_A=15.0m$ is the altitude of section A

$v_A=10.20 m/s$ is the speed of water in section A

$v_B=2.55 m/s$ is the speed of water in section B

We can write Bernoulli's equation:

$p_A + \rho g h_A + \frac{1}{2}\rho v_A^2 = p_B + \frac{1}{2}\rho v_B^2$

where

$\rho=1000 kg/m^3$ is the water density

$p_B$ is the pressure in section B

And solving for pB, we find:

$p_B = p_A + \rho g h_A + \frac{1}{2}\rho (v_A^2-v_B^2)=\\=205,000+(1000)(9.8)(15.0)+\frac{1}{2}(1000)(10.20^2-2.55^2)=400,769 Pa$

Which is

$p_B = 400.8 kPa$

5 0

3 years ago