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4 years ago

6

The average force applied to move an initially stationary golf cart over a period of 25 s is 500 N. What is the change in moment

um of the cart?

1 answer:

Yuki888 [10]4 years ago

6 0

Answer:

12500(kg*m/s)

Explanation:

F=ma=mv/t=p/t

p=F*t=500N*25 s=12500(kg*m/s)

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3 years ago

Read 2 more answers

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

C) 10.20 m/s

D) 400.8 kPa

Explanation:

a)

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

$V=72 m^3$

is filled in a time of

$t = 1 h$

We also know that the volume flow rate of water in the pipe is constant in every point of the pipe, so it can be calculated directly from the two data above.

The volume of the tank, in liter, is

$V=72 m^3 = 72,000 L$

While the time, in seconds, is

$t=1 h = 3600 s$

Therefore, the volume flow rate in Liters per second is:

$Q=\frac{V}{t}=\frac{72,000}{3600}=20 L/s$

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

$Q=\frac{V}{t}=\frac{72.0m^3}{3600 s}=0.02 m^3/s$

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

$v=\frac{Q}{A}=\frac{0.02}{0.00785}=2.55 m/s$

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

$A=\pi r^2 = \pi (0.025)^2=0.00196 m^2$

So, since we have

$Q=Av$

we can find the speed of water in section A:

$v=\frac{Q}{A}=\frac{0.02}{0.00196}=10.20 m/s$

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

$p_B = 400.8 kPa$

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3 years ago

Show that v=u+at is dimensionally correct

Zanzabum

Answer:

Check attachment

Explanation:

Check attachment

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3 years ago