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Yuri [45]
3 years ago
14

What is the momentum of an 18-kg object moving at 0.1 m/s ?

Physics
1 answer:
Anit [1.1K]3 years ago
6 0

Answer:

1.8 m/s

Explanation:

here's the solution : -

momentum = mass × velocity

=》18 × 0.1

=》1.8 m/s

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A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the
ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

\frac {Vf^{2}-Vo^2}{2.a} =X

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s

the speed with the ball hits the ground is 32.46m/s

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3 years ago
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Which units are used to express sound intensity? <br><br> Decibels <br> Hertz<br> Volts <br> Amperes
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Decibels I think that's the answer
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Gestalt therapists focus much of their work on helping clients get in touch with their feelings. Do you feel
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In 5 meters, a person running at 0.8 m/s accelerates at 1.6 m/s2. How fast 16 points
frutty [35]

They were going at a velocity 4.07m/s

<u>Explanation:</u>

Distance s =5 m

initial velocity u= 0.8 m/s

Acceleration a =1.6m/s2

We have to calculate the velocity with which they were going afterwards i.e final velocity.

Use the equation of motion

v^2=u^2+2as\\=0.8^2+2\times 1.6\times 5\\=16.64\\v=4.07 m/s

They were going with a velocity 4.07 m/s afterwards.

5 0
3 years ago
A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re
Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Pressure at reservoir = 10 atm

T_1 = Temperature at reservoir = 300 K

P_2 = Pressure at exit = 1 atm

T_2 = Temperature at exit

R_s = Mass-specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

For isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3

The density of the flow at the exit is 2.2721 kg/m³

4 0
3 years ago
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