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3 years ago

An object speed is 7.2 m/s and its momentum is 360 kg m/s what is mass of object

Physics

1 answer:

MAXImum [283]3 years ago

8 0

Momentum = mass × velocity
360 = mass × 7.2
mass = 360/7.2 = 50 kg

Hope it helped!!

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Rebecca jogs in place for two minutes before her track meet. Rebecca weighs 100 pounds. Which of the following statements is tru

MrRissso [65]

The answer is A because it was 2 minutes and she weighs 100 lbs.... so 2(100)=200.... 200 lbs of work

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3 years ago

Read 2 more answers

A spring is hanging from the ceiling. When a 250 gram of mass is attached to the free end, the spring elongates by 5 cm. The spr

lord [1]

Answer:

k = 49 N/m

Explanation:

Given that,

Mass, m = 250 g = 0.25 kg

When the mass is attached to the end of the spring, it elongates 5 cm or 0.05 m. We need to find the spring constant. Let it is k.

The force due to mass is balanced by its weight as follows :

mg=kx

$k=\dfrac{mg}{x}\\\\k=\dfrac{0.25\times 9.8}{0.05}\\\\k=49\ N/m$

So, the spring constant of the spring is 49 N/m.

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3 years ago

Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz

Ronch [10]

Answer:

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3 years ago

A gas is placed in a storage tank at a pressure of 49.2 atm at 39.0C . As a safety device, a small metal plug in the tank is mad

Amiraneli [1.4K]

Answer:

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=49.2 atm\\T_1=39.0^oC = 312.15 K\\P_2=?\\T_2=198^oC=471.15 K$

Putting values in above equation, we get:

$\frac{49.2atm }{312.15 K}=\frac{P_2}{471.15 K}\\\\P_2=74.26 atm$

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.

4 0

3 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

3 years ago