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ivolga24 [154]
3 years ago
10

what do you mean by momentum? avehicle is runing with a velocity of 5m/s. it the momentum of the vehicle is 5000kg m/s. what is

it mass?
Physics
1 answer:
Fantom [35]3 years ago
8 0
Momentum is simply a quantity that measures the impact of a moving body over something is due to the mass it posseses or the velocity with which it is moving. Mathematically,ut is the product of mass and velocity of a body. It is represented by capital p...(P) P=mv Solution to the problem: we know P=mv For mass,eliminate m from the equation, m=P/v Put values, m=5000/5=1000kg
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Question 7 of 10
jok3333 [9.3K]

Answer:

when a magnet is hanged freely in air it turns in the direction of the north and south while the magnetic north pole faces the south pole of the earth and magnetic south pole faces the north pole if the earth

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3 years ago
If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the 1200.-kg pick-up truck behind him continues tr
Alborosie

Answer: 20.2 m/s

Explanation:

From the question above, we have the following data;

M1 = 800kg

M2 = 1200kg

V1 = 13m/s

V2 = 25m/s

U (common velocity) =?

M1V1 + M2V2 = (M1 + M2). U

(800*13) + (1200*25) = (800+1200) * U

10400 + 30000 = 2000u

40400 = 2000u

U = 40400 / 2000

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5 0
3 years ago
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In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh
horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}

Where,

C_d is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = \frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2

A₂ = Area at the throat

A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0
3 years ago
CNA Secondary and Preparatory School
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Answer:

2:13 2) cm' c6462.

Explanation:

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2 years ago
24) Which of the following equations is correctly balanced? HELP ME ASAP PLEASE
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the correct answer is option C

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