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stepladder [879]
3 years ago
13

A truck has a mass of 3000 kg and a velocity of 10 m/s. Calculate momentum!

Physics
2 answers:
BaLLatris [955]3 years ago
5 0

Given:-

  • Mass of the body (m) = 3000 kg
  • Velocity (v) = 10 m/s

To calculate: Momentum of the body.

We know,

p = mv

where,

  • p = Momentum,
  • m = Mass &
  • v = Velocity.

Thus,

p = (3000 kg)(10 m/s)

→ p = 30000 kg m/s (Ans.)

Evgen [1.6K]3 years ago
4 0

Answer:

\huge{ \boxed{ \sf{30000 \: kg \:  {m/s}^{2}}}}

Explanation:

\text{ \underline{ Given}} :

\star Mass of a truck ( m ) = 3000 kg

\star Velocity of a truck ( v ) = 10 m / s

<u>Finding</u><u> </u><u>the</u><u> </u><u>momentum</u> :

\boxed{ \sf{momentum \: ( \: p \: ) \:  =  \: mass \:   ∗ \:  \: velocity}}

\hookrightarrow{ \sf{momentum \: ( \: p \: ) \:  = 3000 \:  ∗ \: 10}}

\hookrightarrow{ \sf{ \: momentum \: ( \: p \: ) \:  = 30000 \: kg \: m/ {s}^{2} }}

Hope I helped!

Best wishes ! :D

~\sf{TheAnimeGirl}

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Please help need this asap
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Answer:

A)  350 N

B)  58.33 N

C)  35 kg

D)  35 kg

Explanation:

If we use that g = 10 m/s^2, then the acceleration of gravity on the Moon will be 10/6 m/s^2 = 5/3 m/s*2

The weight of the object on Earth is given by:

Weight = mass * g = 35 * 10 = 350 N

The weight of the object on the Moon:

Weight = mass * gmoon = 35 * 5/3 = 58.33 N

The mass of the object on Earth is 35 kg

The mass of the object on the Moon is exactly the same as on the Earth (35 kg) since the mass is a quantity inherent to the object and not to its location.

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The compound magnesium phosphate has the chemical formula Mg3(PO4)2. In this compound, phosphorous and oxygen act together as on
S_A_V [24]

The question is incomplete, the complete question is;

The compound magnesium phosphate has the chemical formula Mg3(PO4)2. In this compound, phosphorus and oxygen act together as one charged particle, which is connected to magnesium, the other charged particle. What does the 2 mean in the formula 5Mg3(PO4)2? A. There are two elements in magnesium phosphate. B. There are two molecules of magnesium phosphate. C. There are two magnesium ions in a molecule of magnesium phosphate. D. There are two phosphate ions in a molecule of magnesium phosphate.

Answer:

There are two phosphate ions in a molecule of magnesium phosphate.

Explanation:

The compound magnesium phosphate is an ionic compound. Ionic compounds always consists of two ions, a positive ion and a negative ion.

In this case, the positive ion is Mg^2+ while the negative ion is PO4^3-.

The subscript, 2 after the formula of the phosphate ion means that there are two phosphate ions in each formula unit of magnesium phosphate.

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2 years ago
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

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