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Gemiola [76]
3 years ago
11

Emily uses a rifle to shoot a bullet at a target. The bullet has a mass of 13 grams. The rifle has a mass of 3,500 grams. When s

he shoots, the bullet accelerates 850 m/s2. According to Newton’s third law of motion, the bullet is exerting the same force on the rifle. Why doesn’t the rifle accelerate 850 m/s2?
Physics
2 answers:
sp2606 [1]3 years ago
4 0
Force exerted by the bullet = mass * acceleration = 0.013 * 850 = 11.05 Newtons.

the rifle exerts same force in opposite direction so we have

11.05 = 3.5 * a
acceleration = 11.05 / 3.5 =  3.16 m /s^-2
algol [13]3 years ago
4 0

Answer:

The mass of the rifle is much greater than the mass of the bullet.

Explanation:

just answered it

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An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

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Formulas in the solutions.
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Answer:

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However, due to the large mass of the Earth, compared to the mass of the Hulk, the downward motion of the Earth due to the reaction force (equal to the force with which the Hulk leaps) is negligibly small, such that the Earth can absorb the Hulk's leap force by reacting mainly locally, at the leap point  by forming a crater, while the Earth in general, remains in the same place

Explanation:

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tatiyna

Answer:

<u>(a) = 4.33</u>

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Explanation:

<h3>Please refer to the attachment below for explanation.</h3>

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