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3 years ago

A 500 kilogram marlin and a 20 kilogram salmon are swimming at the same velocity. How does the momentum of the two fish compare?

Physics

2 answers:

vladimir1956 [14]3 years ago

7 0

Answer:

A. The marlin has more momentum.

ElenaW [278]3 years ago

6 0

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below are the choices that can be found from other sources:

A. The marlin has more momentum.

B. The salmon has more momentum. 

C. Both have the same momentum. 

D. Both have no momentum.

Velocity is speed. Momentum is the product of velocity and mass. Because the Marlin has a much greater mass that means that it's momentum would also be greater.

Lets look at this. momentum = mass x velocity 

lets say that velocity for both fish is at 2 meters per second 
Marlin : momentum = 500 x 2 = 1000 
Salmon : momentum = 20 x 2 = 40 

Therefore the answer is A

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A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of th

Gnoma [55]

Answer:

91.84 m/s²

Explanation:

velocity, v = 600 m/s

acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2

Let the radius of the loop is r.

he experiences a centripetal force.

centripetal acceleration,

a = v² / r

39.2 x r = 600 x 600

r = 3600 / 39.2

r = 91.84 m/s²

Thus, the radius of the loop is 91.84 m/s².

8 0

3 years ago

Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b

zmey [24]

Answer:

The maximum amount of work is $W = 1563.289 \ J$

Explanation:

From the question we are told that

The temperature of the environment is $T = 280\ K$

The volume of container A is $V_A = 2 m^3$

Initially the number of moles is $n = 1.2 \ moles$

The volume of container B is $V_B = 3.5 \ m^3$

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

$W = P_A V_A ln[ \frac{V_B}{V_A} ]$

Now from the Ideal gas law

$P_A V_A = nRT$

So substituting for $P_A V_A$ in the equation above

$W = nRT ln [\frac{V_B}{V_A} ]$

Where R is the gas constant with a values of $R = 8.314 \ J/mol$

Substituting values we have that

$W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]$

$W = 1563.289 \ J$

8 0

3 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him

nataly862011 [7]

Answer:

a) $u_c=0\ m.s^{-1}$ & $m_c.v_c=m_b.v_b\times \cos\theta$

b) $v_c=0.0566\ m.s^{-1}$

c) $p_e=2.9218\ kg.m.s^{-1}$

Explanation:

Given:

mass of the book, $m_b=1.35\ kg$

combined mass of the student and the skateboard, $m_c=97\ kg$

initial velocity of the book, $v_b=4.61\ m.s^{-1}$

angle of projection of the book from the horizontal, $\theta=28^{\circ}$

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

$u_c=0\ m.s^{-1}$

where:

$u_c=$ initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

$m_c.v_c=m_b.v_b\times \cos\theta$

where:

$v_c=$ final velcotiy of the student after throwing the book

$m_c.v_c=m_b.v_b\times \cos\theta$

$97\times v_c=1.35\times 4.61\cos28$

$v_c=0.0566\ m.s^{-1}$

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

$p_e=m_b.v_b\sin\theta$