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Fittoniya [83]
3 years ago
5

A 120 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 91 kg halfback moving at 7.5 m/s. What will be their mutual vel

ocity in meters/second immediately after the collision?
Physics
1 answer:
Vesna [10]3 years ago
3 0

Explanation:

It is given that,

Mass of the tackler, m₁ = 120 kg

Velocity of tackler, u₁ = 3 m/s

Mass, m₂ = 91 kg

Velocity, u₂ = -7.5 m/s

We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,

v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

v=\dfrac{120\ kg\times 3\ m/s+91\ kg\times (-7.5\ m/s)}{120\ kg+91\ kg}

v = -1.5 m/s

Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.

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Wewaii [24]

Answer: The answer is B, The variable that should go on the x- axis is the temperature of water.

Explanation: In an experiment, we must have two variables; the independent variable and the dependent variable.

If the results of the experiment is reported in graphical format, the independent variable is plotted on the x axis while the dependent variable is plotted on the y axis.

In this case, the independent variable is the temperature of the water while the dependent variable is the mass of salt dissolved in 100 mL of water.

Therefore, the variable that should go on the x- axis is the temperature of water.

3 0
2 years ago
Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i
Mice21 [21]

Answer:

r = 5.08 m

Explanation:

The electric force of attraction or repulsion is given by :

F=\dfrac{kq_1q_2}{r^2}

We need to find how far above the electron would the proton have to be if you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it.

So, the force from the proton is balanced by the mass of the electron.

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r=\sqrt{\dfrac{kq_pq_e}{mg}} \\\\r=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times 9.8}} \\\\r=5.08\ m

So, proton have to be at a distance of 5.08 meters above the electron.

3 0
3 years ago
Please help asap 100 POINTS and BRAINLIEST to best answer.
andrew11 [14]

D. :)

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6 0
3 years ago
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K = 1/2 m x v^2

m = mass on the cart

V = velocity imparted to the cart

KA = 1/2 mA x vA^2.......................(1)

KB = 1/2 mB x vB^2........................(2)

Diving equation 1 by equation 2, we get -

KA/KB = mA/mB

= 2

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Option A is correct

6 0
3 years ago
If the man on the left pulls on the object with a force of 500 N and the man on the right pulls on the object with a force of 75
Igoryamba
Force is a vector quantity
so pulling from opposite side will be negative
so
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C is the right answer
becauseause the man on the right applies greater force.
3 0
3 years ago
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