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Papessa [141]
3 years ago
9

Two football players with mass 85 kg and 110 kg run directly toward each other with speeds 4 m/s and 7 m/s respectively. If they

grab each other as they collide, what is the combined speed of the players just after the collision?
Physics
1 answer:
sergejj [24]3 years ago
5 0

Use the conservation of the momentum to answer this question:

m_1\vec v_1+m_2\vec v_2 = m\vec v\implies\\\vec v = \frac{m_1\vec v_1+m_2\vec v_2}{m_1+m_2}\\v = \frac{85kg\cdot 4 m/s+110kg\cdot (-7m/s)}{195kg}=-2.2\frac{m}{s}

The resulting speed of the two bodies will be 2.2 m/s in the (original) direction of the overweight guy (110kg). Their brains not happy about the collision either.

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Answer:do it correct and do your best please yourself do what is right let every body know this is ak200342 talking to you.

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3 years ago
Two positive charges of 6 µC are separated by a distance of 50 cm in air. What is the electric field strength at the midpoint of
attashe74 [19]

Answer:

The electric field strength at the midpoint of the line joining the charges is zero (0)

Explanation:

Given that the two charges are both positive (same charge) and are equal in magnitude that is 6uC. The electric field strength at the midpoint of the line joining the two charges will be equal and opposite in magnitude, therefore they will cancel each other out and the electric field strength at this point will be equal to zero.

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4 years ago
Jim is driving a 2268-kg pickup truck at 19 m/s and releases his foot from the accelerator pedal. The car eventually stops due t
lord [1]

Answer:

The initial kinetic energy of the truck is 409374 J

Explanation:

This problem can be solved in two ways. Let´s solve it first in the easiest way.

The kinetic energy is calculated using this equation:

E = 1/2 · m · v²

Where:

E = kinetic energy

m = mass

v = velocity

Then, the kinetic energy of the truck will be:

E = 1/2 · 2268 kg · (19 m/s)² = 409374 J

And that´s it.

But we can complicate it a bit:

The kinetic energy is the work needed to move an object from rest to a desired velocity. If the object is moving, the work needed to stop it must be of the same magnitude as its kinetic energy (in the opposite direction to the movement).

The equation for work is:

W = F · d

Where:

W= work

F = force

d = distance

We know the magnitude of the force applied to the truck, but we do not know for how much distance that force was applied. The distance can be calculated using the equation for the position of an object moving in a straight line:

x = x0 + v0 · t + 1/2 · a · t²

where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

But we still do not know the time nor the acceleration.

The acceleration can be obtained from the equation of force:

F = m · a

Where

F = force

m = mass

a = acceleration

Then:

900 N = 2268 kg ·a

a = 900 N /2268 kg = 0.397 m/s²

Now, we can calculate the time needed for the truck to stop. We know that at the final time, the velocity is 0. Then, we can use the equation for velocity to obtain that time:

v = v0 + a · t

Where:

v = velocity at time t

v0 = initial velocity

a = acceleration

t = time

Then:

v = 19 m/s - 0.397 m/s² · t

0 = 19 m/s - 0.397 m/s² · t

-19 m/s / -0.397 m/s² = t (acceleration is negative because it is opposite to the direction of the movement)

t = 47.86 s (The truck stoped at 47.86 s after releasing the foot from the accelerator pedal)

With the time and acceleration, we can calculate the traveled distance.

x = 0 m + 19 m/s · 47.86 s - 1/2 · 0.397 m/s² · (47.86s)²

x = 454.66 m (without rounding the acceleration nor the time, the value will be 454.86 m)

Now, we can calculate the work done to stop the truck which will be of the same magnitude as the kinetic energy:

W = 900 N · 454.66 m = 409194 J

(if you do all the calculations without rounding, you will get the same value as we calculated above using the equation of kinetic energy, 409374 J).

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The answer is then D. 

Coefficient is found behind the compounds, the charges are written above. 
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Answer:

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