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3 years ago

Determine the velocity of the running man with respect to car A.

Physics

2 answers:

erastova [34]3 years ago

4 0

Answer: A) -55 km/h

Explanation: The running man is traveling at -55 km/hr with respect to Car A. If the man were standing still, the car would be passing at 50. Since he is traveling at 5 in the opposite direction, that direction is positive, to him. His velocity now adds to the car velocity (50 + 5), but negative to account for the opposite direction.

Hope this helps!

olasank [31]3 years ago

3 0

Answer:

A) -55 km/

Explanation:

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Draw a circuit that contains 2 batteries, three lights in parallel and a switch that controls the whole circuit.

Naddik [55]

Draw a circuit that contains 2 batteries, three lights in parallel and a switch that controls the whole circuit.

6 0

3 years ago

A ball of 10kg falls from rest from a height of 150m, Neglating air resistance, calculate its kinetic energy after falling a dis

GaryK [48]

Answer: $3920\ J$

Explanation:

Given

mass of ball m=10 kg

It is placed at a height of 150 m

It is dropped from the height and allowed to free fall for 40 m

Velocity acquired by the ball during this fall is given by $v^2-u^2=2as$

Insert u=0, a=g

$\Rightarrow v^2-0=2\times 9.8\times 40\\\Rightarrow v=\sqrt{784}\\\Rightarrow v=28\ m/s$

Kinetic energy at this instant

$K.E.=\dfrac{1}{2}\times 10\times 28^2\\\\\Rightarrow K.E.=3920\ J$

3 0

2 years ago

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

levacccp [35]

Answer:

The minimum speed must the car must be 13.13 m/s.

Explanation:

The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.

We know that, mg be the weight of car and rider, which is equal to the centripetal force.

$mg = \dfrac{mv^2}{r}\\v = \sqrt{rg}\\v = \sqrt{17.6\times 9.8}\\v = 13.13\ m/s\\$

So, the minimum speed must the car must be 13.13 m/s.

5 0

2 years ago

We can model the human back as a pivoted rod?

3241004551 [841]

Answer:

So the answer is yes, we can the back be shaped like a spinning rod

spinal column that is approximated by a long and narrow rod,

Explanation:

The bone system of the body is very well modeled in physics, the back has a spinal column that is approximated by a long and narrow rod, this rod is fixed in the lower part to the coccyx and has a weight in the upper part (head), this rod has longitudinal vertical movement and twisting movement around the lower part of the bar.

So the answer is yes, we can the back be shaped like a spinning rod

7 0

3 years ago

Identical balls oscillate with the same period T on Earth. Ball A is attached to an ideal spring and ball B swings back and fort

viva [34]

Answer:

B. Ball B will take longer to complete one cycle

Explanation:

This is simply because the period of a simple pendulum is affected by acceleration due to gravity, while the period of an ideal spring is not.

This can be clearly deduced by observing the formulas for the various systems.

Formula for period of simple pendulum:

T = 2π × $\sqrt{\frac{L}{g} }$

Formula for period of an oscillating spring:

T= 2π × $\sqrt{\frac{m}{K} }$

The period of the simple pendulum is affected by the length of the string and the acceleration due to gravity as shown above. Thus, it will have a different period as the gravitational acceleration changes on the moon. Thus will be a larger period (slower oscillation) as the gravitational acceleration is smaller in this case

The period of the oscillating spring is only affected by the mass of the load an the spring constant as shown above. Thus, it will have a period similar to the one it had on the earth because the mass of the ball did not change as the setup was taken to the moon.

All these will make the ball on the spring (Ball B) oscillate faster than the ball swinging on the string (Ball A)

6 0

2 years ago