describe how the students position in relation to a reference point and how the boys body could move as he walks to class

A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2, which force is closest to the net force a

natulia [17]

65234. i say this because the net force is stopping the jig

3 0

3 years ago

A battery supplies 0.71 A to three resistors connected in parallel. The current through the first resistor is 0.27 A and the cur

Yuki888 [10]

Answer:

0.2 A

Explanation:

Current: This is the rate of flow of charge in a circuit. The S.I unit of charge is Ampere (A).

Note: When resistors are connected in parallel, the total current is equal to the sum of the individual current in each resistor.

It = I1 + I2 + I3......................... Equation 1

Where,

It = Total current, I1 = current in the first resistor, I2 = current in the second resistor, I3 = current in the third resistor.

Given: It = 0.71 A, I1 = 0.27 A, I2 = 0.24 A.

Substitute into equation 1

0.71 = 0.27+0.24+I3

0.71 = 0.51+I3

I3 = 0.2 A.

Hence the current through the third resistor = 0.2 A

3 0

3 years ago

- A thin film of oil * (n = 1.45) on a puddle of water, producing different colors. What is the minimum thickness of a place whe

Darya [45]

Answer:

There will be a phase change at the first interface and no phase change at the second interface:

If the film is 1/4 wavelength thick this restriction will hold

The wavelength of the light in oil is 545 nm / 1.45 = 376 nm

376 nm / 4 = 94 nm

"D" is correct

4 0

3 years ago

A 1.2 x10 3 kilogram automobile in motion strikes a 1.0 x 10 -4 kilogram insect as a result the insect is accelerated at a rate

Mariana [72]

According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
$F=ma$
where F is the magnitude of the force, m is the mass of the object and a its acceleration.

In this problem, the object is the insect, with mass $m=1.0 \cdot 10^{-4} kg$ . The acceleration of the insect is $a=1.0 \cdot 10^2 m/s^2$ , therefore we can calculate the force exerted by the car on the insect:
$F=ma=(1.0 \cdot 10^{-4} kg)(1.0 \cdot 10^2 m/s^2)=0.01 N$

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.

6 0

3 years ago

Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin

Sergio [31]

Answer:

Explanation:

Potential energy on the surface of the earth

= - GMm/ R

Potential at height h

= - GMm/ (R+h)

Potential difference

= GMm/ R - GMm/ (R+h)

= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.

Extra energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / ( R + h)

8 0

3 years ago