The weight of an object is the force pulling the object:

Mr. Llama walked from his house to the bus stop. The bus stop is 2 miles from his house. He returned back to his house from the

Hoochie [10]

Answer:

Displacement of Mr. Llama: Option D. 0 miles.

Explanation:

The magnitude of the displacement of an object is equal to the distance between its final position and its initial position. In other words, as long as the initial and final positions of the object stay unchanged, the path that this object took will not affect its displacement.

For Mr. Llama:

Final position: Mr. Llama's house;
Initial position: Mr. Llama's house.

The distance between the final and initial position of Mr. Llama is equal to zero. As a result, the magnitude of Mr. Llama's displacement in the entire process will also be equal to zero.

7 0

4 years ago

Which causes the thermocline?

Igoryamba

<span>Thermocline is a layer between warm water from the ocean’s surface and cool water from below the ocean. In here, the temperature decreases rapidly from the warmer layer to the colder layer. A thermocline forms due to the heat of the sun heating the ocean’s surface. Because of the difference in density between warm and cooler ocean water, cooler ocean water sinks and warmer ocean water floats. This is caused due to the heat and mass transfer between particles of the ocean. The answer is letter C. The sun’s radiation does not extend below a certain depth; therefore, deeper ocean water is colder than surface water.</span>

4 0

3 years ago

Read 2 more answers

A proton enters a uniform magnetic field of strength 2 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’

sattari [20]

Answer:

Magnetic force, $F=9.6\times 10^{-17}\ N$

Explanation:

It is given that,

Magnetic field, B = 2 T

Velocity of the proton, v = 300 m/s

Charge on the proton, $q=1.6\times 10^{-19}\ C$

The magnetic field is oriented perpendicular to the proton’s velocity. The magnetic force on the charged particle is given by :

$F=qvB\ sin\theta$

The magnetic field is oriented perpendicular to the proton’s velocity, $\theta=90^{\circ}$

$F=1.6\times 10^{-19}\times 300\times 2$

$F=9.6\times 10^{-17}\ N$

So, the magnitude of the force that the proton experiences while it moves through the magnetic field is $9.6\times 10^{-17}\ N$ . Hence, this is the required solution.