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2 years ago

An electric immersion heater is put at the bottom of a large tank of water. The water next to the heater becomes warm

Physics

1 answer:

djyliett [7]2 years ago

4 0

Answer:

when the water is heated with immersion heater, the water becomes less dense due to which the warm water rises up and the cooler water fills it's space.

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Newton’s Second Law of motion defines force in terms of mass and acceleration, F= ma. Which of the following is false? The term

NikAS [45]

Answer:

The forces creating the net force must lie in the same direction.

Explanation:

newton's second law states that the net force acting on the body is equal to the product of mass and the acceleration of the body.

If there are several forces acting on the body in different directions, then we have to find teh net force by using the vector sum and then find the acceleration.

It is not necessary that all the forces acting in the same direction.

if they are in different directions then we have to find the net force by t=using the formula for the vector sum.

5 0

3 years ago

A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s. She then rides back along the same route from the

vitfil [10]

Answer:

12.31 m/s

Explanation:

If we recall from the previous knowledge we had about speed,

we will know that:

speed = distance/ time.

As such:

The average speed of the rider bicycle is

average speed = total distance/ total time

Mathematically, it can be computed as:

$v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}$

$v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}$

$v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}$

$v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}$

$\mathbf{v_{avg} =12.31 \ m/s}$

8 0

2 years ago

Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated

Artemon [7]

Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN

Explanation: To find the answer we need to know more about the Newton's law of gravitation.

<h3>What is Newton's law of gravitation?</h3>

Gravitation is the force of attraction between any two bodies.
Every body in the universe attracts every other body with a force.
This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
Mathematically we can expressed it as,

$F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2$

<h3>How to solve the problem?</h3>

Here, we have given with the data's,

$M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m$

Thus, the force of attraction between these two bodies will be,

$F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN$

Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.

Learn more about the Newton's law of gravitation here:

brainly.com/question/28045318

#SPJ4

6 0

1 year ago

Indicate on the chart whether you would classify each model as representing an element, compound, or mixture.

Ne4ueva [31]

Answer:

1 compound

2 mixture

3 elements

4 0

3 years ago

Read 2 more answers

A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu

malfutka [58]

Answer:

Acceleration, $a=-2.48\ m/s^2$

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}$

$a=-2.48\ m/s^2$

So, the acceleration on the rough ice $-2.48\ m/s^2$ and negative sign shows deceleration.

8 0

3 years ago