Energy stores in food would be considered what

Answer:

A because that's the answer

Answer:

  1500 J

Explanation:

Apparently, we're to assume that no energy is lost to friction or other effects. Then all of the energy the train has is due to work being done on it.

  1500 J of work was done on the train

Answer:

Normative

Explanation:

It is normal occurance but not unnatural

Answer:

HOPE THIS HELPS!!

Explanation:

Velocity is speed as well as distance mentioned.

The car is traveling at a constant speed of 20m/s but it may not be traveling in any particular direction. It would be taking many turns (left and right) or it would be moving in a circular manner.

So,it is at constant speed but not constant velocity.

The maximum speed the mass can have before it breaks is 2.27 m/s.

The given parameters:

The maximum speed the mass can have before it breaks is calculated as follows;

[tex]T = ma_c\\\\Mg = \frac{Mv^2}{r} \\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v_{max} = \sqrt{0.525 \times 9.8} \\\\v_{max} = 2.27 \ m/s[/tex]

Thus, the maximum speed the mass can have before it breaks is 2.27 m/s.

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Answer:

I hope it will help you...

#CARRYONLERANING

Ignore my comment, that information is unnecessary.

Assuming no friction between the skis and the water, the total work performed on the skier is due only to the pulling force (T, for tension). By Newton's second law, the net force exerted on the skier is

∑ F = T = ma

where m = 73.5 kg is the mass of the skier, and a is their acceleration.

Under constant acceleration, the skier is pulled a distance x such that

(12.2 m/s)² - (5.2 m/s)² = 2ax

and solving for x gives

x = (60.9 m²/s²) / a

Then the total work performed on the skier is

W = Tx

W = ma • (60.9 m²/s²) / a

W = (73.5 kg) (60.9 m²/s²)

W ≈ 4480 J

An earthquake with a magnitude of 5.5 releases 4.5 × 10¹² J of energy.

The magnitude of an earthquake is a number that characterizes the relative size of an earthquake and is based on the measurement of the maximum motion recorded by a seismograph.

If the magnitude of the earthquake is M = 5.5, we can calculate the amount of energy released (E) using the following expression.

[tex]M = \frac{2}{3} log(\frac{E}{E_0} )\\\\E = E_0 \times 10^{\frac{3}{2}M } = 10^{4.4} \times 10^{\frac{3}{2}(5.5) } = 4.5 \times 10^{12} J[/tex]

where,

An earthquake with a magnitude of 5.5 releases 4.5 × 10¹² J of energy.

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Answer:

yes , true the magnitude of the reaction force.

We want to find the energy that we need to transform 2kg of ice at 0°C to water at 20°C.

We will find that we must give 840,000 Joules.

First, we must change of phase from ice to water.

We use the specific latent heat of fusion to do this, this quantity tells us the amount of energy that we need to transform 1 kg of ice into water.

So we need 336,000 J of energy to transform 1kg of ice into water, and there are 2kg of ice, then we need twice that amount of energy:

2*336,000 J = 672,000 J

Now we have 2kg of water at 0°C, and we need to increase its temperature to 20°C.

Here we use the specific heat, it tell us the amount of energy that we need to increase the temperature per mass of water by 1°C.

We know that:

specific heat of capacity of water = 4200 J/kg°C

This means that we need to give 4,200 Joules of energy to increase the temperature by 1°C of 1kg of water.

Then to increase 1°C of 2kg of water we need twice that amount:

2*4,200 J = 8,400 J

And that is for 1°C, we need to give that amount 20 times (to increase 20°C) this is:

20*8,400 J = 168,000 J

Then the total amount of energy that we must give is:

E = 672,000 J + 168,000 J = 840,000 J

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The ratio of time of flight for inelastic collision to elastic collision [tex](t_A :t_B)[/tex] is 1:2

The given parameters;

Considering inelastic collision, the final velocity of the system is calculated as;

[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\m_1u_1 + 0 = v(m_1 + m_2)\\\\v= \frac{m_1u_1}{m_1 + m_2} \ -- (1)\\\\[/tex]

The time of motion of the system form top of the table is calculated as;

[tex]v = u + gt\\\\v = 0 + gt\\\\v = gt\\\\t= \frac{v}{g} \\\\t_A = \frac{m_1u_1}{g(m_1 + m_2)} \ \ ---(2)[/tex]

Considering elastic collision, the final velocity of the system is calculated as;

[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\m_1u_1 + 0 = m_1v_1 + m_2v_2\\\\m_1 u_1 = m_1v_1 + m_2v_2[/tex]

Apply one-directional velocity

[tex]u_1 + (-v_1) = u_2 + v_2\\\\u_1 -v_1 = 0 + v_2\\\\v_1 = v_2 -u_1[/tex]

Substitute the value of [tex]v_1[/tex] into the above equation;

[tex]m_1u_1 = m_1(v_2 - u_1) + m_2 v_2\\\\m_1u_1 = m_1v_2 -m_1u_1 + m_2v_2\\\\2m_1u_1 = m_1v_2 + m_2v_2\\\\2m_1u_1= v_2(m_1 + m_2)\\\\v_2 = \frac{2m_1u_1}{m_1+ m_2} \ --(3)[/tex]

where;

[tex]v_2[/tex] is the final velocity of the block after collision

Since the bullet bounces off, we assume that only the block fell to the ground from the table.

The time of motion of the block is calculated as follows;

[tex]v_2 = v_0 + gt\\\\v_2 = 0 + gt\\\\t = \frac{v_2}{g} \\\\t_B = \frac{v_2}{g} \\\\ t_B = \frac{2m_1u_1}{g(m_1 + m_2)} \ \ ---(4)[/tex]

The ratio of time of flight for inelastic collision to elastic collision is calculated as follows;

[tex]\frac{t_A}{t_B} = \frac{m_1u_1}{g(m_1 + m_2)} \times \frac{g(m_1 + m_2)}{2m_1u_1} \\\\\frac{t_A}{t_B} = \frac{1}{2} \\\\t_A:t_B = 1: 2[/tex]

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Answer:

B) the average distance from the Earth to the Sun

Explanation:

Answer:

Earth's mass is 5.98 x 1024 kg and has a radius of 6.37 x 106 m. 8.06 N/kg ... what would be the weight of a 70.0-kg student on the surface of Jupiter?Explanation:

Answer:

Though pitch and frequency are not equivalent, they are correlated. This means that as one goes up, the other does as well. A higher frequency produces a higher pitch, and a lower frequency produces a lower pitch.

Explanation:

Answer:

Friction

Explanation:

As the toy cars rolls away, more friction is created. The more friction there is, the more friction on surface rubs against another which creates friction which in-term slows it down. Hope this helps.

Answer:

d. action force and the reaction force act on two different objects.

Hope you could understand.

If you have any query, feel free to ask.

In an action-reaction pair, action force and the reaction force act on two different objects.

Newton's Third Law of Motion states that "For every action, there is an equal and opposite reaction, action and reaction act on two different bodies."

It means that equal force are act on reaction force that is it is also known as reciprocal law of motion.

When considering the options,

So, In an action-reaction pair, action force and the reaction force act on two different objects. Thus, Option B is the correct answer.

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Answer:

A

Explanation:

According to Newton third law of motion,  what must be true of the magnitude of the reaction force is that the reaction force will be equal and opposite to the action force of 110 newtons

From Newton third law of motion which state that in every action of force, there will be equal and opposite reaction.

If a 75 kg box collides with a stationary 35 kg box with a force of 110 N, the mass 35 kg will produce a reaction force which is equal and opposite to the action force 110 newton received from mass 75 kg.

Therefore, what must be true of the magnitude of the reaction force from mass 35 kg is that the mass will produce an equal and opposite force equal to 110 Newtons.

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The two forces involved in this collision are the colliding force and the reaction force which are equal in magnitude but opposite in direction.

According to Newton's third law, action and reaction are equal and opposite. This law is very important in describing head - on collisions between objects and is particularly apt in describing the collision between the two boats.

In this collision, two forces are important these are the action and the reaction. The action is the colliding force which the boat that was hit experiences. The reaction force is the force that this boat experiences as a result of hitting the boat in front of it. The both forces are equal in magnitude but opposite in direction.

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The mass of the beach ball is 0.33 kg

From the question,

We are to determine the mass of the beach ball.

Using the formula

ρ = mv

Where ρ is the momentum

m is the mass

and v is the velocity

From the given information,

ρ = 3 kg m/sec

v = 9 m/sec

Putting the parameters into the formula, we get

3 = m × 9

∴ m = 3 ÷ 9

m = 0.33 kg

Hence, the mass of the beach ball is 0.33 kg

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Answer:

the radio wave. - The frequency ranges between 300 gigahertz (GHz) to 30 hertz (Hz).

Explanation:

Answer:

hi your answer to this question is 220'

Explanation:

hope this helped you

(a) The time taken for the projectile to reach the maximum height is 32.65 s.

(b) The horizontal range of the projectile is 9,049.1 m.

The given parameters:

The time taken for the projectile to reach the maximum height is calculated as follows;

[tex]v_f = u- gt\\\\0 = 320 - 9.8t\\\\9.8t = 320\\\\t = \frac{320}{9.8} \\\\t = 32.65 \ s[/tex]

The horizontal range of the projectile is calculated as follows;

[tex]R = \frac{u^2 sin(2\theta)}{g} \\\\R = \frac{320^2 \times sin(2\times 30)}{9.8} \\\\R = 9,049.1 \ m[/tex]

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Answer:

50 Joules

Explanation:

Answer:

8 seconds

Explanation:

Since the carspeed is in km/h, we need equal units, so we will make 100.0m 0.1000km.

Then we need to find how long it takes the car to travel 0.1km

We can use the formula distance=speed * time and get

0.1=45 * time

Therefore we get .002222... hours

Multiplying this by 3600 (to get seconds, 60x60), we get 8 seconds

Answer:

8m/s

Explanation:

speed = 45km/h

Distance = 100m

we have to find time = ?

Formula for speed is = Distance/ Time

Here Distance is given in 'm' so we need to convert speed value in 'm'

So to convert km/h in m formula is divide 45km/h by 3.6 then the km/h value gets converted in m so now the value is 12.5 m/s

So now,

speed = Distance/Time

we have to find Time

Then,

Time = Distance/ Speed

= 100/12.5

= 8m/s

Answer:

1. Body Tube

2.Revolving Nose Piece

3.Low Objective

4.Med Objective

5.High Objective

6.Stage clips

7.Diaphragm

8.Light source

9.Ocular lens

10. Arm

11.Stage

12.Coarse Adjustment knob

13.Fine Adjustment knob

14. base

Answer:

v = 3 m/s

Explanation:

P is the momentum

m is the mass

v is the velocity

P = m•v

300 = 100•v Divide both sides per 100

300/100 = v

v = 3 m/s

[tex]\text{Given that,}\\\\\text{Momentum, p = 300 kg m s}^{-1}\\\\\text{Mass, m = 100 kg}\\\\\text{We know that,}\\\\p = mv\\\\\implies v = \dfrac pm = \dfrac{300}{100} = 3 ~ \text{m s}^{-1}[/tex]

Answer:

Yes.

Explanation:

Displacement is a vector quantity, meaning it has to be a straight graph jointing from starting point to final point and has a direction as well as magnitude and it can also be negative as well as positive and zero.

Displacement is also changes in position (points) of distance. A distance is scalar quantity meaning it only has magnitude and does not have direction. Since distance is scalar quantity, it cannot be negative. Also graph of distance can be a curve graph or any graph.

So when does displacement equal to distance? If an object is moving in straight path or line in one/fixed direction, both displacement and distance are same.

Because if a distance is a straight path and the displacement is simply a straight line (or ray) jointing from starting point to end point, both distance and displacement are both straight path or rays. Since distance and displacement have same graph, we can conclude that both are same in values.

If we go by calculus, given x = 2t as example of straight graph where x stands for position and t stands for time. We can find the displacement from b to a by using the following formulas:

Displacement

[tex]\displaystyle \large{s=\int\limits^b_a {v} \, dt}[/tex]

Distance

[tex]\displaystyle \large{l=\int\limits^b_a {|v|} \, dt}[/tex]

v stands for velocity which we can find from:

Velocity

[tex]\displaystyle \large{v=\frac{dx}{dt}}[/tex]

If you have learnt calculus, first, differentiate x = 2t with respect to t.

[tex]\displaystyle \large{v=2}[/tex]

Then substitute v = 2 in s and l to find find if both displacement and distance are equal in straight path.

Displacement

[tex]\displaystyle \large{s=\int\limits^b_a {2} \, dt}\\\displaystyle \large{s=[2t]\limits^b_a}\\\displaystyle \large{s=2b-2a = 2(b-a)}[/tex]

Distance

[tex]\displaystyle \large{l=\int\limits^b_a {|2|} \, dt}\\\displaystyle \large{l=\int\limits^b_a {2} \, dt}\\\displaystyle \large{l=[2t]\limits^b_a}\\\displaystyle \large{l=2b-2a=2(b-a)}[/tex]

Since both displacement and distance are equal when integrating on straight graph, we can conclude that an object moving in straight fixed point has same distance and displacement.