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igor_vitrenko [27]
3 years ago
9

An object with more mass has more kinetic energy than an object with less

Physics
1 answer:
Ber [7]3 years ago
4 0

Explanation:

The kinetic energy of an object is given in terms of its mass and object as :

K=\dfrac{1}{2}mv^2

An object with more mass has more kinetic energy. An object with less  mass, if both objects are moving at the same speed, in opposite directions, will have less kinetic energy. The square of v will give positive number.

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During what era did the French and Indian War occur?
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7 0
3 years ago
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wite a paragraph explaining the difference between things that have matter and things that don't have matter.
mihalych1998 [28]

Answer: Matter is how much space or opacity an object takes up. In short anything that take up space. Things like balls, trees, and even people are all made of matter. Mass is how much matter a object has. Air also has mass also, even though we can't see it. Things like cars, buildings, even planets have mass.

Explanation: Paragraphs are sometimes 4-6 sentences.

6 0
3 years ago
Driving along a boring stretch of interstate in Illinois, you start experimenting using the average speed equation you learned i
astra-53 [7]
The average speed would be 33.29m/s.
The average speed equation is:

Average speed =  \frac{total distance}{total time}

First you will need to solve for the distance you traveled in each scenario. So we can solve this by getting the product of speed and the time traveled. 

Scenario 1:
Speed = 29m/s
Time = 120s
Distance = ?

Distance = (29m/s)(120s)
               = 3,480m

Scenario 2
Speed = 35m/s
Time = 300s
Distance = ? 

Distance = (35m/s)(300s)
               = 10,500m

Now that you have the distance of both, you can solve for your average speed. 

Average speed = \frac{total distance}{total time}
                                = \frac{3,480m+10,500m}{120s+300s}
                                = \frac{13,980m}{420s}
                                = 33.29m/s
5 0
3 years ago
A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra
kkurt [141]

Answer:

V_{ft}= 317 cm/s

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

P_i = P_f

Where:

P_i=M_cV_{ic} + M_tV_{it}

P_f = M_cV_{fc} + M_tV_{ft}

Now:

M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}

Where M_c is the mass of the car, V_{ic} is the initial velocity of the car, M_t is the mass of train, V_{fc} is the final velocity of the car and V_{ft} is the final velocity of the train.

Replacing data:

(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}

Solving for V_{ft}:

V_{ft}= 3.17 m/s

Changed to cm/s, we get:

V_{ft}= 3.17*100 = 317 cm/s

b) The kinetic energy K is calculated as:

K = \frac{1}{2}MV^2

where M is the mass and V is the velocity.

So, the initial K is:

K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2

K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2

K_i = 22.06 J

And the final K is:

K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = 19.61 J

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0
3 years ago
A star is moving away from an observer at 1% of the speed of light. At what wavelength would the observer find an emission line
Ivan

Answer:

  λ = 5940 Angstroms

Explanation:

This is an exercise of the relativistic Doppler effect

        f’= f  √((1- v / c) / (1 + v / c))

Where the speed in between the strr and the observer is positive if they move away

Let's use the relationship

         c = λ f

         f = c /λ

We replace

              c /λ’ = c /λ  √ ((1- v / c) / (1 + v / c))

              λ = λ’ √ ((1- v / c) / (1 + v / c))

Let's calculate

             v = 0.01 c

             v = 0.01 3 10⁸

             v=  3 10⁶ m / s

             λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]

             λ = 6000 √ [0.99 / 1.01]

             λ = 5940 Angstroms

6 0
3 years ago
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