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3 years ago

You eat an apple, so what energy transformations will it go through?

Physics

2 answers:

olga nikolaevna [1]3 years ago

5 0

The energy transform into the energy of ATP

Valentin [98]3 years ago

3 0

I suppose if you are taking physics the answers should be Potential Energy.

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How do I do number 8, plz respond quick, I have a big unit test on this and I’m rocking a 65 science average rn. Thank you for t

xxTIMURxx [149]

Answer:

3 is the correct answer for number 8

6 0

3 years ago

Read 2 more answers

A 10 kg migratory swan cruises at 20m/s. A calculation that takes into ac-count the necessary forces shows that this motion requ

IgorC [24]

Answer:

Part A:

Distance=864000 m=864 km

Part B:

Energy Used=ΔE=8638000 Joules

Part C:

$\frac{\triangle m}{m}=0.004998=0.49985\%$

Explanation:

Given Data:

v=20m/s

Time =t=12 hours

In Secs:

Time=12*60*60=43200 secs

Solution:

Part A:

Distance = Speed**Time

Distance=v*t

Distance= 20*43200

Distance=864000 m=864 km

Part B:

Energy Used=ΔE= Energy Required-Kinetic Energy of swans

Energy Required to move= Power Required*time

Energy Required to move=200*43200=8640000 Joules

Kinetic Energy= $\frac{1}{2}mv^2$

$K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules$

Energy Used=ΔE=8640000 -2000

Energy Used=ΔE=8638000 Joules

Part C:

Fraction of Mass used=Δm/m

For This first calculate fraction of energy used:

Fraction of energy=ΔE/Energy required to move

ΔE is calculated in part B

Fraction of energy=8638000/8640000

Fraction of energy=0.99977

Kinetic Energy= $\frac{1}{2}mv^2$

Now, the relation between energies ratio and masses is:

$\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2$

$\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977$

$\frac{\triangle m}{m}=0.004998=0.49985\%$

3 0

3 years ago

A change in the gravitational force acting on an object will affect the object

MArishka [77]

It is weight, if I understand your question.

4 0

3 years ago

An engineer is designing a small toy car that will be launched from rest. The engineer wants to maximize the kinetic energy of t

aalyn [17]

The modifications to the car design that would have the greatest effect on increasing the kinetic energy of the car is to increase the mass of the car slightly (option B).

<h3>What is kinetic energy?</h3>

Kinetic energy is the energy possessed by an object because of its motion. The kinetic energy equal (nonrelativistically) to one half the mass of the body times the square of its speed.

According to this question, an engineer is designing a small toy car that will be launched from rest. The engineer wants to maximize the kinetic energy of the car when it is launched by a compressed spring.

However, he can only make one adjustment to the initial conditions of the car. Considering the fact that the mass of an object is directly proportional to the kinetic energy.

This suggests that the modifications to the car design that would have the greatest effect on increasing the kinetic energy of the car is to increase the mass of the car slightly.

Learn more about kinetic energy at: brainly.com/question/12669551

#SPJ1

7 0

1 year ago

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.

Genrish500 [490]

<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

Horizontal distance he must clear = 22 meter

The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

We have equation of motion, $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when displacement = 1.5 meter.

$1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds$

So the car has to cover a distance of 22 meter in 2.119 seconds.

So minimum speed required = 22/0.553 = 39.78 m/s

Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 $m/s^2$

Substituting

$-1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0$

In case of horizontal motion

Horizontal speed of car = V cos 7 = 0.993V

So it has to travel 22 meter in t seconds

0.993Vt = 22, Vt = 22.155 m

Substituting in the equation $4.9t^2-0.122Vt-1.5=0$

We will get $4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds$

Speed required = 22.155/0.926 = 23.93 m/s

Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

7 0

3 years ago