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3 years ago

Which of the following changes would make a heat engine waste more energy as heat

1 answer:

4 0

A decrease in it's operating temperature would make a heat engine less efficient. This is because in order to operate, a heat engine needs to be hot and maintain that temperature.

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Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum

Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

$\frac{1}{2}mv^2 = mgh$

where

m is the mass of the man

v is the speed after jumping

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s$

In the acceleration phase, we know that the initial velocity is

$u=0$

And the force exerted on the floor is 2.3 times the gravitational force, so

$F=2.3 mg$

This means the net force on you is

$F_{net} = F-mg=2.3mg-mg=1.3 mg$

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

$a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g$

Now we can use the following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

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

where s is the quantity we want to find. Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m$

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

$\Delta h = -0.77 m$

This means that the lowest point reached by the center of mass above the floor during the crouch is

$h'=h+\Delta h = 1.0 - 0.77 = 0.23 m$

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0

3 years ago

A soccer ball is kicked from the ground with an initial speed v at an upward angle θ. A player a distance d away in the directio

Nady [450]

Answer:

The speed of player is given by

$V=\frac{2v^{2} 2sin\alpha.cos\alpha-gd }{2vsin\alpha}$

Explanation:

The time of flight for a projectile motion is given by

$T=\frac{2vsin\alpha }{g}$ (i)

where t is the time of flight, v is the initial speed, and α is the angle.

Now the person must also reach the impact point of ball in the same time as above.

Now the total distance D the player needs to cover is basically R horizontal range of projectile minus the distance d, range R is given by,

$R=\frac{2v^{2} sin2\alpha }{g}$

Now the distance the player must cover is given by

D= R-d

D= $\frac{2v^{2} sin2\alpha }{g}$ - d

$D=\frac{2v^{2}sin2\alpha-gd}{g}$ (ii)

Now the average speed of player is given by

$V=\frac{D}{T}$ (iii)

Replacing the values of D and T from eq. (i) and (ii) in eq. (iii).

$V=\frac{\frac{2v^{2} sin2\alpha-gd }{g}}{\frac{2vsin\alpha }{g} }$

$V=\frac{2v^{2} 2sin\alpha.cos\alpha-gd }{2vsin\alpha}$

3 0

3 years ago

Indicate all ways in which the graph used in question 4 would change if the Normal Force applied to the object was increased. Be

Maksim231197 [3]

Answer:

Answer:

PLEASE BRAINLESS ME INEED

# CARRY ON LEARNINGAnswer:

PLEASE BRAINLESS ME INEED

# CARRY ON LEARNING

5 0

3 years ago

A hiker starts at point P and walks 2.0 km due east and then walks at an angle of 30 degrees north of east for 1.4 km.

olchik [2.2K]

Answer:

The magnitude of the hiker’s displacement is 2.96 km

Explanation:

Let the initial displacement of the hiker, = x = 2km

the final displacement of the hiker, = y = 1.4 km

The resultant of the two vectors, According to Pythagorean theorem is the vector sum of the two vectors.

R' = x' + y'

Check the image uploaded for solution;

8 0

3 years ago

PLZ HELP THE CORRECT ANSWER GETS BRAINLEAST!

Bond [772]

The answer is B because the tire is circular. it will have more momentum and of course it can’t be on snow since it’s not smooth enough for the ice. Plus the sled will go slower on a grass hill since sled needs smooth flooring for no friction.

4 0

4 years ago