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ioda
3 years ago
6

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force)

acts on her. The student has a mass of 73 kg, and the height of the water slide is 11.8 m. If the kinetic frictional force does -5.5 × 103 J of work, how fast is the student going at the bottom of the slide?
Physics
1 answer:
salantis [7]3 years ago
3 0

Answer:

<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>

Explanation:

The Non conservative force is defined as a force which do not store energy or get he energy dissipate the energy from the system as the system progress with the motion.

Given are

   <em>  mass of the student 73 kg</em>

<em>      height of water glide 11.8 m</em>

<em>      work done as -5.5*10³ J</em>

Have to find speed at which the student goes down the glide.

According to<em> Law of Conservation of energy</em>,

          K.E =P.E+Work Done

 mv²/2=mgh +W

Rearranging the above eqn for v

v = √2(gh+W/m)

Substituting values,

V =  12.48 m/s.

<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>

 

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You drop your cell phone. Prior to hitting the ground, the phone's kinetic energy will ________ and its potential energy will __
SVETLANKA909090 [29]

Answer:

The kinetic energy of the phone would increase. The gravitational potential energy of the phone would decrease.

Explanation:

The kinetic energy {\rm KE} of an object is proportional to the square of the speed of that object. If air resistance is negligible, the phone would accelerate under gravitational pull and speed up. Hence, the kinetic energy of the phone would increase.

The gravitational field near the surface of the earth is approximately constant. Hence, the gravitational potential energy {\rm GPE} of the phone would be proportional to its height. As the phone approaches the ground, the height of the phone becomes lower and the gravitational potential energy of the phone would decrease.

5 0
1 year ago
A 10kg box is sliding at 50m/s. Find the momentum
MAXImum [283]

Answer:

The momentum of the ball is 500 kg·m/s

Explanation:

The momentum is given by Mass × Velocity

The given parameters are;

The mass of the box = 10 kg

The velocity by which the box is sliding = 50 m/s

Therefore, the momentum of the ball is given as follows;

The momentum of the ball = 10 kg × 50 m/s = 500 kg·m/s

The momentum of the ball = 500 kg·m/s

5 0
3 years ago
A scientist adds different amounts of salt to 5 bottles of water. He then measures how long it takes for the water to boil. what
g100num [7]
The dependent variable is the amount of time it takes for the water to boil. This variable is dependent because is depends on the amount of salt.
4 0
3 years ago
How many Sig Figs (Significant Figures) are in each number?<br><br> 5070.0<br><br> 870.064080
nlexa [21]

∑ Hey, Lethality ⊃

Answer:

5070.0 has 5 significant figures

870.064080 has 9 significant figures

Explanation:

<u><em>Given:</em></u>

<em>How many Sig Figs (Significant Figures) are in each number?</em>

<em>5070.0</em>

<em>870.064080</em>

<u><em>Solution:</em></u>

<em>-------------------------------------------------------------------------------------------------------------</em>

<em>5070.0</em>

<em>5070.0 has 5 significant figures ( 5 , 0 , 7 , 0 , and 0 )</em>

<em>Number of significant figures: 5</em>

<em>-------------------------------------------------------------------------------------------------------------</em>

<em>870.064080</em>

<em>870.064080 has 9 significant figures ( 8, 7 ,0,0, 6,4,0,8 and 0 )</em>

<em>Number of significant figures: 9</em>

<em>-------------------------------------------------------------------------------------------------------------</em>

<em />

<u><em>xcookiex12</em></u>

<em>8/23/2022</em>

5 0
1 year ago
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0
3 years ago
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