A 60 kg student weighs 600 N.
He does a bungee jump.
Calculate the change in gravitational potential energy as the student falls 50 m.
Give the unit.

Will give brainliest!

Answers

Answer 1

Answer:

30 000 J/Nm

Explanation:

60 x 10 x 50

=600 x 50

=30000 J/Nm

Answer 2

The change in gravitational potential energy of the student is 18000 Joule.

What is gravitational potential energy?

The energy that an item has or acquires when its location changes as a result of being in a gravitational field is known as gravitational potential energy. Gravitational potential energy can be defined as an energy that has a connection to gravitational force or gravity.

Given parameters:

Mass of the student: m = 60 kg.

Weighs of the student: W = 600 N.

the student falls h =50 m.

Hence, change in gravitational potential energy of the student =

Weighs of the student × change in height

= 600 N × 30 m

= 18000 Joule.

So, the change in gravitational potential energy of the student is 18000 Joule.

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Related Questions

Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 12.0 mg piece of tape held 0.55 cm above another. (The magnitude of this charge is consistent with what is typical of static electricity.)

Answers

Answer:

    q = 2 10⁻⁸ C

Explanation:

For this exercise we use the translational equilibrium equation

                    F_e -A =

                    F_e = W

the electric force is given by Coulomb's law

                    F_e = [tex]k \frac{q_1q_2}{r^2}[/tex]

in this case they indicate that the loads on the tapes are equal

                    F_e = k q² / r²

we substitute

                    k q² / r² = m g

                    q = [tex]\sqrt{ \frac{mg r^2}{k} }[/tex]

calculate  

                     q = [tex]\sqrt { \frac{ 12 \ 10^{-3} \ 9.8 (0.55 \ 10^{-2})^2 }{9 \ 10^9} }[/tex]    

                     q = [tex]\sqrt{ 3.9526 \ 10^{-16}[/tex]

                     q = 1,999 10⁻⁸ C

                     q = 2 10⁻⁸ C

Which is an example of kinetic energy?

A. The energy stored in
ethanol

B. A ball sitting at the top of a ramp

C. A compressed spring

D. A hockey puck sliding across ice

Answers

D . A hockey puck sliding across ice

D. A hockey puck sliding across ice

The masses of astronauts are monitored during long stays in orbit, such as when visiting a space station. The astronaut is strapped into a chair that is attached to the space station by springs and the period of oscillation of the chair in a friction-less track is measured.
(a) The period of oscillation of the 10.0 kg chair when empty is 0.750 s. What is the effective force constant of the springs?
(b) What is the mass of an astronaut who has an oscillation period of 2.00 s when in the chair?
(c) The movement of the space station should be negligible. Find the maximum displacement of the 100,000 kg sace station if the astronaut's motion has an amplitude of 0.100 m.

Answers

Answer:

a)  k = 701.8 N / m, b)  m_{ast} = 61.1 kg, c)  v ’= -1.3 10⁻⁴ m / s

Explanation:

a) For this exercise let's use the relationship of the angular velocity

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

          k = w² m

the angular velocity is related to the period

          w = 2π / T

we substitute

          k = 4 π²    [tex]\frac{m}{T^2}[/tex]

let's calculate

          k = 4 π²   10 /0.75²

          k = 701.8 N / m

b) now repeat the measurement with an astronaut on the chair

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

where the mass Month the mass of the chair plus the mass of the astronaut

        M = m + [tex]m_{ast}[/tex]

       

          M = k / w²

          w = 2π / T

let's calculate

           w = 2π / 2

            w = π rad / s

           

            M = 701.8 /π²

            M = 71,111 kg

now we use that

          M = m + m_{ast}

          m_{ast} = M - m

          m_{ast} = 71.111 - 10.0

          m_{ast} = 61.1 kg

c) if the astronaut's movement is simple harmonic

          x = A cos wt

therefore the speed is

         v = [tex]\frac{dx}{dt}[/tex]

         v = -Aw sin wt

maximum speed is

          v = - Aw

          v = 0.100 π

          v = 0.31416 m / s

we can suppose that the movement of the space station and the astronaut  is equivalent to division of the same

         

initial instant. Before the move

         p₀ = 0

final instant. When the astronaut is moving

        p_f = M_station v’+ m_{ast} v

the moment is preserved

         p₀ = pf

         0 = M__{station} v ’+ m_{ast} v

         v ’= - [tex]\frac{m_{ast} }{M_{station} } \ v[/tex]

we substitute

         v ’= [tex]\frac{61.1 }{ 100000 } \ 0.31416[/tex]

         v ’= -1.3 10⁻⁴ m / s

the negative sign indicates that the station is moving in the opposite direction from the astronaut

What is the acceleration of a car that goes from 0 MS to 60 MS and six seconds

Answers

a = Δv/Δt
a = v2-v1/t2-t1
a = 60MS - 0MS / 6 seconds
a = 60 MS/ 6 seconds
a = 10 m/s^2

Please help 25 points!

Three waves with frequencies of 1 Hertz (Hz), 3 Hz, and 9Hz travel at the same speed. Which of the following statements is correct?

A. The 1 Hz wave contains the most energy.

B. The crests of all three waves are of equal height.

C. The wavelength of the 9Hz wave is three times that of the 3 Hz wave.

D. The 1 Hz wave has the longest wavelength.

Answers

Answer:

B

Explanation:

The crest of all three waves are of equal height

how many pennies can 4 folds of a paper hold?

Answers

A lot I don’t really under stand the question

A liquid fueled rocket is red on a test stand. The rocket nozzle has an exit diameter of 30 cm and the combustion gases leave the nozzle at a velocity of 3800 m/s and a pressure of 100 kPa, which is the same as the ambient pressure. The temperature of the gases in the combustion area is 2400 C. Find (a) the temperature of the gases at the nozzle exit plane, (b) the pressure in the combustion area, and (c) the thrust developed. Assume that the gases have a speci c heat ratio of 1.3, and a molar mass of 9. Assume that the ow in the nozzle is isentropic.

Answers

Answer:

1. Temperature= 869.35 K

2. Pressure of combustion = 12994.043 kpa

3. Thrust = 127x10⁶N

Explanation:

this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.

1.

The temperature = (273+2400k) - (3800)²/2(4003)

= 2673 - 14440000/8006

= 2673 - 1803.65

= 869.35 K

Approximately 869.4K

2. We first get mach number

= 3800/√1.3(923.8)(869.35)

= 3800/1021.78

= 3.719

Pressure = 100kpa[1+2.07464415]^1.3/0.3

= 12995.043kpa

C. Thrust

Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)

= 12678.621

= 126.781 kN

Thrust is approximately 127kN = 127x10⁶N

QUCIK!! SOMEONE PLEASE HELP! I’LL MARK BRAINLIEST!!

Answers

Answer:

A. v = √2gh

B. No! The final velocity does not depend on the mass of the car.

C. Yes! the final velocity depends on the steepness of the hill

D. 3.28 m/s

Explanation:

A. Determination of the final velocity.

½mv² = mgh

Cancel out m

½v² = gh

Cross multiply

v² = 2gh

Take the square root of both side

v = √2gh

B. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is no mass (m) in the formula.

Thus, the final velocity does not depend on the mass of the car.

C. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is height (h) in the formula.

Thus, the final velocity depends on the steepness of the hill

D. Determination of the final velocity.

Height (h) = 0.55 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = √2gh

v = √(2 × 9.8 × 0.55)

v = √10.78

v = 3.28 m/s

Given that Carbon-14 has a half-life of 5700 years, determine how long it would take for
this reduction to occur.

Answers

Answer:It will take about 3000 years

Explanation:

The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is [01] m. (a) What is the force (N) that the track must exert on the car? (positive is up) (b) What must be the force (N) that the car exerts on a 61 kg passenger?

Answers

This question is incomplete, the complete question is;

The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is 15.8 m.

(a) What is the force (N) that the track must exert on the car? (positive is up)

(b) What must be the force (N) that the car exerts on a 61 kg passenger?

Answer:

a) the force (N) that the track must exert on the car is -6139.14 N

b) the force (N) that the car exerts on a 61 kg passenger is -1170.27 N

Explanation:

Given the data in the question;

Let N represent the force that the track must exerted on the car

Net force on the car Fnet = Mg + N

so

M × a = Mg + N

N = Ma - Mg

N = Ma - M(v²/R)

we substitute

N = (320kg × 9.8m/s²) - ( 320 × ((21.4m/s)² / 15.8 m) )

N = 3136 - ( 320 × 28.9848 )

N = 3136 - 9275.136

N = -6139.14 N

Therefore, the force (N) that the track must exert on the car is -6139.14 N

b) What must be the force (N) that the car exerts on a 61 kg passenger?

Let N represent the force that the car exerts on 61kg passengers

so

Net force of passengers Fnet = mg + N

Ma = Mg + N

N = Ma - Mg

N = Ma - M(v²/R)

N = (61kg × 9.8m/s²) - ( 61 × ((21.4m/s)² / 15.8 m) )

N = 597.8 - ( 61 × 28.9848)

N = 597.8 - 1768.0728

N = -1170.27 N

Therefore, the force (N) that the car exerts on a 61 kg passenger is -1170.27 N

The centripetal force of the track on the car moving in the circular path is [tex]1.465 \times 10^6 \ N[/tex].

The force (N) that the car exerts on a 61 kg passenger is 597.8 N.

Centripetal force of the track

The centripetal force of the track on the car moving in the circular path is calculated as follows;

[tex]F_c = \frac{mv^2}{r}\\\\ F_c = \frac{320 \times 21.4^2}{0.1} \\\\F_c = 1.465 \times 10^6 \ N[/tex]

Normal force of the passenger

The force (N) that the car exerts on a 61 kg passenger is equal to the force the passenger exerts on the car based on Newton's third law of motion.

F = mg

F = 61 x 9.8

F = 597.8 N

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You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, where the lumps collide and stick together. Just before the impact, the velocity of one lump was < 3, 3, -3 > m/s, and the velocity of the other lump was < -4, 0, -4 > m/s. What is the velocity of the stuck-together lump just after the collision

Answers

Answer:

[tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]

Explanation:

[tex]u_1[/tex] = Velocity of one lump = [tex]3x+3y-3z[/tex]

[tex]u_2[/tex] = Velocity of the other lump = [tex]-4x+0y-4z[/tex]

m = Mass of each lump = [tex]30\ \text{g}[/tex]

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

[tex]mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]

The velocity of the stuck-together lump just after the collision is [tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex].

.................,,,,,,,,,,,

Answers

Answer:

B

Explanation:

Motion is movement, the teacher's movement is motion

A student is conducting an experiment to compare the resistivity of two unknown materials by using two wires, each made of one of the materials and each connected in a circuit. The student measures the potential difference across and current in the wires. What must be the same to be able to compare the resistivities using just the potential difference and current measurements?

Answers

Answer:

is there a. b.  c  or d?

Explanation:

Help plsssssssssss I write it 100 time no one answer

Answers

Answer:

1.93×10²⁸ s

Explanation:

From the question given above, the following data were obtained:

Number of electron (e) = 2×10²⁴

Current (I) = 10 A

Time (t) =?

Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:

1 e = 96500 C

Therefore,

2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e

2×10²⁴ e = 1.93×10²⁹ C

Thus, 1.93×10²⁹ C of electricity is passing through the point.

Finally, we shall determine the time. This can be obtained as follow:

Current (I) = 10 A

Quantity of electricity = 1.93×10²⁹ C

Time (t) =?

Q = it

1.93×10²⁹ = 10 × t

Divide both side by 10

t = 1.93×10²⁹ / 10

t = 1.93×10²⁸ s

Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point

Static Friction
Now let’s examine the static case. Remain on the “Force graphs” tab at the top of the window. Make sure the box labeled “Ffriction” is checked at the left of the screen, this will allow us to measure to force of friction experienced by an object as it slides down the ramp.

Draw a free body diagram for an object sitting on the incline at rest, assuming the incline is at the maximum angle BEFORE the object starts to move. Be sure to include friction and stipulate whether it is kinetic or static. ​

Answers

name the element and explain why it is unusual:
a)A liquid metal
b)A non-metal that conducts electricity

A soccer ball was kicked over the edge of a wall and traveled 35 m horizontally at a speed of 5.6m/s. Calculate the vertical height of the wall.

Answers

Answer:

Are you sure it was soccer ball? Or meine hearts

Explanation:

The current flow in the light bulb is 0.5A
a.Calculate the amount of electric charge that flow through the bulb in 2 hour
b.If one election carries a
charge 1.6 x 10-14 c Find the number of election through the bulb in 2 hour?

Answers

Answer:

Explanation:

Given that,

The current in the light bulb, I = 0.5 A

(a) We know that,

Electric current = charge/time

or

Q = It

Put t = 2 hours = 7200 s

So,

Q = 0.5 × 7200

Q = 3600 C

(b) Charge on one electron, [tex]Q=1.6\times 10^{-19}\ C[/tex]

Let there are n electrons flow through the bulb in 2 hours.

I = Q/t

Since, Q = ne

So,

I = ne/t

[tex]n=\dfrac{I\times t}{e}\\\\n=\dfrac{0.5\times 7200}{1.6\times 10^{-19}}\\\\n=2.25\times 10^{22}[/tex]

Hence, this is the required solution.

A 450.0 kg roller coaster is traveling in a circle with radius 15.0m. Its speed at point A is 28.0m/s and its speed at point B is 14.0 m/s. At point A the cart is already moving with circular motion. a) Draw free bodydiagramsfor the cartatpointsAand B(two separate free body diagrams). b) Calculate the acceleration of the cartat pointsAandB(magnitude and direction). c) Calculate the magnitude of the normal force exerted by the trackson the cartat point A. d) Calculate the magnitude of the normal force exerted by the tracks on the cart at point B.

Answers

Answer:

b)  a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N

Explanation:

a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle

b) Let's start at point A

Let's use that the acceleration is centripetal

           a = v² / r

let's calculate

            a = 28² / 15.0

            a = 52.26 m / s²

as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards

Point B

           a ’= 142/15

           a ’= 13.06 m / s²

in this case the acceleration is vertical downwards

c) The values ​​of the normal force

point A

let's use Newton's second law

           ∑ F = m a

           N- W = m a

           N = mg + ma

           N = m (g + a)

           N = 450.0 (9.8 + 52.25)

           N = 2.79 10⁴ N

d) Point B

            -N -W = m (-a)

             N = ma -m g

             N = m (a-g)

             N = 450.0 (14.0 - 9.8)

             N = 1.89 10³ N

TWO forces, one of 12N and another or 24N

act on body in such a way that they make an angle of 90° with each other. Find the resaltant of two forces.

Answers

Answer:

26.833 N

Explanation:

The computation of the resaltant of two forces is shown below:

Given that

Force A = 12N

Force B = 24N

Based on the above information  

Resultant R is

[tex]=\sqrt{A^2 + B^2 + 2AB \times cos \theta}\\\\=\sqrt{144 + 576 + 2\times 24\times 12\times cos90^{\circ}}\\\\=\sqrt{144+576+576\times 0}\\\\=\sqrt{720}[/tex]

=26.833 N

Shanti is riding on a train that is moving at a speed of 90 km/h. He is carrying a power cord for his phone that is 1.2 m long.

Which describes the length of the power cord when Shanti gets off the train?

cannot be determined
less than 1.2 m
more than 1.2 m
equal to 1.2 m

Answers

Answer:

D. equal to 1.2

Explanation:

on edg

The length of the power cord will be equal to 1.2 m.

Describe about the length of power cord? The train is moving at a speed of 90 km /hr. Train was moving but the person in the train can be considered to be at rest. Shanti is the person travelling on the train. Her cord can be used only by her and the cord length of the phone will be 1.2 m.The length can be measured through the distance.The unit of length is meter.As we know the concept of motion and rest, there only the train in motion, shanti was at rest and shanti's power cord were also in the rest. Power cord length will be determined only at the time of manufacturing.If the power cord length to be change then the crimping process.So, the length will not change suddenly.

The length of the power cord when shanti gets off the train is equal to 1.2 m.

The Correct answer is Option D.

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A particle move in the xy plane so that its position vector r=bcosQi +bsinQj+ ctk, where b, Q and c are constants. show that the partial move with constant speed.​

Answers

Answer:

The speed of this particle is constantly [tex]c[/tex].

Explanation:

Position vector of this particle at time [tex]t[/tex]:

[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].

Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:

[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].

Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].

Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:

[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].

The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :

[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].

Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)

Flying insects such as bees may accumulate a small positive electric charge as they fly. In one experiment, the mean electric charge of 50 bees was measured to be +(30±5)pC+(30±5)pC per bee. Researchers also observed the electrical properties of a plant consisting of a flower atop a long stem. The charge on the stem was measured as a positively charged bee approached, landed, and flew away. Plants are normally electrically neutral, so the measured net electric charge on the stem was zero when the bee was very far away. As the bee approached the flower, a small net positive charge was detected in the stem, even before the bee landed. Once the bee landed, the whole plant became positively charged, and this positive charge remained on the plant after the bee flew away. By creating artificial flowers with various charge values, experimenters found that bees can distinguish between charged and uncharged flowers and may use the positive electric charge left by a previous bee as a cue indicating whether a plant has already been visited (in which case, little pollen may remain). What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)?
(a) Because air is a good conductor, the positive charge on the bee’s surface flowed through the air from bee to plant.
(b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee.
(c) The plant became electrically polarized as the charged bee approached.
(d) Bees that had visited the plant earlier deposited a positive charge on the stem.

Answers

Answer:

a) True

Explanation:

There are several possible explanations for this positive charge

* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,

to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct

* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.

When reviewing the different statements we have

a) True, it agrees with the second explanation of the phenomenon

b) False. The earth is a deposit of negative charge

c) false. If this is the case the charge should be negative

d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.

Answer:

Explanation:

.

A storage tank has the shape of an inverted circular cone with height 12 m and base radius of 4 m. It is filled with water to a height of 10 m. Find the work required to empty the tank by pumping all of the water to the top of the tank. (The density of water is 1000 kg/m3. Assume g

Answers

Answer:

Work required to empty the tank by pumping all of the water to the top of the tank = 1674700 Kgm/s^2

Explanation:

Volume of Circular cone = V = (1/3)πr2h

where r is the radius in meters

and h is the height in meters

Substituting the given values in above equation, we get -

V = [tex]\frac{1}{3} * 3.14 * 4^2 * 10 = 167.47[/tex] cubic  meters.

The force required will be equal to the mass of water in the cone

[tex]= 167.47 * 1000[/tex]

= 167470 Kg

Weight = Mass * g

= 167470 * 10

= 1674700 Kgm/s^2

- .
?



y

(っ◔◡◔)っ ♥ chose the answer with the question marks ♥

Answers

Answer:

okay I'm a bit confused but I like the little emoji dudw

Answer:

?

Explanation:

.

your "A" never changes, while your "Y" changes depending on strength of gravity. is it mass or weight? ​

Answers

Answer:

A - mass. B - Weight

Explanation:

This is because weight varies with the strength of gravity. Mass is just the amount of matter in an object

39. What is the change in momentum for a 5,000 kg ship in

outer space that experiences no net force over a 1 hr

period?

Answers

Answer:

Change in momentum is zero.

Explanation:

The following data were obtained from the question:

Mass (m) = 5000 kg

Time (t) = 1 h

Net force (F) = 0

Change in momentum =?

Force = Rate of change of momentum

0 = change in momentum

Change in momentum = 0

We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.

Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 X10^-9 C and the other is given a charge of -18 X 10^-9 C. a. Find the electric force exerted on one sphere by the other. b. The sphere are connected by a conducting wire. After equilibrium has occurred, find the electric force between the two spheres.

Answers

Answer:

Explanation:

Force between two charged conducting sphere

= k x Q₁ x Q₂ / r² ,  k is a constant  Q₁ and Q₂ are charges and   r is distance between them .

= 9 x 10⁹ x 12 x 10⁻⁹ x 18 x 10⁻⁹ / .30²

= 21600 x 10⁻⁹

= 2.16 x 10⁻⁵ N .

b )

After the spheres are joined together , there is redistribution of charge and remaining charge will be equally shared by them .

Charge on each sphere = (12 - 18 ) x 10⁻⁹ / 2

= - 3 x 10⁻⁹ C .

Force = 9 x 10⁹ x 3 x 10⁻⁹ x 3 x 10⁻⁹ / .30²

= 900 x 10⁻⁹ N .

If you have a 0.125 kg lead piece at
20.0°C, how much heat must you
add to melt it? (Remember, you
must warm it to its melting point
first.)
Material
Lead
Melt Pt (°C)
327
L (1/kg)
2.32.104
Boil Pt (°C) Lv (1/kg)
1750 8.59.105
c (1/(kg*c)
128
(Unit = J)

Answers

Answer:

7,812 J

Explanation:

Using the relation:

Q = mcΔθ

Q = quantity of heat

C = specific heat capacity of lead

Δθ = temperature change (T2 - T1)

M = mass of substance

Q = mass * specific heat * Δθ

Q = 0.125kg * 128 * (327 – 20)

Q = 0.125 * 128 * 307

Q = 4912 J

For melting:

Q = mass * Hf

0.125 * (2.32 * 10^4)

= 2,900 J

Total = 4,912 J + 2,900 J = 7,812 J

A spring with a constant of 76 N/m is extended by 0.9 m. How much energy is stored in the extended spring?

Answers

Answer:

[tex]E=30.78\ J[/tex]

Explanation:

The force constant of the spring, k = 76 N/m

The extension in the spring, x = 0.9 m

We need to find the energy is stored in the extended spring. The energy stored in the spring is given by :

[tex]E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 76\times (0.9)^2\\\\E=30.78\ J[/tex]

So, 30.78 J of energy is stored in the spring.

Which statement best compares potential and kinetic energy?
O Objects always have more potentiał energy than kinetic energy.
O Kinetic energy increases and potential energy decreases when the velocity of an object increases
O Only potential energy decreases when an object's height increases.
O Objects always have more kinetic energy than potential energy.

Answers

Answer:

Kinetic energy increases and potential energy decrease when velocity of an object increase.

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