A person applies a constant force of 100 N on an object of mass 20 kg. The object moves horizontally at a constant speed of 0.20 m/s through a distance of 0.80 m. What is the net work done on the object

Answers

Answer 1

Answer:

the net work done on the object is 80 J.

Explanation:

Given;

force applied on the object, F = 100 N

mass of the object, m = 20 kg

speed of the object, v = 0.2 m/s

distance moved by the object, d = 0.8 m

The net work done on the object is calculated as follows;

W = F x d

W = 100 N x 0.8 m

W = 80 J

Therefore, the net work done on the object is 80 J.


Related Questions

if the density of a napthalene ball is 0.02kg.what is the mass of the napthalene ball if it has a volume of 100m³

Answers

Density=mass/volume
Mass=100x0.02=2

A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of (vA)1 = 5 ft/s while connected to the end of a rope. If the rope is pulled inward with a constant speed of vr = 4 ft/s, determine the speed of the box at the instant rB = 1 ft. How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box

Answers

Answer:

W = 1.875 J

Explanation:

For this exercise let's use the relationship between work and kinetic energy

          W = ΔK

The kinetic energy of rotational motion is

         K₀ = ½ I w²

we can assume that the box is small, so it can be treated as a point object, with moment of inertia

          I = m rₐ²

angular and linear velocity are related

          v = w r

          w = v / r

we substitute in the equation, for point A

         K₀ = ½ (m rₐ²) (v / rₐ)²

         K₀ = ½ m v²

For the final point B, as the system is isolated the angular momentum is conserved

initial        L₀ = Io wo

final          L_f = I_f w_f

                L₀ = L_f

                 I₀ w₀ = I_f w_f

               

                (m rₐ²) w₀ = (m  [tex]r_{b} ^2[/tex]) w_f

                 w_f = (rₐ/r_b)² w₀

with this value we find the final kinetic energy

         K_f = ½ I_f w_f²

         K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)²  w₀) ²

         K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]

 

we substitute in the realcion of work

          W = K_f - K₀

          W = ½ m  [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²

          W = ½ m  [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²

           W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2

          W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]

let's calculate

           W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)

           W = 0.625 (3)

           W = 1.875 J

         

An object undergoing simple harmonic motion takes 0.15 s to travel from one point of zero velocity to the next such point. The distance between those points is 30 cm. (a) Calculate the period of the motion. s (b) Calculate the frequency of the motion. Hz (c) Calculate the amplitude of the motion. cm

Answers

Answer:

Explanation:

Point of zero velocity are extreme points situated on either side of equilibrium position .

a )

Time taken to travel between these two points is .15 s

time for half the oscillation = .15 s

Time for full one oscillation = .30 s .

Time period of oscillation = .30 s

b)

frequency = 1 / time period

= 1 / .30s = 3.33 oscillation per second.

c )

Distance between these two point is equal to two times amplitude

2 x amplitude = 30 cm

amplitude = 15 cm

Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10−2 cm , mounted coaxially within a cylindrical anode with a radius of 0.5580 cm . The potential difference between the anode and cathode is 400 V . An electron leaves the surface of the cathode with zero initial speed (vinitial=0). Find its speed vfinal when it strikes the anode.

Answers

Answer:

The final speed will be "[tex]1.185\times 10^7 \ m/sec[/tex]".

Explanation:

The given values are:

Potential difference,

Δv = 400 v

Radius,

r = 0.5580 cm

As we know,

⇒  [tex]W=e \Delta v[/tex]

and,

⇒  [tex]\frac{1}{2}mv^2=e \Delta v[/tex]

then,

⇒  [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]

On substituting the values, we get

⇒     [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]

⇒     [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]

⇒     [tex]=1.185\times 10^7 \ m/sec[/tex]

Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500

Answers

Answer:

8 kV

Explanation:

Here is the complete question

Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?

Solution

Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V

So, Q = CV

= 500 × 10⁻⁶ F × 800 V

= 400000 × 10⁻⁶ C

= 0.4 C

Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV

V = Q/C

= 0.4 C/500 × 10⁻⁶ F

= 0.0008 × 10⁶ V

= 800 V

The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)

= 10 × 800 V

= 8000 V

= 8 kV

Review please help.

Answers

Answer:

1 and 3

Explanation:

because they are going up from 0

A woman shouts at a boy who is underwater what happens to the speed of the sound wave as it moves from the air into the water

Answers

Answer:

B. it increases

Explanation:

As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).

Answer:

B is the correct answer.

Explanation:

Tony ran 600 meters in 60 seconds. What was Tony's speed during the
race?

Answers

10 meters per second.
tony's speed during the race was 10

Balance the equation by choosing the correct coefficient numbers in the drop down menus.
[Select]
SO2 +
[Select]
VH₂ →
[Select]
S +
[ Select]
H20
It is suggested you write this on scratch paper and balance it before choosing your answers :)

Answers

Answer:

SO₂ + 2H₂ —> S + 2H₂O

The coefficients are: 1, 2, 1, 2

Explanation:

SO₂ + H₂ —> S + H₂O

The above equation can be balance as follow:

SO₂ + H₂ —> S + H₂O

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by writing 2 before H₂O as shown below:

SO₂ + H₂ —> S + 2H₂O

There are 2 atoms of H on the left side and 4 atoms the right side. It can be balance by writing 2 before H₂ as shown below:

SO₂ + 2H₂ —> S + 2H₂O

Now, the equation is balanced.

The coefficients are: 1, 2, 1, 2

The magnitude of the force can be determined as? ​

Answers

Answer:

the mass of the object multiplied by the acceleration of the object

Explanation:

N2L states that F = ma (force equals mass times acceleration).

What kind of energy is in a moving skateboard

Answers

Answer:

I guess it is kinetic energy

Answer:

kinetic energy because my dog told me

When a 20 kg explosive detonates and sends a 5 kilogram piece traveling to the right at 105 m/s
what is the speed and direction of the other 15 kilogram piece of the explosive!

Answers

Answer:

speed: 35m/s

direction: left

Explanation:

Assuming the right side is the positive direction:

before explosion:

P = mv = 0

after explosion:

P' = 15P + 5P

(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')

P' = 0.75mv1' + 0.25mv2'

P' = (15kg)v' + (5kg)(105m/s)

P' = 525kg/m/s + (15kg)v1'

P = P'

525kg/m/s + (15kg)v1' = 0

(15kg)v1' = -525kg/m/s

v1' = -35m/s

speed = |-35| = 35m/s

direction is to the left since the right side is the positive direction.

16. Two electric bulbs marked 100W 220V and 200W 200V have tungsten
filament of same length. Which of the two bulbs will have thicker
filament?

Answers

Answer:

The second bulb will have thicker filament

Explanation:

Given;

First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V

Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V

Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m

Resistance of the first bulb:

[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]

Resistance of the second bulb:

[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]

Resistivity of the tungsten filament is given by the following equation;

[tex]\rho = \frac{RA}{L}[/tex]

where;

L is the length of the filament

R is resistance of each filament

A is area of each filament

[tex]A = \pi r^2[/tex]

where;

r is the thickness of each filament

[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]

Therefore, the second bulb will have thicker filament

I WILL GIVE BRAINLIEST TO CORRECT ANSWER
Two boxes are being pulled to the right by a force F across a frictionless table. The force is acting directly on M 2 which is connected to M 1 by a massless cord. Which of the following expressions best represents the acceleration of M 1?

Answers

sorry i don’t know, i really need the points

A carnival ride starts at rest and is accelerated from an initial angle of zero to a final angle of 6.3 rad by a rad counterclockwise angular acceleration of 2.0 s2 What is the angular velocity at 6.3 rad?​

Answers

The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.

Final angular velocity of the carnival ride

The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;

ωf = ωi + 2αθ

where;

ωf  is the final angular velocity of the carnival ride = ?ωi is the initial angular velocity of the carnival ride = 0α is the angular acceleration = 2.0 rad/s²θ is the angular displacement of the carnival ride = 6.3 rad

ωf = 0 + 2(2.0) x 6.3

ωf = 25.2 rad/s

Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.

Learn more about angular velocity here: https://brainly.com/question/6860269

Answer: 5.0 rad/s

Explanation: Because that’s what khan said so try it out.

In which type of circuit does charge move in only one direction?
A. A D.C CIRCUIT
B. AN A.C CIRCUIT
C. A COMBINED CIRCUIT
D. A PARALLEL CIRCUIT

Answers

C. A combined circuit
That’s the answer

a highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of:

Answers

A car making this turn is pulled downward by its own weight, and pushed up by the road at an angle of 45°, so by Newton's second law,

• the net horizontal force on the car is

F = N cos(45°) = m a = m v ² / R

• the net vertical force on the car is

F = N sin(45°) - m g = 0

where

• N = magnitude of the normal force

m = mass of the car

• a = v ² / R = centripetal acceleration of the car

v = tangential speed of the car

• R = 100 m = radius of curvature

• g = 9.8 m/s² = acceleration due to gravity

From the net vertical force equation, we get

N = m g / sin(45°)

and substituting this into the net horizontal force equation and solving for v gives

(m g / sin(45°)) cos(45°) = m v ² / R

v = √(R g cos(45°) / sin(45°)) ≈ 31 m/s

We have that A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of

V=32m/s

From the question we are told

a highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of:

Generally the equation for the Velocity is mathematically given as

[tex]V=\sqrt{rgtan\theta}[/tex]

Therefore

[tex]V=\sqrt{rgtan\theta}\\\\V=\sqrt{100*9.8*tan45}\\\\V=32m/s[/tex]

Therefore

A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of

V=32m/s

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CiCi is hiking in the woods after a rainstorm when she sees a single large mass of rock and soil moving quickly downhill.

Which type of mass movement is this?

A. landslide
B. slump
C. creep
D. mudflow

Answers

The answer is, D. Mudflow

What is the frequency of a monochromatic light used in a diffraction experiment that has a wavelength of 6.38 ✕ 10e-07 m?

Answers

Answer:

[tex]f=4.70\times 10^{14}\ Hz[/tex]

Explanation:

Given that,

The wavelength of light, [tex]\lambda=6.38\times 10^{-7}\ m[/tex]

We need to find the frequency of the light. We know that,

[tex]c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.38\times 10^{-7}}\\\\f=4.70\times 10^{14}\ Hz[/tex]

So, the required frequency of light is equal to [tex]4.70\times 10^{14}\ Hz[/tex].

The moon does not stay at the same distance from the earth.why?​

Answers

Answer:

The moon does not stay at the same distance of the earth because the ortbit of the moon is slightly elliptical. If earth is not tilted at an angle of 66.5°, there will be no change in the season and the earth will have equal length of days and night.

Explanation:

mark me brainlest

A 2.0 kg breadbox on a fric-
tionless incline of angle u 40 is
connected, by a cord that runs over a
pulley, to a light spring of spring con-
stant k 120 N/m, as shown in
Fig. 8-43. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10 cm down the in- cline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box’s acceleration at the instant the box momentarily stops?

Answers

Dhjwbxuzb wm I known kdn wi. Wlzkk n

Which one the answer to this question

Answers

The second bubble is the answer:)

Example of the center of the gravity

Answers

Answer:

The example of the center of the gravity is the middle of a seesaw

Explanation:

I hope this will help you and plz mark me brainlist

Becoming informed about economics helps a person understand the reasons a command economy is ideal. role of government in regulating production. why consumers receive tax revenue. reasons an economy must always be completely regulated. Mark this and return

Answers

Answer:

Role of government in regulating production

Explanation:

The role of government in regulating show , provides the legal and social framework, uphold competition, provides public goods and services.

What is the role of economics in the community?

The community's role in conserving and enhancing common-property resources is well known.

In extra, its role in helping market growth by its power to execute trade agreements among transacting parties belonging to the community network is stressed.

Thus, it provides the legal and social framework, maintains competition, and provides public goods and services.

To learn more about  economics in community click here:

https://brainly.com/question/1344575

Hi please zoom in to see it clearly, uh you don’t have to answer them all but it would be nice !!! (no links please)

Answers

Answer:

3. (a) - is the answer most likely

4: (a)

5 (d or c)

6 (b)

Ya whatever the person above said :))

An artificial satellite circling the Earth completes each orbit in 126 minutes. (a) Find the altitude of the satellite.

Answers

Answer:

Explanation:

Time period of rotation

T = 2πR/ V where R is radius of orbit and V is orbital velocity

Orbital velocity V = √ ( GM/R ) , m is mass of the earth .

T = 2πR √R / GM

T² = 4π²R³ / GM

Putting the values

( 126 x 60 )² = 4 x 3.14² x R³ / 6.67 x 10⁻¹¹ x 5.97 x 10²⁴

57.15 x 10⁶ = 39.44 x R³ / 39.82 x 10¹³

R³ = 577 X 10¹⁸

R = 8.325 x 10⁶ m

= 8325 km

Radius of earth = 6400 km

height of satellite = 8325- 6400 = 1925 km .

Which two chemical equations show double-replacement reactions?
A. C+02 - CO2
B. 2Li + CaCl2 - 2LiCl + Ca
I C. Ca(OH)2 + H2S04 - CaSO4 + 2H20
D. Na2CO3 + H2S - H2CO3 + Na2S

Answers

The two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

What is double replacement reaction?

A double replacement reaction have two ionic compounds that are exchanging anions or cations.

From the given options, we can choose the following based on their exchange of anions or cations.

Ca(OH)2 + H2S04 - CaSO4 + 2H20Na2CO3 + H2S - H2CO3 + Na2S

Thus, the two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

Learn more about double replacement reaction here: https://brainly.com/question/14281077

#SPJ2

Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit

Answers

Answer:

0.03 A

Explanation:

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

Voltage (V) = 12 V

Resistor (R) = 470 Ω

Current (I) =?

From ohm's law, the voltage, current and resistor are related by the following formula:

Voltage = current × resistor

V = IR

With the above formula, we can obtain the current in the circuit as follow:

Voltage (V) = 12 V

Resistor (R) = 470 Ω

Current (I) =?

V = IR

12 = I × 470

Divide both side by 470

I = 12 / 470

I = 0.03 A

Thus, the current in the circuit is 0.03 A

Answer:

0.03 A

Explanation:

Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit

0.03 A

Batteries are not perfect. They can't deliver infinite current. As the current load on a battery gets larger, the voltage output gets smaller.

a. True
b. False

Answers

True Im typing this for charater limit

You put a diode in a microelectronic circuit to protect the system in case an untrained person installs the battery backward. In the correct forward-bias situation, the current is 255 mA with a potential difference of 116 mV across the diode at room temperature (300 K). If the battery were reversed, so that the potential difference across the diode is still 116 mV but with the opposite sign, what would be the magnitude of the current in the diode

Answers

Answer:

The current in the new circuit is 0

Explanation:

A diode is an electronic component that allows the electric current to go only in one direction. If in the first case the current was 255 mA, and the battery was changed ( change in polarity ) no current will flow through the circuit. That change is similar or equivalent to change the diode to the no pass position

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