How many degrees are there between the direction of motion and the force of friction?
A. O
B. 270
C. 90
D. 180

Answers

Answer 1
The answer is b!!!!!
Answer 2

The correct option is (d).

Angle should be 180° between the direction of force and the direction of motion of a body as work done is equal to force multiplied by displacement multiplied by cosx.

Friction is a force that opposes motion. When two objects are in contact, friction is acts in a direction opposite to the motion of the object.A force known as friction prevents relative motion between surfaces that are in touch. One of the more basic features of friction is that it always runs parallel to the surfaces with which it makes contact and opposes any motion or attempted motion of the systems with respect to one another.

What is the direction of friction force?When a force is applied to stop it from moving, frictional force provides resistance. As a result, it always moves in the opposite direction of an item in motion. Friction acts on the right when force is applied to the left.

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Related Questions

which system of units is used by only a small number of countries in the world, including the u.s

Answers

Answer:

imperial system

Explanation:

Answer:

british system

Explanation:

A P E X

The law of universal gravitation describes the relationships between
Group of answer choices

A. unbalanced force, balanced force, and net force.

B. speed, distance, and time.

C. speed, velocity, and acceleration.

D .force, mass, and distance.

Answers

Answer:

D force mass and distance

Whats the word for when "velocity equals 0 and direction changes" Its 14 letters, and I have _ _ _ _ i _ _ _ _ _ h_ i _ _ t

Answers

Answer:

Maximum Height

Explanation:

The maximum height is the highest point reached by a projected body. At this point, final velocity, v of the body is equal to 0; because upward motion is at its peak and the body start to fall again.

Hence, final velocity v is always 0 at this maximum point and the direction of motion changes from upward to downward.

Final velocity at this maximum height is zero because of the directional change experied by the object and the fact that upward motion of the body terminates at this point.

A 4kg object is moving at 5 m/s . A 10-N force is applied in the direction of motion and if the object has a final velocity 10m /s. calculate the distance that the object moves

Answers

Answer:

Distance, S = 15m

Explanation:

Given the following data;

Mass = 4kg

Force = 10N

Initial velocity = 5m/s

Final velocity = 10m/s

To find the distance;

First of all, we would calculate the acceleration

Force = mass * acceleration

10 = 4*acceleration

Acceleration = 10/4

Acceleration = 2.5m/s²

Now to find the distance, we would use the third equation of motion.

[tex] V^{2} = U^{2} + 2aS [/tex]

Where;

V represents the final velocity measured in meter per seconds. U represents the initial velocity measured in meter per seconds. a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

10² = 5² + 2*2.5*S

100 = 25 + 5S

5S = 100 - 25

5S = 75

Distance, S = 75/5

Distance, S = 15m

An object with a mass of 2.0 kg accelerates 2.0 m/s 2when an unknown force is applied to it. What is the amount of force?

Answers

Answer:

4 N

Explanation:

mass = 2 kg

acceleration = 2 m/s^2

Force = mass * acceleration

         = 2 *2

         = 4 N

9. In a closed circuit made of a battery, a wire, a lightbulb, and a switch which part's main responsibility is allowing electricity to flow over a distance?
A: battery
B: light bulb
C: switch
D: wire

Answers

Answer:

d

Explanation:

Answer:

D: wire

Explanation:

Un bloque de 3 kg en reposo se deja libre a una altura de 5 m sobre una rampa curva y sin rozamiento. Al pie de la rampa se encuentra un resorte de constante k = 400 N/m, como se muestra en la fig. El objeto desliza por la rampa y llega a chocar contra el resorte comprimiéndolo una distancia x antes de que quede en reposo momentáneamente. Determinar: a) La velocidad con la que el bloque alcanza al resorte. ____________________ b) La distancia x que el bloque comprime al resorte. __________________ c) La velocidad con la que el bloque es expulsado por el resorte. ____________________ d) La altura que alcanza sobre la parte curva. ________________ e) ¿Alcanzará la misma altura si la rampa no está libre de rozamiento? ___________________

Answers

Answer:

a) La velocidad del bloque cuando llega al resorte es de aproximadamente 9,9 m / s

b) La distancia a la que se comprime el resorte es de aproximadamente 0,86 m

c) La velocidad con la que el resorte expulsa el bloque es de aproximadamente 9,9 m / s

d) La altura que alcanza el bloque es de 5 metros.

e) El bloque no alcanzará la misma altura si la rampa no está libre de fricción

Explanation:

a) Los parámetros dados del bloque son;

La masa del bloque, m = 3 kg

La altura a la que se coloca el bloque, h = 5 m

La constante de resorte, k = 400 N / m

La aceleración debida a la gravedad, g = 9,8 m / s²

La energía potencial de un cuerpo, P.E. = m · g · h

Por tanto, la energía potencial inicial del bloque, P.E. se da como sigue;

P.E. = 3 kg × 9,8 m / s² × 5 m = 147 julios

P.E. = 147 julios

La energía cinética del bloque al pie de la rampa, K.E. = 1/2 · m · v²

Dónde;

v = La velocidad del bloque cuando llega al resorte

Por lo tanto, para el bloque dado tenemos;

K.E. = 1/2 · m · v² = 1/2 × 3 kg × v²

Por el principio de conservación de la energía, tenemos;

El PE. del bloque en reposo a una altura de 5 m = La energía cinética al pie de la rampa. K.E.

∴ P.E. = K.E.

147 J = 1/2 × 3 kg × v²

v² = 147 J / (1/2 × 3 kg) = 98 m² / s²

v = √ (98 m² / s²) = 7 · √2 m / s

v = 7 · √2 m / s ≈ 9,9 m / s

b) La energía recibida por el resorte comprimido, E = 1/2 · k · x²

Dónde;

k = La constante del resorte = 400 N / m

x = La distancia a la que se comprime el resorte

Por el principio de conservación de la energía, tenemos;

La energía recibida por el resorte comprimido, E = La energía potencial inicial del resorte, P.E.

∴ E = 1/2 · k · x² = P.E.

De lo que tenemos;

E = 1/2 × 400 N / m × x² = 147 julios

x² = 147 Julios / (1/2 × 400 N / m) = 0,735 m²

x = √ (0,735 m²) = 0,7 · √ (3/2) m ≈ 0,86 m

La distancia a la que se comprime el resorte = x ≈ 0.86 m

c) La velocidad con la que el resorte expulsa el bloque se indica a continuación;

La energía en el resorte = 1/2 · k · x² = La energía cinética dada al bloque, 1/2 · m · v²

∴ 1/2 · k · x² = 1/2 · m · v²

∴ La velocidad con la que el bloque es expulsado por el resorte, v = La velocidad con la que el bloque llega al resorte = 7 · √2 m / s

La velocidad con la que el resorte expulsa el bloque, v = 7 · √2 m / s ≈ 9,9 m / s

d) La altura que alcanza el bloque también viene dada por la siguiente relación anterior;

P.E. = K.E.

∴ m · g · h = 1/2 · m · v²

v = 7 · √2 m / s

De donde tenemos h = La altura inicial del bloque en la rampa = 5 metros

e) El bloque no alcanzará la misma altura si la rampa no está libre de fricción porque se utilizará energía para superar la fuerza de fricción

a) La velocidad final del bloque es aproximadamente 9.903 metros por segundo.

b) El resorte se deforma 0.858 metros.

c) Por el principio de la conservación de energía y sabiendo la ausencia de fuerzas disipativas, la velocidad del objeto expulsado del resorte es aproximadamente 9.903 metros por segundo.

d) Por el principio de la conservación de energía y si existieran fuerzas disipativas, la altura máxima sería menor a la hallada en el punto a).

a) Conforme a la situación de este problema, la energía cinética traslacional final ([tex]K[/tex]), en joules, es igual a la energía potencial gravitacional inicial ([tex]U[/tex]), en joules.

[tex]U = K[/tex] (1)

Por las definiciones de las energías cinética traslacional y potencial gravitacional expandimos la ecuación anterior:

[tex]m\cdot g\cdot h = \frac{1}{2}\cdot m\cdot v^{2}[/tex] (1)

Ahora despejamos la velocidad de esa ecuación:

[tex]v = \sqrt{2\cdot g\cdot h}[/tex]

Donde:

[tex]m[/tex] - Masa del bloque, en kilogramos.[tex]g[/tex] - Aceleración gravitacional, en metros por segundo al cuadrado.[tex]h[/tex] - Altura inicial del bloque, en metros.[tex]v[/tex] - Velocidad final del bloque, en metros por segundo.

Si sabemos que [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] y [tex]h = 5\,m[/tex], entonces la velocidad final del bloque es:

[tex]v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}[/tex]

[tex]v\approx 9.903\,\frac{m}{s}[/tex]

La velocidad final del bloque es aproximadamente 9.903 metros por segundo.

b) Por el principio de conservación de la energía, la energía cinética traslacional inicial es igual a la energía potencial elástica final, cuyas fórmula es la siguiente:

[tex]\frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)

Where:

[tex]k[/tex] - Constante de resorte, en newtons por metro.[tex]x[/tex] - Deformación del resorte, en metros.

Ahora despejamos la deformación del resorte:

[tex]x = \sqrt{\frac{m}{k} }\cdot v[/tex] (3)

Si sabemos con [tex]k = 400\,\frac{N}{m}[/tex], [tex]m = 3\,kg[/tex] y [tex]v \approx 9.903\,\frac{m}{s}[/tex], entonces la deformación del resorte es:

[tex]x = \sqrt{\frac{3\,kg }{400\,\frac{N}{m} } }\cdot \left(9.903\,\frac{m}{s} \right)[/tex]

[tex]x \approx 0.858\,m[/tex]

El resorte se deforma 0.858 metros.

c) Por el principio de la conservación de energía y sabiendo la ausencia de fuerzas disipativas, la velocidad del objeto expulsado del resorte es aproximadamente 9.903 metros por segundo.

d) Por el principio de la conservación de energía y si existieran fuerzas disipativas, la altura máxima sería menor a la hallada en el punto a).

Invitamos cordialmente a ver este problema sobre el principio de conservación de la energía: https://brainly.com/question/16582988

If Mars were 10 times closer to the Sun, then the Sun would attract Mars with
times
(less or more) force.

Answers

Answer:

more

Explanation:

The velocity does not change greatly between Mars and Venus —explained Dong— but Venus's closer proximity to the sun boosts the density by almost a factor of 4.5. This would mean that atmosphere on Mars would be lost even more rapidly than at its current position.

Please answer quick! Thanks guys!

Answers

Answer:

wire By please mark my answer in brain list

Answer:

Remove wire G, if the wire after it is F. Remove wire A as C is connecting to it already.

PLEASE PLEASE HELPPP!
Select all that apply!

Answers

Answer:

1 AND 3

Explanation:

how high is the pole from which the lamp fell freely if it hits the ground at a speed of 15 m/s​

Answers

Answer & Explanation:

The acceleration due to gravity is about 10 m/s^2.

That means a the speed of a falling object will increase by 10 m/s every second that it falls.

We know that the lamp hits the ground at 15 m/s.

That means the lamp has been falling for

[tex]\frac{15 m/s}{10 m/s^{2} } = 1.5 s[/tex]

Now we use the formula  [tex]h = \frac{1}{2} gt^{2}[/tex]  to calculate the height, where

h is the height

g is the acceleration due to gravity and

t is the time taken to fall a height of h.

h = 1/2 x 10 x 1.5^2 = 11.25 m

a clay ball with a mass of 0.35 kg hits another 0.35 kg ball at rest and the two stick together. The first balls intial speed us 4.2m/s.

What are the balls final speed?​

Answers

Answer:

4.2 m/s

the balls final speed is V (m/s)

because the clay ball hits another and the two stick together

=> 0,35.4,2 + 0,35.4,2 = (0,35 + 0,35).V

<=> 2,94 = 0,7V

<=> V = 2,94/0,7 = 4,2

For a clay ball with a mass of 0.35 kg, the ball's final speed is mathematically given as

V2= 4.2

What is the balls final speed?​

Question Parameter(s):

A clay ball with a mass of 0.35 kg

Hits another 0.35 kg ball at rest and the two stick together.

The first ball's initial speed is 4.2m/s.

Generally, the equation for the conservation of momentum is mathematically given as

M1v1=m2v2

Therefore

0.35*4.2 + 0.3*5.4.2 = (0.35 + 0.35)*V

2,94 = 0,.7V

V2 = 2.94/0.7

V2 = 4.2

In conclusion

V2= 4.2

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Why is physical activity important?
O It helps keep you healthy.
It helps control your weight
It helps imcove your balance and coordination
All of the Above
at least one activity you can enjoy to help you get your 60 min

Answers

Answer:

All of the Above

walking

What is the charge on an object that experiences a force of 5 Newtons in an electric field of 50 Newtons per coulomb?

Answers

Answer:

Explanation:

F = qE

F is the force in Newtons

q is the test charge

E is the electrical field produced by the source charge

[tex]5=q(50)\\q=1*10^-^1Coulombs[/tex]

The magnitude of charge on the object is of 0.1 C.

Given Data:

The magnitude of electric field is, E = 50 N/C.

The magnitude of Electric force is, F = 5 N.

The region where any particle feels the effect of electric force on itself, due to other charged entities, is known as Electric field. And the relation between the electric force and electric field is given as,

F = E × q

Here,

q is the magnitude of charge on the object.

Solving as,

q = F/E

q = 5 / 50

q = 0.1 C

Thus, we can conclude that the magnitude of charge on the object is of 0.1 C.

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How would you describe your appetite over the past couple of weeks? Have your eating habits altered in any way?

Answers

Answer:

no im in a rehab program so we eat the sam shi every day

Explanation:

brainlest?

Explanation:

During lockdown, I ate LOADS OF JUNK FOOD. TRUST ME!!

I gained 14 kg of weight and became 68 kg

just due to junk food.I am 13 btw and at that time I was 5 feet 4 so definitely fat. all those pizzas and burgers and all.then I realised that I have gained weight. I have to REDUCE MY WEIGHT. I am quite determined on what I say. it took me 3 months or so and I finally got back to my original weight. How? just by going for a stroll everyday, eating healthy and constantly working out. eating less is not the solution,eating less calories is the solution.Also, by eating healthy food,I grew taller. I am 5 feet 6 now. So yes, definitely it helps put alot and keep eating healthy food and you will definitely be healthy.

Thank You.

Rain and wind place which type of load on structures?
A) Dynamic load
B) Soft load
C) Gradual load
D) Static load

Answers

It’s a dynamic load.
Hopes this helps :)
answer:a is the answer

Is the potential energy on the surface of Earth zero?If not,what will be it's value?

Answers

Answer:

Potential energy does not have an absolute measurement. It is always relative to some reference point. Gravitational potential always increases when you go up and decreases when you go down. But the choice of a zero point is arbitrary.

If you’re dropping objects onto the ground, then choosing the ground as a zero point makes the calculations easier. But you could just as easily make the zero point the top of Mt. Everest, and all the answers would turn out the same. Up still means more energy and down means less, but now the PE of an object at sea level would have a negative sign, but not as negative as an object ten meters above sea level.

So everything would still work fine.

In fact, planetary astronomers take this idea to extremes. Instead of the top of Mt. Everest, they set the zero point for potential energy as infinitely high—so far “up” that gravity is so weak that going “up” another kilometer doesn’t gain you any energy. Then the value for potential energy everywhere else in the universe anywhere near a planet has a negative sign, but just as before, all the answers in relative terms turn out fine.

Explanation:

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Why does the Moon appear red during a lunar eclipse?

1.)Because the sunlight bends as it passes through the Earth's atmosphere

2.)Because the sunlight bends as it passes through the Moon's atmosphere

Answers

Answer:

Image result for Why does the Moon appear red during a lunar eclipse?

A lunar eclipse takes place when the sun, Earth and moon line up in space. The moon passes through Earth's shadow. ... Bottom line: The moon can look red during a total lunar eclipse because of sunlight that's filtered and refracted by Earth's atmosphere

Someone plz help I’m on a time limit :(

What patterns occur in the Earth-Sun-Moon system that allow us to identify the cause of
eclipses?

Answers

Answer:

Earth's rotation and orbit and the Moon's orbit cause observable patterns (Day and night, seasons, phases of the Moon, tides).

calculate the ground pressure of a 90kg gas cylinder the diameter of the cylinder is 1.1 m

Answers

Answer:

P = 928.09 Pa

Explanation:

Given that,

Mass of a gas cylinder, m = 90 kg

The diameter of the cylinder, d = 1.1 m

Radius, r = 0.55 m

We need to find the ground pressure. The pressure is equal to force per unit area. So,

[tex]P=\dfrac{F}{A}\\\\P=\dfrac{mg}{\pi r^2}\ (A\ is\ area\ of\ base)\\\\P=\dfrac{90\times 9.8}{\pi \times (0.55)^2}\\\\P=928.09\ Pa[/tex]

So, the ground pressure is 928.09 Pa.

2. A large collection of stars, gas, and dust is called a _____?

Answers

They are bound together by gravity so it’s called a galaxy

A wave has a period of 2 seconds and a wavelength of 4 meters. Calculate its frequency and speed.
Note: Recall that the frequency of a wave equals 1/period & the period of a wave equals 1/frequency.

Answers

Answer:

frequency = 0.5 /s

speed = 2m/s

Explanation:

frequency = 1/period = 1/2 = 0.5 /s

speed = frequency × wavelength

= 0.5 × 4 = 2m/s

Determine the torque
produced by a perpendicular force of 75
N at the end of a 0.2 m wrench.

Answers

Answer:

Explanation:

the answer is C.)

The torque produced by a perpendicular force of 75 N at the end of a 0.2 m wrench is 15 Nm.

What is torque?

The force which causes the object to rotate about any axis is called torque. In math form, it is equivalent to the product of force and perpendicular distance.

Torque is a twisting or turning force that frequently results in rotation around an axis, which may be a fixed point or the center of mass. The ability of something rotating, such as a gear or a shaft, to overcome turning resistance is another way to think of torque.

Given:

The perpendicular force on the wrench, P = 75 N,

The length of the wrench, r = 0.2 m,

Calculate the torque of the wrench as shown below,

The torque of the wrench = r × P × sin a

Here, the force is perpendicular hence, a = 90°

The torque of the wrench = 0.2 × 75 × sin90

The torque of the wrench = 15 × 1

The torque of the wrench = 15

Thus, the torque on the wrench is 15 Nm.

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A 0.225 kg sample of tin initially at 97.5°c is dropped into 0.115 kg of water. the initial temperature of the water is 10.0°c. if the specific heat capacity of tin is 230 j/kg • °c, what is the final equilibrium temperature of the tin-water mixture

Answers

Answer:

The final equilibrium temperature of the tin-water mixture is approximately 18.468 °C

Explanation:

The parameters of heat energy transfer from the tin to the water are given as follows;

The mass of the sample of tin, m₁ = 0.225 kg

The initial temperature of the tin, T₁ = 97.5 °C

The mass of the water into which the tin is dropped, m₂ = 0.115 kg

The initial temperature of the water, T₂ = 10.0 °C

The specific heat capacity of tin, c₁ = 230 J/(kg·°C)

The specific heat capacity of water, c₂ = 4,200 J/(kg·°C)

Let 'T' represent the final equilibrium temperature of the tin-water mixture, we have;

The heat lost by the tin, ΔQ[tex]_{tin}[/tex] = The heat gained by the water ΔQ[tex]_{water}[/tex]

∴  ΔQ[tex]_{tin}[/tex] = ΔQ[tex]_{water}[/tex]

Where;

ΔQ[tex]_{tin}[/tex] = m₁·c₁·(T₁ - T)

ΔQ[tex]_{water}[/tex] = m₂·c₂·(T - T₂)

By substitution, we have;

ΔQ[tex]_{tin}[/tex] = 0.225 kg × 230 J/(kg·°C) × (97.5°C - T)

ΔQ[tex]_{water}[/tex] = 0.115 kg × 4,200 J/(kg·°C) × (T - 10.0°C)

From ΔQ[tex]_{tin}[/tex] = ΔQ[tex]_{water}[/tex], we have;

0.225 kg × 230 J/(kg·°C) × (97.5°C - T) = 0.115 kg × 4,200 J/(kg·°C) × (T - 10.0°C)

∴ 5,045.625 J - 51.75 J/°C × T = 483 J/°C × T - 4,830 J

5,045.625 J + 4,830 J = 534.75 J/°C × T

∴ 534.75 J/°C × T = 9,875.625 J

T =  9,875.625 J/(534.75 J/°C) = 18.4677419 °C ≈ 18.468 °C

The final equilibrium temperature of the tin-water mixture, T ≈ 18.468 °C.

A rock has 750 J of potential energy as it sits on a ledge. If the rock were pushed off the ledge, how much would it have just before it hit the ground?

Answers

Answer:

the 750 j will have potential energy is 375

Explanation:

750/2 is 375

Write down the use of light pipe.???​

Answers

Light pipes are used to illuminate small areas. For example, light from a small LED to a switch indicator or a small graphic. Typically, a light pipe would go from a circuit to the area that requires illumination.

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A snail travels down the sidewalk at 25 mm/min for 4 minutes How far does the snail travel?​

Answers

The snail would travel 100mm

Each minutes is 25mm and it took 4 minutes

4 x 25 = 100

or

25+25+25+25=100

The snail would travel 100mm

Each minutes is 25mm and it took 4 minutes

4 x 25 = 100

A horizontal net force F is exerted on an object at rest. The object starts at x=0 m and has a speed of 4.0 m/s after moving 4.0 m along a horizontal frictionless surface. The net force F as a function of the object's position x is shown below

Answers

Explanation:

The solution is be found in the attachment.

A horizontal net force F is exerted on an object at rest. The object starts at x = 0 m and has a speed of 4.0 m/s after moving 8.0 m along a horizontal frictionless surface. The net force F as a function of the object's position x is shown below.

What is the mass of the object?

Answer: 30 kg

Which statement describes the difference between ionic bonds and Van der
Waals forces?
A. Ionic bonds involve the sharing of electrons between atoms, while
Van der Waals forces involve the attraction of nonpolar molecules.
B. Ionic bonds involve the transfer of electrons between ions, while
Van der Waals forces involve the sharing of electrons between
atoms.
C. lonic bonds involve the transfer of electrons between ions, while
Van der Waals forces involve the attraction of nonpolar molecules.
D. Ionic bonds involve the attraction of nonpolar molecules, while
Van der Waals forces involve the transfer of electrons between
ions.

Answers

Answer:

C. lonic bonds involve the transfer of electrons between ions, while

Van der Waals forces involve the attraction of nonpolar molecules.

Explanation:

Just did it



A steel ball bearing is released from a height H and

rebounds after hitting a steel plate to a height H.

What is true about the collision with the steel plate?

Answers

Answer:

ELASTIC collision

kinetic energy is conservate

Explanation:

As the ball bounces to the same height, it can be stated that the impact with the floor is ELASTIC.

As the floor does not move the conservation of the moment

            po = pf

            -mv1 = m v2

- v1 = v2

So the speed with which it descends is equal to the speed with which it rises

Therefore the kinetic energy of the ball before and after the collision is the same

Answer:

CORRECT (SELECTED)

It is elastic since kinetic energy was conserved.

Explanation:

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