5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.

Can someone solve by showing the steps?

Answers

Answer 1

This question involves the concepts of work done and the frictional force.

a. Work done by the person is "692.82 N".

b. Work done by the frictional force is "490.5 N".

a.

Work done by the person can be given by the following formula:

[tex]W=FdCos\theta[/tex]

where,

W = work done by the person = ?

F = Force applied by the person = 80 N

d = distance traveled = 10 m

θ = angle between force and motion = 30°

Therefore,

[tex]W=(80\ N)(10\ m)Cos30^o[/tex]

W = 692.82 N

b.

Work done by the frictional force is given by the following formula:

[tex]W_f=fd\\W_f=\mu mgd[/tex]

where,

[tex]W_f[/tex] = work done by the frictional force = ?

μ = coefficient of friction = 0.5

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]W_f=(0.5)(10\ kg)(9.81\ m/s^2)(10\ m)[/tex]

[tex]W_f=490.5\ N[/tex]

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Related Questions

Find the dimension of the gravitational constant in this equation F=Gm1m2/r¹r²

Answers

The gravitational force acting between the two bodies is given by:

F=G

r

2

m

1

m

2

G=

m

1

m

2

Fr

2

The dimension of the force is [MLT

−2

]

=

[M][M]

[MLT

−2

][L

2

]

=M

−1

L

3

T

−2

What is the relationship between balancing equations and the law of conservation of matter

Answers

Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. Therefore, there must be the same number of atoms of each element on each side of a chemical equation.

A mass undergoes SHM with amplitude of 4 cm. The energy is 8.0 J at this time. The mass is cut in half, and the system is again set in motion with amplitude 4.0 cm. What is the energy of the system now?

Answers

Hi there!

[tex]\large\boxed{E_{total} = 8.0 \text{ J}}[/tex]

For a mass undergoing SHM, the total energy of the system is given as:

[tex]ME = \frac{1}{2}kA^2[/tex]

Where:

k = Spring constant (N/m)

A = amplitude (m)

There is no quantity of mass in the equation, so the total mechanical energy of the system is NOT impacted by the object's mass.

Thus, the energy of the system will still be 8.0 J.

A 1-kg mass at the Earth's surface weighs how much

Answers

Answer:

the answer is weight=10N

Answer:

[tex]\boxed {\boxed {\sf 9.8 \ Newtons}}[/tex]

Explanation:

Weight is also called the force of gravity. This force acts on all objects at all times, pulling them down toward the center of the Earth.

It is calculated by multiplying the mass by the acceleration due to gravity.

[tex]F_g=mg[/tex]

The mass of the object is 1 kilogram. This scenario is occurring on Earth, so the acceleration due to gravity is 9.8 meters per second squared.

m= 1 kg g= 9.8 m/s²

Substitute the values into the formula.

[tex]F_g= 1 \ kg *9.8 \ m/s^2[/tex]

Multiply.

[tex]F_g= 9.8 \ kg*m/s^2[/tex]

Convert the units. 1 kilogram meter per second squared is equal to 1 Newton, so our answer of 9.8 kilogram meters per second squared is equal to 9.8 Newtons.

[tex]F_g= 9.8 \ N[/tex]

A 1 kilogram mass at Earth's surface weighs 9.8 Newtons.

Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.120 rev/s . As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight.

Answers

Answer:

Explanation:

0.120 rev/s(2π rad/rev) = 0.24π rad/s

At the highest point of the arc, gravity must supply the required centripetal acceleration. As the normal force is 1/4 of his normal weight, then 3/4 of gravity acceleration must be used as centripetal acceleration

0.75g = ω²R

R = 0.75(9.81) / (0.24π)²

R = 12.942198...

R = 12.9 m

The radius of the circle is equal to 12.95m which is rotating with an angular velocity of 0.120 rev/s.

What is vertical circular motion?

A body spins in a vertical circle so that its motion at different points is different from the motion of the body is said to be vertical circular motion.

The velocity and tension vary in maximum magnitude from the lowest to the highest position because of the effect of the gravitational force of the earth.

Given, the angular velocity of the Ferris wheel, ω = 0.120 rev/s

ω = 0.120 rev/s × 2π rad/rev

ω= 0.7536 rad/s

If r is the radius of the circle and 'm' is the mass of the jack.

From newton's second law of motion, the net force will be equal to

mg - N = mrω²

mg - (mg/4) = mrω²

r = 3g/4ω²

r = 3×9.81 / (4× 0.7536)

r = 12.95 m

Therefore, the radius of the circle in which the jack travels is equal to 12.95m.

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Your question is incomplete, most probably the complete question was,

Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.120 rev/s. As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight. What is the radius of the circle in which Jack travels?

Is electrical energy the same or different from energy it takes to play a soccer game

Answers

Answer:

Diffrent

Explanation:

Electrical energy is electrical charges moving. When you play soccer kinetic energy is used. Kinetic energy is the movement of atoms, objects, and electrons.

Children in a tree house lift a small dog in a basket 3.85 m up to their house. If it takes 201 J of work to do this, what is the combined mass of the dog and basket

Answers

Answer:

Explanation:

The work will equal the increase in potential energy.

PE = mgh

m = PE/gh = W/gh = 201/(9.81(3.85)) = 5.32 kg

Two trucks leave at different times (from the same place) headed for the same city. Both trucks arrive at the same time. Based on this information, which of the following sentences is true? Select one:
a. The trucks travelled the same distance in the same amount of time.
b. The trucks were traveling at the same average speed.
c. The trucks travelled different distances.
d. The truck that left later was travelling faster.

Answers

Answer:

d. The truck that left later was travelling faster

Explanation:

Both trucks travelled from the same place to the same place, meaning they both travelled the same distance;

They both arrive at the same time, but the second truck left later so it took less time to travel the distance than the first truck;

The only variable that can account for this difference is speed;

The one that left later, therefore, must have been going faster.

1.25 is the closest to 1.04 or not I want to answer please. I think it's true, but I want to prove it scientifically, please.

Answers

Answer/Explanation:

It False, because if You Round Both of them..

1.25= 1.30

1.04= 1.00

it's like, 1 dollar and 4 cents; compared to 1 dollar and 25 cents. Obviously 25 cents is a lot more than 4 cents.

Which of the vectors in the graph below is the negative of the vector v
A. a
B. d
C. c
D. b

Answers

Answer:

Option c.

Explanation:

Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?​

Answers

Answer:

d = -33.1 m and Vf = -25.5 m/s

Explanation:

Given:

a = -9.8 m

t = 2.6 s

Vᵢ = 0 m/s

To Find:

d = ?

Vf = ?

Now,

d = Vᵢ × t + 0.5 × a × t²

d = (0 m/s) × (2.60 s) + 0.5 × (-9.8 m/s²) × (2.60 s)²

d = -33.1 m (- indicates direction)

Vf = Vᵢ + a × t

Vf = 0 + (-9.8 m/s²) × (2.60 s)

Vf = -25.5 m/s (- indicates direction)

Thus, d = -33.1 m and Vf = -25.5 m/s

-TheUnknownScientist 72

A 2.0 kg particle moving along the z-axis experiences the
force shown in (Figure 1). The particle's velocity is
3.0 m/s at x = 0 m.
A) At what point on the x axis does the particle have a turning point?

Answers

At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Given that a 2.0 kg particle moving along the z-axis experiences the  force shown in a given figure.

Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.

If the particle's velocity is  3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

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A man is whirling a 0.25 kg ball on a 1.5 m long string at 3 m/s. Find the centripetal acceleration of this ball.

Question 2 options:

0.5 m/s2

13.5 m/s2

6 m/s2

2 m/s2

Answers

The centripetal acceleration of this ball is equal to 12 [tex]m/s^2[/tex]

Given the following data:

Diameter = 1.5 mSpeed, V = 3 m/s.Mass = 0.25 kg

Radius = [tex]\frac{Diameter}{2} = \frac{1.5}{2} = 0.75 \;meters[/tex]

To find the centripetal acceleration of this ball:

The acceleration of an object along a circular track is referred to as centripetal acceleration.

Mathematically, the centripetal acceleration of an object is given by the formula:

[tex]A_c = \frac{V^2}{r}[/tex]

Where:

Ac is the centripetal acceleration.r is the radius of the circular track.

V is the velocity of an object.

Substituting the given parameters into the formula, we have;

[tex]A_c = \frac{3^2}{0.75}\\\\A_c = \frac{9}{0.75}\\\\A_c = \frac{9}{0.75}[/tex]

Centripetal acceleration = 12 [tex]m/s^2[/tex]

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a load of 600 N is lifted using a first class lever applying an effort of 350 N. If the distance between the fulcrum and the effort is 60 cm and the distance between the load and the fulcrum is 30cm, calculate it's efficiency​

Answers

Explanation:

(a) A machine is a device by which we can either overcome a large resistive force at some point by applying a small force at a convenient point and in a desired direction or by which we can obtain a gain in the speed.

(b) An ideal machine is a machine whose parts are weightless and frictionless so that which there is no dissipation of energy in any manner. Its efficiency is 100%, i.e. the work output is equal to work input.

The thermometer bulb should have____
a) high heat capacity
b) No heat capacity
c) Small heat capacity
d) Varying heat capacity

Answers

high heat capacity .

HELP PLS!!

A 3 kg mass is raised a distance of 14 m above the earth by a vertical force of 93 N.
a
The gain in potential energy of the mass, to 3 significant figures, is:

Answers

Hi there!

We know that:

U (Gravitational Potential Energy) = mgh

Where:

g = acceleration due to gravity (m/s²)

m = mass (kg)

h = height/displacement (m)

Plug in the values:

U = 3 × 9.8 × 14 = 412 J

A gold doubloon 6.1 cm in diameter and 2.0mm thick is dropped over the side of a Pirate Ship. When it comes to rest on the ocean floor at a depth of 770m how much has its volume changed​

Answers

The volume of a material is the total amount of matter that it can contain. The volume of the given coin has been determined to be 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex]. Since the gold doubloon do not absorb water, then its volume remains constant at the ocean floor.

The volume of the gold doubloon can be determined by;

volume = [tex]\pi r^{2}[/tex] + h

where r is the radius of the coin and h is its thickness.

Such that; diameter = 6.1 cm (61 mm) and h = 2.0 mm

r = [tex]\frac{diameter}{2}[/tex]

 = [tex]\frac{61}{2}[/tex]

r = 30.5 mm

Thus,

volume of the coin = [tex]\frac{22}{7}[/tex] x [tex](30.5)^{2}[/tex] x 2

                               = 5847.2857

Therefore, the volume of the gold doubloon is 5847.3 [tex]mm^{3}[/tex]. This can also be expressed as 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex].

Since the gold doubloon is  not miscible with water, thus its volume at a depth of 770 m at the ocean floor is the same as its initial volume.

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Check Pic please, need help immediately ​

Answers

It’s d


I did this already

If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds

Answers

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

According to above question ~

Current (I) = 4 Amperes

Time (t) = 3 seconds

Charge (q) = ?

Let's find the charge (q) by using formula ~

[tex]I = \dfrac{q}{t} [/tex]

[tex]4 = \dfrac{q}{3} [/tex]

[tex]q = 4 \times 3[/tex]

[tex]q = 12 \: \: coulombs[/tex]

Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds

A small 1240-kg SUV has a wheelbase of 3.2 m. If 67% of its weight rests on the front wheels, how far behind the front wheels is the wagon's center of mass

Answers

Answer:

Explanation:

Let d be the distance to the center of mass from the front wheels

Sum moments about the front wheel contact point to zero

1240(9.8)[d] - 1240(9.8)(1 - 0.67)[3.2] = 0

1240(9.8)[d] = 1240(9.8)(1 - 0.67)[3.2]

                d = (1 - 0.67)[3.2]

                d = 1.056 m

The second hand on a clock is 3.00 cm long. What is the speed of the outermost tip of that second hand

Answers

In 60 minutes or 3600 seconds, the tip of the minute hand traverses the circumference of a circle with radius 3.00 cm, so it moves with a tangential speed of

(3.00 cm)/(3600 s) ≈ 0.00083 cm/s = 8.3 μm/s

Convert :

36°C = ... °F
373 K = ... °C


Question easy​

Answers

Answer:

36 C= 96.8 F

373 K= 99.85

Explanation:

C to F: (36 x 1.8) + 32

         = 64.8 +32

         = 96.8 F

K to C: C= K- 273.15

           C= 373-273.15

           C= 99.85

____

= 36°C

=( 36 × 9/5 ) + 32

=(36 ÷ 5 × 9) + 32

=(7,2 × 9) + 32

= 64,8 + 32

= 96,8°F

______

______

= 373 K

= 373 - 273

= 100°C

[tex] \boxed { \sf semoga \: membantu \: :v }[/tex]

Convection currents occur when _________ energy transfers between two parts of a fluid

Answers

Answer:

heat

Explanation:

The angle of incidence (5 points)



must equal the angle of reflection


is always less than the angle of reflection


is always greater than the angle of reflection


may be greater than, less than, or equal to the angle of reflection

Answers

Answer:

D.) the same.

They are traveling in the same vacuum so there is no resistance and no outside influences. They will travel at the same speed as each other will little to no variance in their speed.

Explanation:

hope this helps. . . <3

good luck!    uωu

How would you best define the word drug?

A: Something that makes you tired
B: Something that can kill you
C: Something that effects your body and mind
D: Stored for energy



someone help

Answers

Answer:

C

Explanation:

Definition of drug: a medicine or other substance which has a physiological effect when ingested or otherwise introduced into the body

A man applies a force of 540 N to the barrow in a direction 75 from the horizontal. He moves the barrow 30 m along the level ground. Calculate the work he does against friction? ​

Answers

The work done by the man against friction is 4,192.86 J.

The given parameters;

force applied, F = 540 Nangle of inclination, θ = 75⁰horizontal distance, x = 30 m

The work done by the man against friction is calculated as follows;

[tex]W = F \times d \times cos(75)\\\\W = 540 \times 30 \times cos(75)\\\\W = 4,192.86 \ J[/tex]

Thus, the work done by the man against friction is 4,192.86 J.

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What is a list of all the states of matter?

Answers

Answer:

3

Explanation:

state of matter are solid

liquid and

gases

Answer:

3

Explanation:

state of a matter are solid liquid and gas

An object is dropped from a vertical height of 1.89 m above the balcony level. What is the object’s speed when it is 2.20 m below the balcony level if 10.0% energy is lost due to the air resistance? Does it matter when to apply 10% loss before V calculations or after? [8.49m/s] [yes it does, 0.9Energy result in √0.9Velocity]

Answers

Answer:

The 10% must come off the energy, not the velocity

Explanation:

Change in height h = 1.89 + 2.20 = 4.09 m

Potential energy = mgh = m(9.81)(4.09) = 40.1m J

Friction losses 40.1m(0.10) = 4.01m J

Kinetic energy at end 40.1m - 4.01m = 36.1m J

KE = ½mv²

v = √(2KE/m)

v = √(2(36.1m)/m = √72.2

v = 8.4983...

v = 8.50 m/s

a boat's engine can give it a velocity of 25m/s. if the boat heads east across a river which of the following due south with a velocity of 8.5m/s; what is the resultant velocity of the boat? (remember you must find both a magnitude and direction!)

Answers

Answer:

Explanation:

v = √(25² + 8.5²) = 26.40549... = 26 m/s

θ = arctan(8.5/25) = 18.77803... = 19° S of E

a stone weight 490 N on a planet whose acceleration due to gravity is half that of the Earth. Calculate (1) the mass of a stone​

Answers

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Acceleration due to gravity on that planet is half than that of om earth ~

that is ~

[tex] \dfrac{9.8}{2} \: \: m/ {s}^{2} [/tex]

[tex]4.9 \: \: m/s {}^{2} [/tex]

And it's weight is equal to ~

[tex]490 \: \: N[/tex]

now, As we know ~

[tex]f = m \times g[/tex]

here ,

f = Weight = 490 Newtons

m = mass of stone = ?

g = Acceleration due to gravity on that planet = 4.9 m/s²

let find the mass of stone ~

[tex]490 = m \times 4.9[/tex]

[tex]m = \dfrac{490}{4.9} [/tex]

[tex]m = 100 \: \: kg[/tex]

Mass of the stone is 100 kg ~

I hope it helped

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