3. An object of mass 8 kg is sliding down a friction lessinclined plane of length 11 m that makes an angleof 70 deg with the horizontal. Calculate thework done by gravitational force as the objectslides from the top of the inclined plane to theground. (1 point)A. 099.045 JB. O1135.872 JC.810.391 JD.499.917

3. An Object Of Mass 8 Kg Is Sliding Down A Friction Lessinclined Plane Of Length 11 M That Makes An

Answers

Answer 1

Given

Mass of the object, m=8 kg

Length of the inclined plane, l=11 m

Angle of inclination,

[tex]\theta=70^o[/tex]

To find

Calculate the work done by gravitational force as the object slides from the top of the inclined plane to the

ground.

Explanation

The height of the ramp

[tex]h=lsin\theta=11sin70^o[/tex]

The work done by gravitational force,

[tex]\begin{gathered} W=mgh \\ \Rightarrow W=8\times9.8\times11sin70^o \\ \Rightarrow W=810.390J \end{gathered}[/tex]

Conclusion

The work done is C.810.391


Related Questions

How much time would it take a car to go from a speed of 10m/s to 30m/s if it acceleratesat a rate of 4.0m/s/s?

Answers

Given,

The initial velocity of the car, u=10 m/s

The final velocity of the car, v=30 m/s

The acceleration of the car, a= 4.0 m/s²

From one of the equations of the motion, we have

[tex]v=u+at[/tex]

Where t is the time duration.

On rearranging the above equation,

[tex]t=\frac{v-u}{a}[/tex]

On substituting the known values,

[tex]t=\frac{30-10}{4.0}=5\text{ s}[/tex]

Therefore the time that the car takes to accelerate is 5 s

Imagine that someone is looking out of the top floor window of a skyscraper with a brick in his hand at the same instant someone else is looking out if the window on the floor below also holding a brick if both bricks were dropped at the same instant would the distance between them increase decrease or remain the same over time why

Answers

The distance between them remain the same over time both of them are accelerating - because of gravity

What is acceleration ?

acceleration: the rate at which the speed and direction of a moving object vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration for all other motions.

Both of them are accelerating - because of gravity - and the one that you dropped first has been accelerating for longer - and is therefore going faster than they one that you dropped later.

Until the two objects both reach their terminal velocities - assuming the objects are identical - their speeds will eventually be the same - and from that point onwards - the distance between them won’t change

The distance between them remain the same over time both of them are accelerating - because of gravity

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Shown in the figure, are electric field lines around two point charges. The two charges might be which of these? A)two protons B)a proton and an electron C)two electrons D)two croutons E)two neutrons

Answers

Given:

A figure of the electric field lines.

To find:

The nature of the charges.

Explanation;

The electric field lines always start at a positive charge. And the electric field lines always end at a negative charge. If there is a single charge, the field lines may start at infinity or end at infinity.

Electric field lines also do not form closed loops. They do not cross each other.

In the figure, the field lines start at the blue charge and end at the charge represented by the color yellow.

Therefore the blue charge is positive and the yellow charge is negative.

Final answer:

Therefore the correct answer is option B.

A 1 kg apple and a 2 kg grapefruit are 2 m apart. What is the force of gravity between each fruit?

Answers

According to Newton's law of gravitation, any particle of matter in the universe attracts any other with a force varying directly as the product of their masses and inversely as the square of the distance between them. It is expressed as

F = Gm1m2/r^2

where

G = gravitational constant and its value is G = 6.673 x 10^-11 Nm^2/kg^2

m1 and m2 are the masses of the bodies

r is the distance between the bodies

From the information given,

m1 = 1

m2 = 2

r = 2

By substituting these values into the formula,

F = (6.7 x 10^-11 x 1 x 2)/2^2

F = 33.5 x 10^- 12

F = 3.35 x 10^-11

The second option is correct

Round the answer for each problem to the correct number of significant figures.

a. (7.31 × 104) + (3.23 × 103)

b. (8.54 × 10–3) – (3.41 × 10–4)

c. 4.35 dm × 2.34 dm × 7.35 dm

d. 4.78 cm + 3.218 cm + 5.82 cm

e. 38,736 km ÷ 4784 km

Answers

After solving each problem to correct the number of significant figures :

(a) 76330

(b) 8199 × 10⁻³

(c) 74.81 dm.

(d) 13.8 cm

(e) 8.09 km

What is an Equation?

Equations are mathematical expressions that have two algebraic expressions on either side of an equals (=) sign. The expressions on the left and right are shown to be equal to one another, demonstrating this relationship. L.H.S. = R.H.S. (left hand side = right hand side) is a fundamental mathematical formula.

According to the question,

(a) (7.31 × 10⁴) + (3.23 × 10³)

(73100) + (3230) = 76330

(b) (8.54 × 10⁻³) – (3.41 × 10⁻⁴)

= 8199 × 10⁻³

(c) 4.35 dm × 2.34 dm × 7.35 dm

= 74.81 dm.

(d) 4.78 cm+3.218 cm+5.82 cm

= 13.8 cm

(e) 38,736 km ÷ 4784 km

= 8.09 km

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In which direction does the tension force on the pirate point when a gust of wind rises up?

Answers

Answer: For downwind sailing, with the sail oriented perpendicular to the wind direction, the pressure increase on the upwind side is greater than the pressure decrease.

Explanation: this is all i know

12. How could extreme heat (resulting from Climate Change) affect human andanimal life?

Answers

ANSWER:

The answer is given in the step by step of the question

STEP-BY-STEP EXPLANATION:

Extreme heat can affect human and animal life in the following ways:

• Animals are sensitive to changes in temperature, as much or more than humans. Think how bad you feel when you have a fever. They also suffer this type of discomfort as a consequence of climate change.

,

• Climate change is pushing many species to the limit. They lack water to drink or suffer from temperatures in which they are not comfortable.

,

• Provocation of forest fires damaging the habitat of animals and humans

Two sounds of 610Hz and 620Hz are played together. The frequency you hear will be ___. The beat frequency is ___

Answers

Explanation

The beat frequency is equal to the complete value of the alteration in the frequency of the two waves,it can be calculated by using the expression

[tex]f_{beat}=\lvert{{f_1}-f_2}\rvert[/tex]

so

Step 1

a)let

[tex]\begin{gathered} f_1=610\text{ Hz} \\ f_2=620\text{ Hz} \end{gathered}[/tex]

b) replace and calculate

[tex]\begin{gathered} f_{beat}=\lvert{{f_1}-f_2}\rvert \\ f_{beat}=\lvert610\text{ Hz-620Hz}\rvert \\ f_{beat}=10\text{ Hz} \end{gathered}[/tex]

so

the beat frequency is 10 HZ

Step 2

the frequency yuo will detect is the difference of the frequency, so , teh frequency beat, so

the frequency you hear will be 10 Hz

I hope this helps you

how many joules does a lamp marked 12 volts, 24 w consumed in an hour . And also what is the current?

Answers

Given data:

* The voltage given is 12 volts.

* The value of the power given is 24 Watt.

* The time till which the lamp is used is,

[tex]\begin{gathered} t=1\text{ hr} \\ t=60\times60\text{ s} \\ t=3600\text{ s} \end{gathered}[/tex]

Solution:

(a). The energy consumed by the lamp in time t is,

[tex]E=P\times t[/tex]

where P is the power of lamp, t is the time, and E is the energy consumed,

Substituting the known values,

[tex]\begin{gathered} E=24\times3600 \\ E=86400\text{ J} \\ E=86.4\times10^3\text{ J} \\ E=86.4\text{ kJ} \end{gathered}[/tex]

Thus, the energy consumed by the lamp in 1 hour is 86.4 kJ.

(b). The power of the lamp in terms of voltage and current is,

[tex]\begin{gathered} P=VI \\ I=\frac{P}{V} \end{gathered}[/tex]

where P is the power, V is the voltage and I is the current,

Substituting the known values,

[tex]\begin{gathered} I=\frac{24}{12} \\ I=2\text{ A} \end{gathered}[/tex]

Thus, the current flowing through the lamp is 2 A.

Choose all the answers that apply.Asteroids:only orbit planetsonly orbit the sunare made of frozen gas and icetake millions of years to orbit the sun sometimes come very close to Earth

Answers

ANSWER:

5th option: sometimes come very close to Earth

STEP-BY-STEP EXPLANATION:

Asteroids are small rocky objects that orbit the Sun. Although asteroids orbit the Sun like planets, they are much smaller.

Therefore, of the options the only totally true option is "sometimes come very close to Earth"

So the correct answer is 5th option: sometimes come very close to Earth

A traffic light hangs from a pole as shown in (Figure 1). The uniform aluminum pole AB is 7.20 m long and has a mass of 11.0 kg . The mass of the traffic light is 22.0 kg .
-Determine the tension in the horizontal massless cable CD.
-Determine the vertical component of the force exerted by the pivot A on the aluminum pole.
-Determine the horizontal component of the force exerted by the pivot A on the aluminum pole.

Answers

There is a 409.30 N force in the horizontal massless cable CD.

The vertical component of the pivot A's force on the aluminum pole, which is  323.73N.

The force that the pivot A applies to the aluminum pole has a horizontal component of 409.30 N.

Force is any influence to the motion of the body .

Basically the product of mass and acceleration.

Length of the pole = 7.20 m

Mass of the pole = m = 11.0 kg

Mass of the traffic light = M = 22.0 kg

Let the length of the rod AD be L

Also [tex]Lsin\alpha = 3.8[/tex]

[tex]\alpha = 37[/tex]°

[tex]Sin\alpha = 0.602[/tex]

L = [tex]\frac{3.8}{0.602}[/tex]

L = 6.312 m

In order to determine the tension (T) in the cable, the free body diagram will give the detailed information.

On balancing all the forces in both the 'x' and 'y' direction, i.e. summation of all the forces in 'x' and 'y' direction must equal to zero.

∑[tex]F_{x}[/tex]  = 0

[tex]R_{x}[/tex] - T = 0

[tex]R_{x}[/tex]  = T

∑[tex]F_{y}[/tex]  = 0

[tex]R_y- W_p - W_l = 0[/tex]

[tex]W_p[/tex] = 9.81 * 11 = 107.91 kg

[tex]W_l\\[/tex] = 9.81 * 22 = 215.82 kg

[tex]R_y\\[/tex] = [tex]W_p + W_l[/tex]

[tex]R_y\\[/tex] = 107.91 + 215.82

[tex]R_y\\[/tex] = 323.73 N

The vertical component of the force exerted by the pivot A on the aluminum pole is 323.73 N

Taking moment along x-axis

M = 0

[tex]Th-mgCos\alpha* \frac{1}{2} - mglCos\alpha[/tex] = 0

T = [tex]\frac{glCos\alpha }{h} *\frac{m}{2}*M[/tex]

T = 409.30 N

The tension in the horizontal massless cable CD is 409.30 N

[tex]R_{x}[/tex]  = T = 409.30 N

The horizontal component of the force exerted by the pivot A on the aluminum pole is 409.30 N.

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When a stone and a feather is thrown down from a height under free fall. Which of the two will first reach the ground???​

Answers

Answer:

The stone

Explanation:

It will be the first to fall because of its mass which is heavier than that of the feather .

The potential energy of two atoms in a diatomic molecule is approximated by U(r)=a/r12−b/r6, where r is the spacing between atoms and a and b are positive constants.
Find the force F(r) on one atom as a function of r .
Express your answer in terms of the variables a , b , and r .

Answers

The force (r) on one atom as a function of r is  a/r¹³ - b/r⁷.

What is the electric force between charged molecules?

The electric force between charged molecules is determined by applying Coulomb's law, which states that the electric force between tow charged molecules is directly proportional to the product of their charges and inversely proportional to the square of their distance.

F = kq²/r²

where;

q is the magnitude of the chargesr is the distance between the charges

The electric potential energy of the atoms is given as;

U = Fr

From the given parameters;

Fr = a/r¹² − b/r⁶

F = 1/r( a/r¹² − b/r⁶)

F = a/r¹³ - b/r⁷

Thus, the force on one atom as a function of r is determined by applying Coulomb's law of electrostatic force. The force is also expressed in terms of the variables a , b , and r .

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A projectile is fired at an upward angle of 55.0 ∘ from the top of a 155 m cliff with a speed of 165 m/s .
a)What will be its speed when it strikes the ground below? (Use conservation of energy and neglect air resistance.)

Answers

A projectile is fired at an upward angle and the speed of the object when it strikes the ground below will be 434.5 m/s.

What is Projectile motion?

A projectile is an object or particle that is thrown toward the surface of the Earth and moves along a curved route only under the influence of gravity. Galileo demonstrated that this curving path was a parabola, but in the unique situation where it is hurled straight up, it may also be a straight line.

According to the question,

[tex]h=v_0_yt+1/2gt^2[/tex]

-155 m = (165 × sin 55°)t - 0.5(9.8)t²

-155 = 135.16t - 4.9 t²

4.9 t² - 135.16t - 155 = 0

t = 27.5 seconds.

Now, the speed of the object when it strikes the ground will be,

[tex]v_f=v_i+gt[/tex]

= 165 + (9.8)(27.5)

[tex]v_f[/tex] = 434.5 m/s.

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What is the mass of 12 m3 of methylated spirit whose relative density is 0.8 ? ( Hint the density of water = 1000 kgm-3 ) A. 9600 kg B.9400 kg C. 8600 kg

Answers

ANSWER

A. 9600 kg

EXPLANATION

Given:

• The volume of the substance, V = 12m³

,

• The relative density of the substance, 0.8

,

• The density of water, 1000 kg/m³

Unknown:

• The mass of the substance, m

The relative density of a substance is defined as,

[tex]\rho_{relative}=\frac{\rho_{substance}}{\rho_{water}}[/tex]

And the density of a substance is,

[tex]\rho=\frac{m}{V}[/tex]

Let's solve the first formula for the density of the substance,

[tex]\rho_{substance}=\rho_{relative}\cdot\rho_{water}=0.8\cdot1000\operatorname{kg}/m^3=800\operatorname{kg}/m^3[/tex]

Then, solve the second formula for m,

[tex]m=\rho\cdot V=800\operatorname{kg}/m^3\cdot12m^3=9600\operatorname{kg}[/tex]

Hence, the mass of this substance is 9600 kg

A small mailbag is released from a helicopter that is descending steadily at 2.82 m/s.(a) After 3.00 s, what is the speed of the mailbag?v = m/s(b) How far is it below the helicopter?d = m(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.82 m/s?v = m/sd = m

Answers

a)

When the package is released from the moving helicopter, the package and the helicopter has a common speed. The package is in freefall. We would calculate the speed of the helicopter after a given time t by applying the formula,

v = vo + gt

where

vo is the initial velocity of of the package and it is equal to the speed of the helicopter

v is the final velocity of the package after time t

g is th acceleration due to gravity

From the information given

vo = 2.82

t = 3

g = 9.8

Thus,

v = 2.82 + 9.8 * 3 = 2.82 + 29.4

v = 32.22 m/s

After 3.00 s, the speed of the mailbag is 32.22 m/s

b) We want to calculate the distance covered by the mailbag in 3 s. We would apply the formula which is expressed as

s = vot + 1/2 x g x t^2

where

s is the distance

vo = 2.82

g = 9.8

t = 3

Thus,

s = 2.82 x 3 + 1/2 x 9.8 x 3^2 = 8.46 + 44.1

s = 52.56 m

Since we want to calculate the distance from the helicopter, we would calculate the diatance that the helicopter also travelled downwards in 3 s by applying the formula for calculating distance which is expressed as

distance = speed x time

Thus

distance = 2.82 x 3 = 8.46 m

Difference in distance = 52.56 - 8.46 = 44.1

The package is 44.1 m from the helicopter

c) If the helicopter is moving upwards, it would be thrown out and it would attain a certain height before it starts descending. The height is calculated by the formula,

h = vo^2/2g

By substituting the values,

h = 2.82^2/2 x 9.8 = 0.406 m

When the mail bag attains this height, it will start moving downwards. At this height, the final velocity is zero. We would calculate the time taken to attain this height by applying the formula,

v = vo - gt

v = 0

Thus,

0 = 2.82 - 9.8 x t

9.8t = 2.82

t = 2.82/9.8 = 0.288

The time left for freefall within the first 3 seconds is

3 - 0.288 = 2.712 s

The height attained by the mailbag in 2.712s is calculated by the formula,

h = gt^2/2

h = 9.8 x 2.712^2/2 = 36.04 m

Distance travelled by helicopter by ascending upward in 3 s is

distance = 2.82 x 3 = 8.46

Height of mailbag from final position after 3 seconds is

36.04 - 0.406 = 35.634

Difference in distance = 35.634 + 8.46 = 44.094

The package is 44.094 m from the helicopter

For the velocity of the mailbag after 3 seconds,

v = - vo + gt

v = - 2.82 + 9.8 x 3 = - 2.82 + 29.4

v = 26.54 m/s

Rtz coordinate system

Answers

Due to the car goes around the trajectory without sliding, then. it is necessary that centripetal force equals the friction force.

Then, we have:

Fc = Fr

where,

Fc = m(v^2/R) centripetal force

Fr = μN = μ*m*g friction force

Then, by replacing the precious expressions we obtain:

m(v^2/R) = μ*m*g

And by solving for v:

[tex]v=\sqrt[]{\mu\cdot g\cdot R}[/tex]

where,

μ = 1.00

g = 9.8 m/s^2

R = 70.0 m

By replacing we get:

[tex]undefined[/tex]

What is the shortest distance in which you can stop, after the brakes are applied, without the groceries sliding off the seat? The static and kinetic coefficients of friction are, respectively, 0.65 and 0.45. Assume that the surface of the seat is horizontal.

Answers

The shortest distance in which you can stop, after the breaks are applied is 80.38 m.

What is the shortest distance you can stop?

The shortest distance in which you can stop is calculated by applying the principle of conservation of energy and work energy principle.

K.E = ¹/₂mv²

where;

K.E is your kinetic energy

m is your mass

v is your speed

The work done by force of friction before you stop is calculated as follows;

W = Ffx

where;

Ff is the frictional forcex is the shortest distance you can stop

W = (μmg)x

where;

μ is coefficient of kinetic friction

W = K.E

(μmg)x = ¹/₂mv²

(μg)x = ¹/₂v²

x = (v²) / (2μg)

x = (32²) / (2 x 0.65 x 9.8)

x = 80.38 m

Thus, the shortest distance in which you can stop, after the breaks are applied is determined by applying the principle of conservation of energy.

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The complete question is below:

Your are driving at 32 m/s, what is the shortest distance in which you can stop, after the brakes are applied, without the groceries sliding off the seat? The static and kinetic coefficients of friction are, respectively, 0.65 and 0.45. Assume that the surface of the seat is horizontal.

Given v = 145 sin (200πt + 3π/2), what is the period?Question 12 options:10 msec5 sec.03 sec.005 sec

Answers

For a sine or cosine function of the form:

[tex]A\sin (B(x+C))+D[/tex]

the perfiod is given by:

[tex]\frac{2\pi}{B}[/tex]

In this case we have that:

[tex]B=200\pi[/tex]

Then we have:

[tex]\frac{2\pi}{200\pi}=\frac{1}{100}=0.01=10\times10^{-3}[/tex]

Therefore, the period is 10 ms

Two different masses have equal, non-zero kinetic energies. The momentum of the smaller mass isSelect one:A) smaller than the momentum of the larger mass.B) There is not enough information to answer the question.C) equal to the momentum of the larger mass.D) zero.E) Larger than momentum of the larger mass.

Answers

Let the mass of smaller object be m and mass of larger object be M.

As the kinetic energies are equal,

[tex]\frac{p_m^2}{2m}=\frac{p^2_{_M}}{2M}[/tex]

On solving further,

[tex]\frac{p_m}{p_M}=\text{ }\sqrt[]{\frac{m}{M}}[/tex]

Hence, option A is correct.

Three non-parallel forces which can be represented both in magnitude and direction by the three sides of a triangle, taken in order, can: A. make a body oscillate B. make a body rotate C. keep a body in equilibrium D. move a body in a straight line E. always produce vertical motion of body​

Answers

Answer:

If the triangle representing the 3 non-colinear forces is closed

then then (C) the body is in equilibrium:

Adding vectors  A + B + C = 0 means that the resultant of adding the three vectors A, B, and C is zero

What conditions must be met in order for work to be done?

A. The applied force must make the object move.

B. The output force must be greater than the input force.

C. At least part of the applied force must be in the same direction as the movement of the object.

D. The work must be greater than the momentum.

Answers

At least part of the applied force must be in the same direction as the movement of the object must be met in order for work to be done.

What are the conditions to work?The following are the two prerequisites for working: To do the work, the body must be subjected to a force, or F 0. The body must move in the direction of the applied force, or S 0, as a result of the applied force.

There must be a force used. The displacement is the distance over which the force must act. The displacement must be a component of the force.

A legal term known as a condition precedent refers to an event that must occur before a certain contract is regarded as being in effect or before either party is obliged to fulfill any obligations.

Therefore, the correct answer is option C. At least part of the applied force must be in the same direction as the movement of the object.

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Every 5 seconds, the crest of a wave in the ocean travels 25 meters. What is the speed of the wave?

Answers

Given

Every 5 seconds

One wave travel 25m

Speed of waves

Explanation

[tex]\begin{gathered} v=\frac{25m}{5s} \\ v=5\text{ m/s} \end{gathered}[/tex]

The answer would be 5 m/s

A block of known mass hanging from an ideal spring of known spring constant is oscillating vertically. A motion detector records the position, velocity, and acceleration of the block as a function of time. Which of the following indicates the measured quantities that are sufficient to determine whether the net force exerted on the block equals the vector sum of the individual forces?A. Acceleration only B. Acceleration and position only C. Acceleration and velocity onlyD. Acceleration, position, and velocityPart 1. “Whether The net force exerted on the block equals the vector sum of the individual forces” really means “Newton’s Second Law”. The problem wants you to make measurements to show that the net force equals mass times acceleration. How would you find the force exerted by the spring? How do you find the force exerted by gravity? Part 2. Connect your answer to the previous question with the right answer. Clearly explain which quantities must be measured (between acceleration, velocity, and position) and explain what each quantity is used for to show Newton’s second law.

Answers

Answer:

B. Acceleration and position only

Explanation:

We need to identify the measurements that show that the net force is equal to the sum of the force exerted by the spring and the force of gravity, so we want to know if the following equation is satisfied

[tex]\begin{gathered} F_{net}=F_s-mg \\ ma=k(\Delta x)-mg \end{gathered}[/tex]

Where m is the mass, a is the acceleration, k is the spring constant, Δx is the stretched, and g is the gravity. The mass m, the spring constant k, and the gravity g are known. So, the measurement quantities that we need are the acceleration and the position.

So, the answer is

B. Acceleration and position only

Part 1.

How would you find the force exerted by the spring?

The force exerted by the spring is equal to k(Δx ), so to find Δx, we need to identify the position.

How do you find the force exerted by gravity?

The force exerted by gravity is calculated as mass times gravity, so it is known.

Part 2.

We need to measure Acceleration and position.

The acceleration to calculate the net force because by the second law of newton Fnet = ma

The position to calculate the force exerted by the spring.

A bungee jumper jumps off a bridge and bounces up and down several times.She finally comes to rest 30 m below the bridge from which she just jumped.If her mass is 50 kg and the spring constant of the bungee cord is 10 N/m,how much energy was lost due to air resistance while she was bouncing?(Recall that g = 9.8 m/s2)A. 9200 NB. 7330 Nc. 10,200 ND. 8605 N

Answers

Given data

*The given height is h = 30 m

*The given mass is m = 50 kg

*The spring constant of bungee cord is k = 10 N/m

*The value of the acceleration due to the gravity is g = 9.8 m/s^2

The net change in potential energy is calculated as

[tex]\begin{gathered} \Delta U_p=mgh \\ =(50)(9.8)(30) \\ =14700\text{ J} \end{gathered}[/tex]

The spring stretch is calculated by using the relation as

[tex]\begin{gathered} F=mg \\ kx=mg \\ x=\frac{mg}{k} \\ =\frac{(50)(9.8)}{10} \\ =49\text{ m} \end{gathered}[/tex]

The energy stored in spring is calculated as

[tex]\begin{gathered} U_s=\frac{1}{2}kx^2 \\ =\frac{1}{2}(10)(49)^2 \\ =12005 \end{gathered}[/tex]

The energy was lost due to air resistance while she was bouncing is calculated as

[tex]\begin{gathered} \Delta E=\Delta U_p-U_s_{}_{} \\ =14700-12005 \\ =2695\text{ J} \end{gathered}[/tex]

Hence, the energy was lost due to air resistance while she was bouncing is 2695 J

1. Consider the example problem, but with the lower pressure reduced to 100 Pa. How much work would be done in a single cycle?1. 29,970 J2. 0 J3. -30,030 J4. 27,000 J

Answers

Answer:

29970 J

Explanation:

Given:

[tex]\begin{gathered} P_A=P_B=10^5Pa \\ P_c=P_D=100\text{ Pa\lparen The new lower pressure\rparen} \\ V_A=V_D=0.1m^3 \\ V_B=V_C=0.4m^3 \end{gathered}[/tex]

Work done across AB

[tex]\begin{gathered} W_{AB}=P_{AB}(V_B-V_A) \\ \\ W_{AB}=10^5(0.4-0.1) \\ \\ W_{AB}=0.3\times10^5 \\ \\ W_{AB}=30000J \end{gathered}[/tex]

Work done across BC = 0 J (Since there is no change in volume)

Work done across DA = 0 J (Since there is no change in volume)

Work done across CD

[tex]\begin{gathered} W_{CD}=P_{CD}(V_D-V_C) \\ \\ W_{CD}=100(0.1-0.4) \\ \\ W_{CD}=100(-0.3) \\ \\ W_{CD}=-30J \end{gathered}[/tex]

Work done in one cycle:

[tex]\begin{gathered} W_{cycle}=W_{AB}+W_{BC}+W_{DA}+W_{CD} \\ \\ W_{cycle}=3000J+0+0-30 \\ \\ W_{cycle}=29970J \end{gathered}[/tex]

29970J of work will be done in a single cycle

Examine the heating curve for water below. Answer each question andcomplete the table to review your understanding of heating curves

Answers

Explanation:

When the temperature is increasing, the kinetic energy change, and when the graph is horizontal (the water is changing phases), the potential energy changes.

So at A, C, and E, there is a change in kinetic energy and at B and D there is a change in potential energy.

Then, we use a specific heat capacity of 4.2 J/g°C when the water is at A, C, and E. We use latent heat of 334 J/g for melting and latent heat of 2260 J/g for evaporation.

Answer:

Therefore, the complete table is

A 60.0 kg skier with an initial speed of 14 m/s coasts up a 2.50 m high rise

Find her final speed right at the top, in meters per second, given that the coefficient of friction between her skis and the snow is 0.38?

Answers

The final speed (in meters per second), given that the coefficient of friction between her skis and the snow is 0.38 is 14.65 m/s

How do I determine the final velocity?

First, we shall obtain the force. This can be obatined as follow:

Mass (m) = 60 KgAcceleration due to gravity (g) = 9.8 m/s²Normal reaction (N) = mg = 60 × 9.8 = 588 NCoefficient of kinetic friction (μK) = 0.38Force (F) = ?

F = μKN

F = 0.38 × 588

F = 223.44 N

Next, we shall obtain the acceleration of the skier. This can be obtained as follow:

Mass (m) = 60Force (F) = 223.44 NAcceleration (a) = ?

a = F / m

a = 223.44 / 61

a = 3.724 m/s²

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

Initial speed (u) = 14 m/sDistance (s) = 2.5 mDeceleration (a) = 3.724 m/s²Final speed (v) =?

v² = u² + 2as

v² = 14² + (2 × 3.724 × 50)

v² = 196 + 18.62

v² = 214.62

Take the square root of both sides

v = √214.62

v = 14.65 m/s

Thus, the the final speed is 14.65 m/s

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The final speed right at the top will be 14.65 m/s and the coefficient of friction between her skis and the snow is 0.38.

What is velocity?

The ratio between displacement to time is referred to as the velocity of the object. It is a vector quantity with SI unit meter per second (m/s).

According to the question, the given information is :

Skier's mass, m = 60 kg

The acceleration due to gravity, g = 9.8 m/s²

Initial Speed, u = 14 m/s

Distance, s = 2.5 m

Deceleration is given as, a = 3.724 m/s².

Normal reaction force (N) will be equal to mg.

N = 60 × 9.8

N = 588 N

The coefficient of kinetic friction is given as, μ = 0.38

By using the formula,

F = μKN

F = 0.38 × 588

F = 223.44 N

Now, let's calculate the acceleration of the skier :

f = ma

a = f/m

a = 223.44/60

a =  3.724 m/s².

Now, calculate the velocity,

v² = u² + 2as

v² = 14² + 2 (3.724) (50)

v² = 196 + 18.62

v² = 214.62

v = 14.65 m/s.

Hence, the final speed right at the top will be 14.65 m/s.

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which force is used when a rock fell

Answers

Answer:

When something falls, it falls because of gravity. Because that object feels a force, it accelerates, which means its velocity gets bigger and bigger as it falls. The strength with which the Earth pulls on something in the form of gravity is a type of acceleration. Earth pulls on everything the exact same amount. A rockfall is a type of fast-moving landslide that happens when rock or earth falls, bounces, or rolls from a cliff or down a very steep slope. Rockfalls start from high outcrops of hard, erosion-resistant rock that become unstable for a variety of reasons. When a rock is thrown upward, if we exempt the air drag then the only force that is acting on the rock is the gravitational force, also known as the weight of the rock. The acceleration due to gravity is always acting downward.

Answer: The force is Gravity

Explanation:

Gravity is what makes us stay to the ground and if a rock fell it would go to the ground just like if i where to jump i would land on the floor/ground  i wont be floating because of gavity.

Have a good day.

Multiple Choice: A car with a mass of 825 kg moves along the roadway at aspeed of 15 m/s to the east. What impulse is required to decrease the speed of theboat to 10 m/s east?

Answers

Given:

The mass of the car is m = 825 kg

The intial speed of the car is

[tex]v_i\text{ = 15 m/s}[/tex]

towards east.

The final speed of the car is

[tex]v_f\text{ = 10 m/s}[/tex]

To find the impulse of the car.

Explanation:

The impulse can be calculated by the formula

[tex]Impulse\text{ = mv}_f-mv_i[/tex]

On substituting the values, the impulse will be

[tex]\begin{gathered} Impulse\text{ = \lparen825}\times10\text{\rparen-\lparen825}\times15\text{\rparen} \\ =\text{ -4125 kg m/s} \end{gathered}[/tex]

The impulse will be 4125 kg m/s due

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