Andre really likes his new car, and he knows it has a certain amount of mechanical energy. Which types of energy are included in the mechanical
energy of the car? Choose the two that apply.
A. electrical energy from the battery
B. kinetic energy from any movement the car has
C. potential energy based on its position
D. thermal energy from when fuel burns in the engine

Answer:

The work done by you, W = 9800 J

Explanation:

Given data,

The displacement of the car, S = 5 m

Let the mass of the car be, m = 2000 kg

Let the coefficient of static friction be 0

Hence, no frictional force is acting on the tire and ground.

Let, you push down the car to provide more friction, μₓ = 0.1 (you can change the value)

In order to provide the static friction, you push down the car is equal to the static friction force,

                                      Fₓ =  μₓ · η

Where,

                                 η - normal force acting on the car (mg)

Substituting the values,

                                   Fₓ = 0.1 x 2000 x 9.8

                                       = 1960 N  

Therefore work done,

                                   W = Fₓ x S

                                        = 1960 N x 5 m

                                        = 9800 J

Hence, the work done by you, W = 9800 J

Answer:

Answer: •Currents occur at the surface of and deep within the ocean.

             •The Coriolis effect contributes to the direction of current flow.

             •Wind direction, the spin of Earth, and differences in temperature  

Explanation:

Answer:

Centripetal force is the force that is necessary to keep an object moving in a curved path and that is directed inward towards the center of rotation.

Explanation:

Definition of centripetal force:

Centripetal force is the force that is necessary to keep an object moving in a curved path and that is directed inward towards the center of rotation.

Example of centripetal force

A string on the end of which a stone is whirled about exerts a centripetal force on the stone.

The diagram is shown below

Where

The centripetal forces acting towards the centre C that is [tex]\vec {AC}[/tex]

and the direction is from A to C.

And the stone is moving in a circular motion with center as C.

Answer:

endothermic reactions

Answer:

endothermic reactions

Explanation:

endothermic reactions means the reaction gets super hot while the surrounding area gets colder. (absorbs energy)

exothermic means the surrounding area gets hot while it gets colder. (releases energy)

The power is 833.3 W

Explanation:

First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

[tex]W=(mg)h[/tex]

where

mg = 1250 N is the weight of the barbell

h = 2 m is the change in height

Substituting,

[tex]W=(1250)(2)=2500 J[/tex]

Now we can calculate the power, which is equal to the work done per unit time:

[tex]P=\frac{W}{t}[/tex]

where

W = 2500 J is the work done

t = 3 s is the time taken

Substituting,

[tex]P=\frac{2500}{3}=833.3 W[/tex]

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Answer:

30 kg

Explanation:

Newton's second law:

F = ma

60 N = m (2 m/s²)

m = 30 kg

Answer:

[tex]30\:g[/tex]

Explanation:

Isaac Newton's second law of motion

[tex]\displaystyle ma = F_{net}[/tex]

[tex]\displaystyle \frac{60}{2} = m \\ \\ 30 = m[/tex]

* Now, that bar above 18 is what is known as bar notation, indicating repetition in the digits.

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Answer:

13 to 19 years old

Explanation:

Because these 6 years there is the word TEEN 13, 14, 15, 16, 17, 18, and 19. but in 20 there is no teen.

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The mechanical advantage of the pulley is 3.56

Explanation:

The Mechanical Advantage (MA) of a pulley system is given by

[tex]MA=\frac{Load}{Effort}[/tex]

where

Load is the weight of the object lifted

Effort is the force applied in input

For the pulley in this problem, we have:

Effort = 55 N

While the load is the weight of the box of mass m = 20.0 kg:

[tex]Load = mg = (20.0 kg)(9.8 m/s^2)=196 N[/tex]

Substittuing, we find the MA of the pulley:

[tex]MA=\frac{196}{55}=3.56[/tex]

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Answer:

No you would not be an entrepreneur

Explanation:

An entrepreneur is someone who creates a business, and also organize the business affairs. The entrepreneur takes on the biggest financial risk.

Answer:

No, taking a job with Amazon does not make one an entrepreneur

Explanation:

An entrepreneur is an individual that sets up a business or businesses (with the aim of making profit). In any organization, the person or persons that start up the business or businesses are the entrepreneurs while the people employed are the employee of the business/businesses.

Entrepreneurs usually create jobs for the employee and not the other way round (even when the idea could come from the employee).

From the above, it can be deduced that a person who gets a job at Amazon is an employee while the owner of Amazon is the entrepreneur.

The final velocity of the rock is 19.8 m/s

Explanation:

The motion of the rock is a free fall motion (subjected only to the force of gravity), with constant acceleration [tex]g=9.8 m/s^2[/tex] towards the ground. Therefore, we can use the following suvat equation:

[tex]v^2-u^2=2as[/tex]

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the vertical displacement

For the rock in this problem,

u = 0 (initial velocity is zero)

[tex]a=g=9.8 m/s^2[/tex]

s = 20 m (height of the cliff)

Solving for v,

[tex]v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(20)}=19.8 m/s[/tex]

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Answer:

The speed of the block after it has moved 3M if the surface in contact have a coefficient of kinetic friction of 0.15​  is 1.7s m/sec

Explanation:

Given:

mass of the block = 6.0 kg

Force with which the block is pulled = 12 N

Kinetic friction \mu= 0.15

Distance travelled  s = 3 m

To Find:

speed of the block after it has moved 3 metres =?

Solution :

W know that the friction formula is

[tex]f_k = \mu m g[/tex]

Substituting the values,

[tex]f_k = (0.15)(6)(10)[/tex]

[tex]f_k= 9 N[/tex]

Now Acceleration is Given by

[tex]a=\frac{F -f_k}{m}[/tex]

[tex]a=\frac{12 - 9}{60}[/tex]

[tex]a=\frac{3}{6}[/tex]

a= 0.5 [tex]m/s^2[/tex]

Initial velocity is u = 0

Also we know that,

[tex]v^2 - u^2=2as[/tex]

So the equation becomes

[tex]v^2 =2as[/tex]

[tex]v=\sqrt{2as}[/tex]

Substituting the values,

[tex]v=\sqrt{2as}[/tex]

[tex]v=\sqrt{2(0.5)(3)}[/tex]

[tex]v= \sqrt{3}[/tex]

v= 1.73 m/s

Answer:

D. 0.29

Explanation:

The final velocity of the candy is 57.7 m/s

Explanation:

The motion of the candy is a free fall motion, since it is subjected only to the force of gravity, so it is a uniformly accelerated motion and therefore we can use the following suvat equation:

[tex]v^2-u^2=2as[/tex]

where

v is the final velocity

u is the initial velocity

[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity

s is the vertical displacement

In this problem, we have:

u = 0 (the candy is dropped from rest)

s = 170 m (the vertical displacement is the height of thr monument)

Solving for v, we find the velocity of the candy as it hits the  ground:

[tex]v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(170)}=57.7 m/s[/tex]

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The work done is 6958 J

Explanation:

The work done by the man to raise the weight is the equal to the increase in gravitational potential energy of the weight, so we can write:

[tex]W=mg \Delta h[/tex]

where

(mg) is the weight, with m being the mass and g the acceleration of gravity

[tex]\Delta h[/tex] is the change in height of the weight

In this problem, we have

mg = 710 N (weight)

[tex]\Delta h = 0.65 m[/tex]

Therefore, the work done is

[tex]W=(710)(9.8)=6,958 J[/tex]

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The work done is 461.5 J.

Explanation:

According to Kinetics, the work done is calculated as the product of force which has done work, and the displacement it has imposed. Hence,

                     [tex]\text {Work done}=F \times s[/tex]

In the problem, the given data, F = 710 N and s = 0.65 m

By applying the given values to the work done equation, we get as follows,

                     [tex]\text { Work done } = 710 \times 0.65 = 461.5 \mathrm{J}[/tex]

The unit is same as the energy, work done defines when 1 N force acts in a 1 m distance along the force’s direction.

Answer: Acted on by equal forces in opposite direction

Explanation:

Newton's First Law says that every body continue in its state of rest or constant speed on a straight line unless being acted upon by an external force.

Answer:

Acted on by equal forces in opposite direction

Explanation:

Answer:

The distance the hiker walked is, d = 13 km

The displacement of the hiker is, S = 9.4 km

Explanation:

Given data,

The displacement of towards east, d₁ = 1 km

The displacement of towards north, d₂ = 2 km

The displacement of towards east, d₃ = 4 km

The displacement of towards north, d₄ = 6 km

The total distance the hiker walked

                           d = d₁ + d₂ + d₃ + d₄

                              = 1 + 2 + 4 + 6

                              = 13 km

The distance the hiker walked is, d = 13 km

The resultant displacement of the hiker, S

                                S = √( A² + B² + 2 A B cosФ)

Where,

                         A = d₁ + d₃ = 5 km

                         B = d₂ + d₄ = 8 km

                         Ф = angle between A and B = 90°

Substituting in the displacement equation

                          S = √( 5² + 8²)

                              = 9.4 km

Hence, the displacement of the hiker is, S = 9.4 km

The new gravitation force at the new location is 40 N

Explanation:

The weight of the astronaut is given by the equation

[tex]F=mg[/tex] (1)

where

m is the mass of the astronaut

g is the acceleration of gravity

The acceleration of gravity at a certain distance [tex]r[/tex] from the centre of the Earth is given by

[tex]g=\frac{GM}{r^2}[/tex]

where G is the gravitational constant and M is the Earth's mass. So we can rewrite eq.(1) as

[tex]F=\frac{GMm}{r^2}[/tex]

When the astronaut is on the Earth's surface, [tex]r=R[/tex] (where R is the Earth's radius), so his weight is

[tex]F=\frac{GMm}{R^2}=640 N[/tex]

Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is

[tex]r'=3R+R=4R[/tex]

Therefore, the new weight is

[tex]F'=\frac{GMm}{(4R)^2}=\frac{1}{16}\frac{GMm}{R^2}=\frac{F}{16}[/tex]

Which means that his weight has decreased by a factor 16: therefore, the new weight is

[tex]F'=\frac{640}{16}=40 N[/tex]

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Final answer:

The gravitational force on the astronaut 3 times farther from the Earth's surface than the radius would be 40 N, as at this distance the gravitational force is reduced to 1/16th of its original value.

Explanation:

The question asks: If an astronaut weighs 640 N on the earth's surface, what is the gravitation force between him and the earth if he is 3 times the distance from the Earth's surface? This requires understanding of Newton's law of universal gravitation, which can be expressed as F = G(m1m2)/r², where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass.

As the astronaut moves to a distance 3 times farther from the surface of the Earth, the distance (r) from the center of the Earth becomes 4r, because the initial distance from the surface to the center of the Earth (the radius of the Earth, or 1r) is included. According to the inverse-square law, if the distance increases by a factor of n, the force decreases by a factor of n2. Thus, at 3 times the distance from the surface, or 4 times the radius of the Earth, the gravitational force becomes 1/16th of what it was at the surface.

Therefore, the new gravitational force is 640 N / 16 = 40 N.

Answer:

Radiation is the answer to your question

Answer:

The acceleration of the car is [tex]1.432 m/s^{2}[/tex]

Explanation:

We are assuming rightward coordinate positive and all quantities are along this direction

We know,

[tex]a = \frac{dv}{dt}[/tex]

where a - acceleration, v=velocity, t=time and x=displacement

multiply by dx in both sides

[tex]adx = \frac{dv}{dt} dx[/tex]

but we know [tex]\frac{dx}{dt} = v[/tex]

Therefore,

[tex]adx = vdv[/tex]

Here we integrate both sides with proper limits

x ranges from 0 to 110 as v ranges from 29 to 34

p = 0, r = 110, q = 29, s = 34

[tex]a\int\limits^r_pdx = \int\limits^s_q {v} \, dv[/tex]

a is given as constant thus can be pulled out of the integration

[tex]a[110-0] = [\frac{34^{2}}{2}- \frac{29^{2}}{2}][/tex]

Therefore,

Accelaration of the car is [tex]1.432 m/s^{2}[/tex]

Note:

Here moving to the right doesn't mean anything significant other than the fact that all quantities are pointing in that direction. Therefore obtained acceleration is also towards the right

If you know equation of motion for constant acceleration as

[tex]v^{2} -u^{2} = 2as[/tex] you can plug in values in this equation to obtain value of a

v - final velocity

u - initial velocity

s - displacement

Final answer:

The appearance of a ship's top first as it approaches the horizon shows the Earth's curvature, as the lower parts disappear first due to the spherical shape.

Explanation:

When a ship comes into view on the horizon, the top appears before the rest of the ship, demonstrating that the Earth is spherical. This is because as the ship sails away from an observer, the curvature of the Earth causes the lower parts of the ship to disappear from view first, much like it would drop behind a hill. This phenomenon, which mariners like Columbus would have been familiar with, is a direct consequence of the Earth's spherical shape.

The higher masts of the ship remain visible for a while longer because they are the last to dip below the horizon as the ship follows the Earth's curvature. This effect proves that the Earth is not flat because if it were, the entire ship would simply appear smaller but would stay in full view as it moves away.

Moreover, the use of lookouts in the ship's mastheads also illustrates the Earth's curvature. On a spherical Earth, lookouts posted higher up can see further over the horizon than those at deck level. Conversely, if the Earth were flat, there would be little to no advantage of having lookouts at higher positions.

The current through a 25 ohm resistor connected to a 5.0 V power supply is 0.2 A. Option A is correct.

Ohm's Law:

The voltage of the two points between in a conductor is directly proportional to the current flows through the conductor.

V = IR

Where,

V = voltage  = 5.0 V

I = current  = ?

R = resistance -  25  ohm

Put the values in the equation,

5 = 25 x I

I = 0.2 A

Therefore,  the current through a 25 ohm resistor connected to a 5.0 V power supply is 0.2 A.

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900 grams of lead can be added max to the tin so that it does not sink.

Explanation:

According to the Archimedes principle, an object will not sink in a medium until the density of the object exceeds the density of the medium.

The medium we have is water.

The density of medium(water) is given as ρ° = 1 g/cm³

Density of an object is simply its mass contained per unit volume.

Density = mass/volume

Volume of the tin = v = 1000 cm³

Mass of the tin = m = 100 g

Mass of lead that can be added without sinking = x

Total mass after adding lead in the tin = Mt = m + x

Total density after lead is added to the tin is given as:

ρt = (m+x)/v

Now, according to Archimedes principle, lead can be added to the tin until the density of tin is equal to the density of water.

ρt = ρ°

Mt/v = ρ°

(m + x)/v = ρ°

⇒ (100 + x)/1000 = 1

⇒ 100 + x = 1000

⇒ x = 1000 - 100

x = 900 g

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Answer:

I think D.

Injured cells stimulate pain, swelling, redness, and heat- which initiates bradykinin-which stimulates the release of histamine- which increases neutrophils and macrophages- leads to phagocytosis of foreign invaders.

Answer:

Injured cells stimulate bradykinin- which initiates the release of histamine- which causes pain, swelling, redness, & heat- increases neutrophils & macrophages- leads to phagocytosis of foreign invaders

Explanation:

Took the test

1) The ball's speed is 3 m/s

2) The final speed of the raft is 0.6 m/s

3) The collision is inelastic

Explanation:

1)

The momentum of an object is given by

[tex]p=mv[/tex]

where

p is the momentum

m is the mass of the object

v is its velocity

For the ball in this problem we have:

p = 12 kg m/s

m = 4 kg

Solving for v, we find its velocity (and so its speed):

[tex]v=\frac{p}{m}=\frac{12}{4}=3 m/s[/tex]

2)

We can solve this part by applying the law of conservation of momentum: in fact, the total momentum of an isolated system (=no external forces) must be conserved. Therefore we can write:

[tex]p_i = p_f[/tex] (1)

where

[tex]p_i = 0[/tex] is the total initial momentum (the swimmer and the raft are at rest at the beginning)

[tex]p_f = mv + MV[/tex] is the total final momentum, where

m = 75 kg is the mass of the swimmer

M = 500 kg is the mass of the raft

v = 4 m/s is the final velocity of the swimmer

V is the final velocity of the raft

And substituting into (1) we find:

[tex]0=mv+MV\\V=-\frac{mv}{M}=-\frac{(75)(4)}{500}=-0.6 m/s[/tex]

Where the negative sign indicates that the raft moves in the opposite direction to the swimmer: so, the speed of the raft is 0.6 m/s.

3)

In a collision between two objects, if the system is isolated the total momentum of the system is always conserved during the collision. However, this is not true for the total kinetic energy: in fact, due to the presence of internal frictions, part of the kinetic energy can be converted into thermal energy or other forms of energy.

Therefore, there are two types of collision:

- Elastic collision: in an elastic collision, also the total kinetic energy of the objects is conserved

- Inelastic collision: in an inelastic collision, the total kinetic energy is not conserved. The most extreme case is the perfectly inelastic collision, in which the two objects stick together after the collision, and in this case there is the maximum loss of kinetic energy.

Since in this problem the two objects stick together, the collision is inelastic.

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Explanation:

The cross section per unit length is uniform, so ρ is constant.

The center of mass is therefore:

x_avg = (∫₀ᴸ x ρ dL) / (∫₀ᴸ ρ dL)

x_avg = (∫₀ᴸ x dL) / (∫₀ᴸ dL)

x_avg = (½ L²) / (L)

x_avg = ½ L

True: when there is an angle between force and displacement, [tex]W=Fd cos \theta[/tex]

Explanation:

The work done by a force when pushing/pulling an object is given by the equation

[tex]W=Fd cos \theta[/tex]

where

F is the magnitude of the force

d is the displacement of the object

[tex]\theta[/tex] is the angle between the direction of the force and of the displacement

We have two extreme cases:

- When the force is parallel to the displacement, then [tex]\theta=0^{\circ}[/tex] and [tex]cos \theta = 1[/tex], so the work done is maximum and simply becomes

[tex]W=Fd[/tex]

- When the force is perpendicular to the displacement, then [tex]\theta=90^{\circ}[/tex] and [tex]cos \theta =0[/tex], so the work done is zero.

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The acceleration of the ball after leaving the hand is [tex]9.8 m/s^2[/tex] downward

Explanation:

In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.

After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is

[tex]F=mg[/tex]

where m is the mass of the ball and g is the acceleration of gravity ([tex]g=9.8 m/s^2[/tex]), acting in the downward direction.

According to Newton's second law, the acceleration of the ball is given by

[tex]a=\frac{\sum F}{m}[/tex]

where

[tex]\sum F[/tex] is the net force acting on the ball

After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:

[tex]a=\frac{mg}{m}=g=9.8 m/s^2[/tex]

This means that the acceleration of the ball remains [tex]9.8 m/s^2[/tex] downward for the entire motion, after leaving the hand.

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Final answer:

The acceleration of the ball after it is thrown upward is a constant 9.8 m/s^2 downward due to gravity.

Explanation:

After you throw a ball upward with a speed of 14m/s, the acceleration of the ball after it leaves your hand is determined by the force of gravity. Regardless of the initial speed given to the ball, once it is not in contact with your hand, there is no longer any force being applied to it in the upward direction. Therefore, the only force acting on it is the force of gravity pulling it back toward the Earth's surface.

This force causes the ball to have a constant downward acceleration of approximately 9.8 m/s2, which is the standard acceleration due to gravity near the Earth's surface (often represented by the symbol g).

The mass of the ball is 6.67 kg

Explanation:

The momentum of an object is given by:

[tex]p=mv[/tex]

where

p is the momentum

m is the mass of the object

v is its velocity

It must be noted that momentum is a vector quantity, so it has both magnitude and direction (but here only the magnitude is given).

For the ball in this problem, we have:

v = 1.5 m/s is its velocity

p = 10.0 kg m/s is its momentum

Solving the formula for m, we find the mass of the ball:

[tex]m=\frac{p}{v}=\frac{10}{1.5}=6.67 kg[/tex]

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Answer:

2100 J

Explanation:

Work = force × distance

W = Fd

W = (140 N) (15 m)

W = 2100 J

Answer:

[tex]\displaystyle 2100\:J[/tex]

Explanation:

[tex]\displaystyle FD = W → 2100 = [15][140][/tex]

Force by Distance

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