If all of the forces acting on an object are balanced, then:
Check all that apply.
I
A. the direction the object is moving in will not change.
B. the acceleration of the object will be 0 m/s2
I
c. the object will not be in motion.
D. the velocity of the object will be 0 m/s.

Answer:

slower

Explanation:

Due to the fact that it is falling, the object experiences an acceleration due to gravity, so this acceleration will make the car go faster every moment, so the speed at the top will be lower than the speed at the bottom.

The situation can also be thought of as a transaction of potential energy (energy due to the mass of an object and its position on the ground) and kinetic energy (energy due to movement). At the top is the point where the object has the greatest potential energy, and as it moves down this potential energy is exchanged for kinetic energy. The kinetic energy increases while going down the ramp, so it will have greater speed as it falls.

The answer is: the speed at the top of the ramp will be slower than the speed at the bottom.

The average speed of the car is 28 mph

Explanation:

The average speed of an object is equal to the ratio between the total distance covered by the object (regardless of its direction) and the time taken. Therefore:

[tex]speed = \frac{d}{t}[/tex]

where

d is the total distance

t is the time taken

The car in this problem travels 5 miles north and 2 miles south, so the total distance covered is

d = 5 + 2 = 7 miles

While the time taken is

t = 1/4 h = 0.25 h

Therefore, the average speed is

[tex]speed = \frac{7}{0.25}=28 mph[/tex]

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The minimum power is 490 W

Explanation:

The work that needs to be done by the robot in order to climb the shelves is equal to the gain in gravitational potential energy of the robot + object, therefore:

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

where

m = 15 kg + 5 kg = 20 kg is the total mass of the system

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

[tex]\Delta h = 20 m[/tex] is the change in height

Substituting,

[tex]W=(20)(9.8)(20)=3,920 J[/tex]

Now we can calcualte the power (wattage) that the robot must have, using the equation

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

where

W = 3,920 J is the work done

t = 8 s is the time interval

Substituting,

[tex]P=\frac{3920}{8}=490 W[/tex]

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1) The initial height of the roller coaster must be 5.1 m

2) The height of the roller coeaster must be 5.5 m

Explanation:

1)

We can solve this problem by using the law of conservation of energy.

In fact, in absence of frictional force, the total mechanical energy must be conserved, so we can write:

[tex]KE+PE = KE'+PE'[/tex]

where

KE = 0 is the initial kinetic energy, at the top (the car starts from rest, so it has zero speed and so zero kinetic energy)

PE = mgh is the initial potential energy, where

m = 8000 kg is the mass of the car

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

h is the initial height

[tex]KE'=\frac{1}{2}mv'^2[/tex] is the kinetic energy of the car at the bottom, where

v' = 10 m/s is the speed of the car at the bottom

PE' = 0 is the potential energy at the bottom (where h = 0)

Re-writing the equation with all the values and solving for h, we find the initial height of the track:

[tex]h=\frac{v^2}{2g}=\frac{10^2}{2(9.8)}=5.1 m[/tex]

2)

In this case, we also have to take into account the energy lost due to friction. So the equation of conservation of energy becomes

[tex]KE+PE = KE'+PE'+W[/tex]

where

W = 32,000 J is the work done by friction

Re-arranging the equation and solving again for h, we find:

[tex]mgh = \frac{1}{2}mv'^2 + W\\h=\frac{v^2}{2g}+\frac{W}{mg}=5.1+\frac{32,000}{(8000)(9.8)}=5.5 m[/tex]

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After the initial push, the rock will keep moving forever at constant velocity (constant speed in a straight line)

Explanation:

We can answer this question by using Newton's first law of motion:

"An object at rest (or in motion at constant velocity) will stay at rest (or will keep moving at constant velocity) unless acted upon unbalanced forces" (Law of inertia)

In this problem, we have a rock in a place very far from any force that can act on it. This means that there are no unbalanced force acting on it, so the rock will keep its state of motion forever.

In this situation, the rock is initially thrown by the astronaut. After the initial push, which accelerates the rock up to a certain velocity, there will be no more forces acting on the rock. This means that the rock will continue moving at a constant velocity forever, so at a constant speed in a straight line.

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Answer:

Kinetic Energy

Explanation:

In relation to the question, as the sheriff rides a horse, the sheriff is in motion and a body in motion is said top possess kinetic energy.

Kinetic energy is that energy of the body that it gains due to the motion or speed of the body which it gains and maintains.

In other words, we can say that the amount of work needed to accelerate an object from its position of rest and set it into motion at some certain velocity.

Answer:

C

Explanation:

Explanation:

Use the Ideal Gas Law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is the universal gas constant,

and T is absolute temperature.

First, identify which variables in the problem are constant (note that R is always a constant).

Next, solve for the constants and use that equation to write a proportion.

Finally, plug in the given values and solve for the unknown variable.

For example, if n and R are the constants:

nR = PV/T

Therefore:

P₁V₁/T₁ = P₂V₂/T₂

The magnitude of the air drag is 784 N

Explanation:

An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:

[tex]F_D = mg[/tex]

where

[tex]F_D[/tex] is the magnitude of the air drag

m is the mass of the object

g is the acceleration of gravity

In this problem, we have

m = 80 kg is the mass of the airman

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

Substituting into the formula, we find:

[tex]F_D = (80)(9.8)=784 N[/tex]

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Answer:

(C) The positive charge of the alpha particles and the negative charge of the electrons  are the properties Rutherford used in his experiment

Explanation:

In his scattering experiment, scientist Ernest Rutherford used the property of positively charged alpha particles and negatively charge electrons. He performed this experiment by passing some of the alpha particles through a gold foil. The result was that the some alpha particles scattered while some passed through the gold foil without collision.

He concluded that the alpha particles are centrally positively charged and needed a large amount repelling force. This experiment of Rutherford is also known as Rutherford model of atom. This experiment helped him in doing so many other discoveries.

-271.96 °C

We are given;

we are required to calculate the temperature of the gas.

Moles = Mass ÷ Molar mass

Molar mass of nitrogen gas = 28.0 g/mol

Therefore;

Moles of N₂ = 92.5 g ÷ 28.0 g/mol

                   = 3.304 moles

According to the ideal gas equation;

PV = nRT , where n is the number of moles and R is the ideal gas constant, 0.082057 L.atm/mol.K

Rearranging the formula;

T = PV ÷ nR

  = ( 5.17 atm × 0.0625 L) ÷ (3.304 moles × 0.082057)

  = 1.19 K

But, °C = K - 273.15

Therefore;

T = 1.19 K - 273.15

  = -271.96 °C

Thus, the temperature of the gas will be -271.96 °C

Answer:

The combination of foods and beverages that constitute an individual’s complete dietary intake over time is known as Eating pattern.

So, option A is the correct answer.

Explanation:

The Eating pattern  is also called Dietary Pattern. Normal eating pattern may be three meals a day, may increased or changed according to the individual, and may refer to overeating.

Dietary guidelines are nutritional guidelines which advice an individual, greater than 2 years, consume a healthy and nutritional diet.

Mediterranean diet is the diet typical to Mediterranean countries, in which there is high consumption of vegetables, fruits, seeds, legumes, nuts and olive. It consists of moderate amounts of carbohydrates and proteins and a moderate to high amount of fats.

A healthy intake is a healthy diet in which the energy from sugars is less than 10% of the total energy intake.

Hence, Option A is the correct answer.

Answer:

[tex]3.9m/s^{2}[/tex]

Explanation:

Using second law of motion

[tex]a =\frac {m1 * g - \frac {T}{r}}{m1 + 0.5 * m2}[/tex] where m1 is mass of block, m2 is mass of flywheel, g is acceleration due to gravity whose value is taken as [tex]9.81 m/s^{2}[/tex], T is torque and r is radius

Substituting 5.5 Kg for m1, 13 Kg for m2, 0.33 m for r, 2.5 Nm for T we obtain

[tex]a = \frac {5.5 \times 9.81 - \frac {2.5}{0.33}}{(5.5 + 0.5 \times13)}=3.9m/s^{2}[/tex]

To find the acceleration of the 5.5 kg block, we consider both gravitational force and frictional torque acting on a 13 kg flywheel. The calculated acceleration of the block is approximately 2.73 m/s².

To determine the magnitude of the acceleration of the 5.5 kg block as it falls, we need to consider several forces and torques involved in the system. These include the gravitational force on the block, the tension in the rope, the rotational inertia of the flywheel, and the frictional torque opposing the motion.

First, let’s identify the key forces:

The rotational inertia (I) of the flywheel (a disk) is given by:

I = 0.5 * mflywheel * r² = 0.5 * 13 kg * (0.33 m)² = 0.70785 kg·m²

Applying Newton’s second law for rotation, we have:

∑τ = I * α → τtension - τfriction = I * α

Here, τtension is the torque due to the tension in the rope, which is equal to T * r. Hence, we have:

T * r - τfriction = I * α

Also, the linear acceleration (a) of the block is related to the angular acceleration (α) of the flywheel by the equation:

a = α * r

Combining these equations, we get:

T * r - τfriction = I * (a / r)

Solving for T in terms of a:

T = (I * a / r²) + (τfriction / r)

Newton’s second law for the falling block gives us:

mblock * g - T = mblock * a

Substitute T from the earlier equation into this one and solve for a:

mblock * g - [(I * a / r²) + (τfriction / r)] = mblock * a

Rearranging terms to isolate a:

a = [mblock * g - (τfriction / r)] / [mblock + (I / r²)]

Substitute the known values:

a = [5.5 kg * 9.8 m/s² - (2.5 Nm / 0.33 m)] / [5.5 kg + (0.70785 kg·m² / (0.33 m)²)]

a ≈ 2.73 m/s²

Thus, the magnitude of the acceleration of the block as it falls is approximately 2.73 m/s².

1) The acceleration is [tex]0.96 m/s^2[/tex]

2) The time taken is 39.6 s

Explanation:

1)

Since the motion of the truck is a uniformly accelerated motion (=constant acceleration), 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 distance covered

For the truck in this problem,

u = 0 (it starts from rest)

v = 38 m/s

s = 755 m

Solving for a, we find the acceleration:

[tex]a=\frac{v^2-u^2}{2s}=\frac{38^2-0}{2(755)}=0.96 m/s^2[/tex]

2)

For this part we can use the following suvat equation

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

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken for the velocity to change from u to v

In this problem,

u = 0

v = 38 m/s

[tex]a=0.96 m/s^2[/tex]

Solving for t,

[tex]t=\frac{v-u}{a}=\frac{38-0}{0.96}=39.6 s[/tex]

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The truck's acceleration, starting from rest to a velocity of 38 m/s over 755 m, is approximately 0.956 m/s². It takes about 39.75 seconds to cover this distance.

a. To determine the acceleration of the truck, we can use the kinematic equation:

v^2 = u^2 + 2as

where:

v = final velocity (38 m/s),

u = initial velocity (0 m/s since the truck was initially at rest),

a = acceleration,

s = displacement (755 m).

Rearranging the equation to solve for acceleration (a), we get:

a = (v^2 - u^2) / (2s)

Substitute the given values:

a = (38^2 - 0^2) / (2 * 755)

a = 1444 / 1510

a ≈ 0.956 m/s^2

b. To find the time taken (t), we can use the kinematic equation:

v = u + at

Since the truck started from rest (u = 0), this simplifies to:

t = v / a

Substitute the values:

t = 38 / 0.956

t ≈ 39.75 s

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The question probable may be:

A Truck accelerates to a velocity of 38 m/s over 755 m of road.

a. If the truck was initially at rest, what was its acceleration?

b. How long willthe truck take to travel this distance?

Renewable resources can be replaced as quickly as they are used, while non-renewable resources are being used up faster than they can be made by nature.

Renewable resources are those that can be replaced by natural processes as quickly as humans use them. Examples include solar power, wind power, and geothermal power. On the other hand, non-renewable resources are consumed or used up faster than they can be made by nature. Examples include fossil fuels such as coal, natural gas, and oil. These resources take millions of years to form and are being used up at a much faster rate than they can be replenished.

The object on Earth has greater mass

Explanation:

The weight of an object is given by

[tex]W=mg[/tex]

where

W is the weight

m is the mass of the object

g is the strength of the gravitational field at the location of the object

The strength of the gravitational field on Jupiter ([tex]g_J[/tex]) is much larger than that on Earth ([tex]g_E[/tex]), so we can write

[tex]g_J > g_E[/tex]

The weight of the first object on Jupiter is

[tex]W_J = m_1 g_J = 50 N[/tex] (1)

where [tex]m_1[/tex] is the mass of the first object, while the weight of the second object on Earth is

[tex]W_E = m_2 g_E = 50 N[/tex] (1)

while [tex]m_2[/tex] is the mass of the second object.

By dividing eq.(1) by (2), we get

[tex]\frac{m_1 g_J}{m_2 g_E}=1\\\frac{m_1}{m_2}=\frac{g_E}{g_J}[/tex]

And since [tex]g_J > g_E[/tex], this means that [tex]m_2 > m_1[/tex], so the object on Earth has greater mass.

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Answer:

it will cover 195km in 3 hours

The work done by the brakes is [tex]-3.0\cdot 10^5 J[/tex]

Explanation:

According to the work-energy theorem, the work done by the brakes on the car is equal to the change in kinetic energy of the car. Therefore:

[tex]W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]

where

W is the work done

m is the mass of the car

v is the final speed

u is the initial speed

For the car in this problem, we have:

m = 1500 kg

u = 20 m/s

v = 0 (the car comes to a halt)

Substituting, we find the work done:

[tex]W=0-\frac{1}{2}(1500)(20)^2=-3.0\cdot 10^5 J[/tex]

And the work is negative because the brakes apply a force in the opposite direction to the motion of the car.

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Answer: equal in magnitude but opposite in direction to the force that it exerts.

Explanation:

Answer:

The second object exert a force that is equal in magnitude and opposite in direction.

Explanation:

Newton's third law of motion states that for every action there is a reaction which is equal in magnitude to the action but acts in opposite directions as the action.

The magnitude of the electric field must be [tex]5.59\cdot 10^{-11} N/C[/tex]

Explanation:

In order for the electron to be in equilibrium, the force of gravity acting on the electron must be equal to the force due to the electric field.

The force of gravity on the electron located on the Earth's surface is:

[tex]F_G = mg[/tex]

where

[tex]m=9.11\cdot 10^{-31} kg[/tex] is the electron mass

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

The force due to the electric field is

[tex]F_E = qE[/tex]

where

[tex]q=1.6\cdot 10^{-16}C[/tex] is the electron charge

E is the magnitude of the electric field

Since the two forces must be balanced,

[tex]F_G = F_E[/tex]

So we find:

[tex]mg=qE\\E=\frac{mg}{q}=\frac{(9.11\cdot 10^{-31})(9.81)}{1.6\cdot 10^{-19}}=5.59\cdot 10^{-11} N/C[/tex]

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Option - B and Option - C are correct.

We have Andre's new car.

We have to determine what types of Energies are included in the Mechanical energies of the car.

The energy possessed by an object due to its motion or its position is called Mechanical Energy.

According to the question -

The Mechanical energy of the new car would include the following two types of energy -

       [tex]$E(K) = \frac{1}{2} mv^{2}[/tex]

        U(h) = mgh

Hence, Option B and C are correct.

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Answer:

Action - Pulling up the train.

Reaction - Friction on the locomotive

Explanation:

Locomotive is pulling the train upwards ,

Which is the action force applied by the locomotive,

As a reaction locomotive will be pulled by the train which is the reaction of pulling

Now, considering it as a action on locomotive , friction force will act on it as a reaction upwards which will result to move it upwards.

For train action is pulling up by locomotive and reaction will be friction acting on it downwards.

The weight of the object is 679.1 N

Explanation:

The weight of an object is given by:

[tex]W=mg[/tex]

where

W is the weight

m is the mass of the object

g is the acceleration of gravity

For an object near the Earth's surface, the acceleration of gravity is

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

The mass of the object in this problem is

m = 69.3 kg

Therefore, its weight is

[tex]W=(69.3)(9.8)=679.1 N[/tex]

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1) Yes, it is a conservative force

2) The potential energy at (5m, 5m) is 20 J

Explanation:

1)

A force is defined to be conservative if the work done by the force when moving an object does not depend on the path taken, but only on the initial and final position of the object.

Let's verify if this condition is met for the force in this problem:

- For going from (0 m, 0 m) to (5 m, 5 m), the work done is 20 J (A)

Then we can go from (0 m, 0 m) to (5 m, 5 m) through a different path:

- from (0 m, 0 m) to (5 m, 0 m) (work done: 25 J)

- from (5 m, 0 m) to (5 m, 5 m) (work done: -5 J)

Total work done in the second path: 25 J + (-5 J) = 20 J --> same as (A)

We can also go from (0 m, 0 m) to (5 m, 5 m) through a different path:

- from (0 m, 0 m) to (0 m, 5 m) (work done: 35 J)

- from (0 m, 5 m) to (5 m, 5 m) (work done: -15 J)

Total work done in the third path: 35 J + (-15 J) = 20 J --> same as (A)

So, the force is conservative, since the work done does not depend on the path taken.

2)

For a conservative force, the change in potential energy of the object moved by the force is equal to the work done by the force in moving the object from A to B.

Here have:

Point A: (0m, 0m)

Point B: (5m, 5m)

So, the change in potential energy is equal to the work done from A to B:

[tex]W=\Delta U = U_B-U_A[/tex]

where

[tex]U_B[/tex] is the potential energy in point B

[tex]U_A[/tex] is the potential energy in point A

In this problem, the potential energy at the origin (point A) is zero, so

[tex]U_A = 0[/tex]

Also we know that the work done is

W = 20 J

So, we find

[tex]U_B = W+U_A = 20 + 0 = 20 J[/tex]

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The force exerted is not conservative as different amounts of work are being done to move the particle along different paths between the same points. Potential energy isn't defined in the usual sense because the force isn't conservative, but if taken as the average of the works done, the potential energy at (5,5) could be taken as 20J.

A conservative force is defined as one in which the work done on a particle by the force as the particle moves from one point to another is independent of the path taken. Specifically, in the case of a conservative force, the work done is only dependent on the initial and final positions. In the given scenario, different amounts of work are being done to move the particle along different paths between the same points. This indicates that the force is not a conservative one.

The potential energy at a point in a field due to a conservative force is equal to the work done by the force to move the particle from a reference point (in this case, the origin point) to that point. Since the force in this scenario isn't conservative, potential energy isn't defined in the usual sense. Nevertheless, if we insist on measuring it, we could take the average of the works done: for instance, between (0,0) and (5,5) the works done are 35+(-15)=20 and -5+25=20, so we could say the potential energy at (5,5) is 20J considering the origin as the reference point with zero potential energy.

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The distance covered is 54.0 m

Explanation:

Since the motion of the car is a uniformly accelerated motion, we can use the following suvat equation:

[tex]s=ut+\frac{1}{2}at^2[/tex]

where

u is the initial velocity

t is the time

a is the acceleration

s is the distance covered

For the car in this problem, we have

u = 0 (it starts from rest)

[tex]a=7.1 m/s^2[/tex] is the acceleration

t = 3.9 s is the time

Substituting, we find s:

[tex]s=0+\frac{1}{2}(7.1)(3.9)^2=54.0 m[/tex]

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The time taken is approximately 26 minutes

Explanation:

The motion of the body in this problem is a uniform motion (= at constant velocity), therefore we can use the following equation:

[tex]v=\frac{d}{t}[/tex]

where

v is the speed

d is the distance covered

t is the time taken

In this problem, we have

v = 75 km/h is the speed

d = 32 km is the distance to be covered

Solving for t, we find the time needed:

[tex]t=\frac{d}{v}=\frac{32}{75}=0.43 h[/tex]

Converting into minutes,

[tex]t=0.43 h \cdot 60 = 25.8 min \sim 26 min[/tex]

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The car is considered as an open system, because an external force is applied to the car that causes constant acceleration of car. Hence, option (B) is correct.

The system isolated from its surrounding is known as closed-system. While, in an open system, the flow of information between the system and surrounding takes place, to cause any change in the system.

As per the concept of open and closed system:

Hence, the car must be considered an an open system, because the acceleration of car is related with the force, which is being applied from the outside of frame of reference.

Thus, we can conclude that considering only car is an open system because external force is applied to the car that causes it to accelerate. Hence, option (B) is correct.

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3) The boulder travelled for a distance of 10 m

4) The work done on the car is zero

Explanation:

3)

The work done by a force on an object is given by:

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

where

F is the magnitude of the force

d is the displacement

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

In this problem we have:

F = 24 N

W = 240 J

[tex]\theta=0^{\circ}[/tex], assuming that the force is applied in the same  direction as the displacement

Therefore, we can find d, the displacement of the boulder:

[tex]d=\frac{W}{F cos \theta}=\frac{240}{(24)(cos 0)}=10 m[/tex]

4)

As in the previous exercise, the work done is

The work done by a force on an object is given by:

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

In this problem, we have:

F = 9000 N is the force applied

d = 0 is the displacement, since the car has not moved

Therefore, the work done on the car is

[tex]W=(9000)(0)(cos 0)=0[/tex]

So, no work is done on the car.

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