The picture below shows a person swinging a toy plane attached to a string in
uniform circular motion.
Which vector points in the direction of the centripetal acceleration of the
plane?

Answers

Answer 1

According to the image A vector points in the direction of the centripetal acceleration of the plane.

The correct option is C.

What is centripetal acceleration?

Centripetal acceleration is a property of an object's motion along a circular path. Centripetal force refers to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.

Briefing:

A person is seen in the image swinging a toy plane on a string in a smooth, circular motion. The velocity direction is tangent to the circular orbit and perpendicular to the direction of the position.

The direction of the velocity and the speed may both vary as an object travels in a circular orbit. The velocity's direction is continually shifting. While acceleration is constantly moving uniformly in a circular path toward the recent, tangent at each point indicates the direction of velocity at that place.

As a result, vector A is pointed in the plane's centripetal acceleration's direction.

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The complete question is -

The picture below shows a person swinging a toy plane attached to a string in uniform circular motion.

Which vector points in the direction of the centripetal acceleration of the

plane?

A-B

B-D

C-A

D-C

The Picture Below Shows A Person Swinging A Toy Plane Attached To A String Inuniform Circular Motion.Which

Related Questions

Suppose that the Towngas supply pressure is 8.5 kPa (gauge pressure) andthe total volume is 2.4 m? enters a building from outside where thetemperature is 10 °C and passes into a building where the temperature is 35°C, if the pressure was reduced to 2 kPa. What would be the new volume ofthe gas?

Answers

Given,

The initial pressure of the gas, P₁=8.5 kPa

The initial volume of the gas, V₁=2.4 m³

The initial temperature of the gas, T₁=10 °C=282.15 K

The temperature of the building, i.e., the temperature of the gas after entering the building, T₂=35 °C=308.15 K

The pressure of the gas after entering the building, P₂=2 kPa

Let us assume the new volume of the gas is V₂

From the combined gas law,

[tex]\begin{gathered} \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \\ \Rightarrow V_2=\frac{P_1V_1T_2}{T_1P_2} \end{gathered}[/tex]

On substituting the known values,

[tex]undefined[/tex]

A/An _____ is described as a device that is used to measure potential difference across any part of a circuit.ammeterfuseground fault interruptervoltmeter

Answers

Answer:

[tex]\text{Voltmeter}[/tex]

Explanation: We need to find an instrument that measures potential difference across any part of the circuit, a potential difference is basically voltage difference across a circuit difference and the device used to measure this difference is known as:

[tex]\text{ Voltmeter}[/tex]

Voltmeter used in a circuit

1.Can money by happiness?why and Why not?
2.Are you happy with the possession you have? Yes/No Why? ​

Answers

Answer:

Explanation:

1) No it can't

2)Yes I am

Answer:

yes it's can be, but Don't put money first

Jacob Grena raises a spoon 0.210 m above a table . If the spoon and its contents have a mass of 30.0 g, what is the gravitational potential energy associated with the spoon at that height relative to the table's surface?

Answers

Explanation:

The spoon is raised so it gains Gravitational potential energy. Formula to find Gravitational potential energy;

Gravitational potential energy = mass × Gravitational field strength × height of the body from the surface (table in this scenario)

In symbols; E = m×g×h

Substitute values:

m = 30g = 0.03kg (don't forget to convert grams to kg)

g = 10N/kg

h = 0.210 m

So it's;

0.03kg × 10N/kg × 0.210m = 0.063 Joules

SI Unit of energy is joules

A) A recipe calls for 5.0 qt of milk. What is this quantity in cubic centimeters?Express your answer in cubic centimeters.B) A gas can holds 2.0 gal of gasoline. What is this quantity in cubic centimeters?Express your answer in cubic centimeters.

Answers

A.

[tex]\begin{gathered} 1qt=946.353cm^3 \\ 5qt=\text{?} \\ \text{quantity}=946.353\times5 \\ \text{quantity}=4731.765cm^3 \end{gathered}[/tex]

B.

[tex]\begin{gathered} 1\text{ gal=}3785.41\text{ }cm^3 \\ 2\text{ gal}=\text{?} \\ \text{quantity}=2\times3785.41 \\ \text{quantity}=7570.82cm^3 \end{gathered}[/tex]

Calculate the total capacitance of three capacitors 30µF, 20µF & 12µF connected in parallel across a d.c supply The answer is :Consider that the equivalent capacitance of three capacitors C1, C2 and C3 in parallel is given by:C=C1+C2+C3In this case:C1 = 30µFC2 = 20µFC3 = 12µFReplace the previous values into the formula for C and simplify:C=30μF+20μF+12μF=62μFHence, the total capacitance is 62µFQuestion 5 : Calculate the total charge on the capacitors connected in parallel if the supply voltage is 500V. Sketch a circuit diagram and label this to show how the charges are located

Answers

The circuit diagram is shown below:

From the diagram we notice that the same voltage will flow in every capacitor, this will be helpful later.

We know that this three capacitors are equivalent to a single equivalent capacitor with 62µF capacitance. The charge in this equivalent capacitor is:

[tex]Q_{eq}=(62\times10^{-6})(500)=0.031[/tex]

Now, as we mentioned, the voltage is the same in each capacitor then the charge in each of them is:

[tex]\begin{gathered} Q_1=(30\times10^{-6})(500)=0.015 \\ Q_2=(20\times10^{-6})(500)=0.01 \\ Q_3=(12\times10^{-6})(500)=0.006 \end{gathered}[/tex]

To check if this is correct we need to remember that the charge in the equivalent capacitor is equal to the sum of the charge in each capactior; for this case this conditon is fulfil; therefore we conclude that:

• The charge in the first capacitor is 0.015 C

,

• The charge in the second capacitor is 0.01 C

,

• The charge in the third capacitor is 0.006 C

The diagram with the labels is shown below:

The couple required to hold a triple turn of 1.5cm² area in equilibrium when carrying a current 2A at 70° to a field with 0.15T is?

Answers

The couple or torque required to hold the triple turn is 1.27 x 10⁻⁴ Nm.

What is the couple or torque required?

The couple required to hold the triple turn is calculated as follows;

τ = M x Bsinθ

where;

M is the magnetic moment B is the magnetic field strength

The magnetic moment is calculated as follows;

M = NIA

where;

N is number of turns = 3I is current = 2 AA is the area of the loop = 1.5 cm² = 0.00015 m²

M = (3) x (2) x (0.00015)

M = 0.0009 m²A

The torque or couple required is calculated as;

τ = (0.0009) x (0.15 x sin70)

τ = 1.27 x 10⁻⁴ Nm

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A fuzzy die that has a weight of 1.70N hangs from the ceiling of a car by a massless string. The car travels on a horizontal road and has an acceleration of 2.70m/s^2 to the left. The string makes an angle theta with respect to the vertical, as shown in the figure below. 1) What is the angle theta?

Answers

First we calculate the mass of the fuzzy die

[tex]m=\frac{W}{g}[/tex]

m is the mass, W is the weight, and g is the gravity

m=?

W=1.70N

g=9.8 m/s^2

we susbtitute

[tex]m=\frac{1.70}{9.80}=0.17\text{ kg}[/tex]

Then we calculate the force of x

[tex]Fx=0.17(2.70)=0.468\text{ N}[/tex][tex]Fy=1.70N[/tex]

Then we have the next diagram

Therefore for the angle

[tex]\theta=\arctan (\frac{F_x}{F_y})=arctan(\frac{0.468}{1.70})=15.4\text{ \degree}[/tex]

ANSWER

The angle is 15.4°

What is the force of gravity between two 50.0kg masses that are separated by 0.300m?3.71x10-8N5.59x10-7N2.78x104N1.85x10-6N

Answers

We will have the following:

[tex]\begin{gathered} F=G\frac{m_1m_2}{r^2}\Rightarrow F=\frac{(6.67\ast10^{-11}m^3\ast kg^{-1}\ast s^{-2})(50kg)(50kg)}{(0.3m)^2} \\ \\ \Rightarrow F=1.852777778...\ast10^{-6}N\Rightarrow F\approx1.85\ast10^{-6}N \end{gathered}[/tex]

So, the force is approximately 1.85*10^-6 N.

A Hydrogen atom is a low density hot gas will give out what type of spectrum?A. A uniform spectrum containing all colorsB. A series of emission lines with equal spaces between the colorsC. A series of emission lines spaced in mathematical sequenceD.a uniform spectrum crossed by numerous dark absorption lines

Answers

The emission of photons takes place when an electron from higher energy orbitals jumps to a lower energy orbital.

Therefore the light emitted will correspond to the energy difference between the orbitals.

When the atom emits the photons, they will have energy equal to the energy difference between the orbitals of the Hydrogen. Therefore the spectrum obtained by the hydrogen gas will contain only those lines which correspond to the energy difference of the orbitals.

Therefore the hydrogen will emit a spectrum that contains a series of emission lines spaced in a mathematical sequence.

Therefore the correct answer is option C.

a 298 kg boat is being propelled forward with a force of 2,365 N. What is the acceleration of the boat if it has a resistance force (rewarded) due to wind and water of 878 N? (Write answer as a 2 digit number)​

Answers

The acceleration of the boat is 4.9m/s²

Mass of boat= 298 kg

Forward force= 2365 N

Resistance force= 878N

We need to apply the concept of laws of motion

Net force= Forward force- Resistance force

Net force= 2365-878 N

= 1487 N

Net force= mass x acceleration

2365= acceleration x 298

acceleration = 4.9 m/s²

Therefore, the acceleration of the boat is 4.9 m/s²

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Full working out…….2.A vibrating mass-spring system has a frequency of 0.56 Hz. How much energy ofthis vibration is carried away in a one-quantum change?

Answers

ANSWER

3.7128 x 10⁻³⁴ J

EXPLANATION

The energy carried in a one-quantum change is the product of Planck's constant, h, and the frequency of vibration, f,

[tex]E=hf[/tex]

Planck's constant is 6.63 x 10⁻³⁴ J*s and, in this case, the frequency of vibration is 0.56 Hz. So, the energy carried away is,

[tex]E=0.56Hz\cdot6.63\cdot10^{-34}J\cdot s=3.7128\cdot10^{-34}J[/tex]

Hence, the energy carried away in a one-quantum change is 3.7128 x 10⁻³⁴ J.

7.0 J of work is done to draw a bowstring back. The bow launches an arrow with a mass of 0.09 kg straight upward.
(a) What is the arrow's kinetic energy as it leaves the bow? (Round your answer to one decimal place.)
J
(b) What is the arrow's speed? (Round your answer to one decimal place.)
m/s
(c) What maximum height does the arrow reach? (Round your answer to one decimal place.)
m

Answers

The final answer is

(a) The KE = 3.88 * 10

(b) The speed of the arrow is 2.93 * 10

(c) Height which the arrow attains is 4.38 * 10

When the arrow is drawn from a bowstring back then there will be some work done. The pulling of the bowstring will have a distance moved and the energy used to do this.

Given,

Work done = 7 J

Mass of the arrow = 0.09 kg

The computations for a and b are the following:

To calculate the Kinetic energy,

                  Work done = force x distance

                  7.0 J = force x 0.09

                   force = 7/0.09=77.77 = 7.77 x 10

                   force = m. a = 0.09 x a

                    77.77 / 0.09 = a = 864.1 = 8.64*10²

KE = 0.5 x m x v²

     = 0.5 x 0.09 x 864.1

     = 3.88 * 10

The speed of the arrow can be calculated as

       speed 38.8/(0.5 x 0.09) = v² = 862 sq-root

       Hence, v = 2.93*10¹ m/s is the answer

Height that the arrow reaches

       29.3² = 2gh

      = 29.3²/(2 x 9.8) = 858.49 / (2 x 9.8) = 43.8 m

h = 43.8 m or 4.38 * 10¹ is the answer

Therefore, we can conclude that the kinetic energy of the arrow that leaves bowstring is 3.88*10 with the speed of 29.3 m/s and it reaches the height of 43.8 m.

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In one or two sentences, explain how you would draw the particles in a gas.

Answers

Explanation

In gases the particles move rapidly in all directions, frequently colliding with each other and the side of the container

so

I would draw it with the particles separated moving and colliding with the others

I hope this helps you

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]

itial height at Two balls are thrown vertically from the same • Ball I is launched upward with an initial velocity voj = + 10m/s. Ball 2 is launched downward with an initial velocity vo2 = - 10m/s. same The distance between the two balls after I second from the beginning of motion is:

Answers

Given

vo1 = +10 m/s

vo2 = -10 m/s

Procedure

Using the free fall equations, we have:

[tex]\begin{gathered} x1=v_{o1}t-\frac{1}{2}gt^2 \\ x1=10*1-\frac{1}{2}9.8*1 \\ x1=5.1m \end{gathered}[/tex][tex]\begin{gathered} x2=v_{o2}t-\frac{1}{2}gt^2 \\ x2=-10*1-\frac{1}{2}9.8*1 \\ x2=-14.9m \end{gathered}[/tex][tex]\begin{gathered} x1-x2=5.1-\lparen-14.9) \\ x1-x2=20 \end{gathered}[/tex]

The distance between the balls would be 20m

Why is the (k) a negative value in hooks law.

Answers

Well, we will hav that neither "k" or "x" are negative, the negative sign is external to both values and the reason is that this is written in that way to represent the "direction" of the movement of the system. Form example, when the spring is extended then it will represent that it will pull back, and when it is compacted it represents that it will pull outwards.

[tex]F_s=-kx[/tex]

It is simply a technicallity.

The velocity-time table represents the motion of a rightward-
moving motorcycle.
Magnitude =
Time (s)
0.0
0.5
1.0
1.5
2.0
What is the magnitude (i.e., value) and direction of the
acceleration?
Direction =
Velocity (m/s)
24.0, right
22.0, right
20.0, right
18.0, right
16.0, right
m/s/s
(No -
sign.)
(Tap field to change.)
NE

Answers

The direction of motion of a body or object depends on its velocity. Speed can be thought of as a scalar quantity in its simplest form.  In essence, velocity is a vector quantity. It is the speed at which distance is changing.

What is velocity?

Galileo Galilei, an Italian physicist, is credited with being the first to calculate speed by dividing it by the required amount of time and the distance traveled.

Speed, according to Galileo, is the distance traveled in a set amount of time.

The speed at which an object is moving is referred to as its velocity.

Examples of fast motion include an automobile driving north on a highway or a rocket blasting off.

The equation v = u + at, where v signifies the ultimate speed, can be used to express an object's final velocity, which is equal to its starting velocity plus acceleration times the distance traveled.

Therefore, the direction of motion of a body or object depends on its velocity. Speed can be thought of as a scalar quantity in its simplest form.  In essence, velocity is a vector quantity. It is the speed at which distance is changing.

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Below is a diagram of a 5.0 kg block being dragged to the right, along a horizontal surface. The
coefficient of dynamic friction, is 0.40. Take acceleration due to gravity, g, as 9.81 ms ².
5 kg
30. N
What is the acceleration of the block? Give your answer correct to two significant figures, in
m-s-2, without units.

Answers

The acceleration of the block when the coefficient of friction is 0.40 is 3.924 m/s².

The ratio of the frictional resistive force to the perpendicular force pushing the objects together is known as the coefficient of friction.

Acceleration is the rate at which an object's velocity with respect to time changes.

The block of mass 5 kg is dragged to the right on a horizontal surface.

The coefficient of friction is 0.40.

The acceleration due to gravity is 9.81 m/s².

The force on the block is 30 N.

Now, the force is defined as the product of the mass and acceleration of the object.

Now, using the conservation of force:

F = μmg

ma = μmg

a = μg

a = 0.40 × 9.81

a = 3.924 m/s²

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The chart shows data for four moving objects.
Object
W
X
Y
Z
Which object has the greatest acceleration?
W
Initial Velocity
(m/s)
Ο Ζ
11
10
12
20
Final Velocity
(m/s)
29
34
40
28
Change in
Time (s)
6
12
7
8

Answers

Object Y has the greatest acceleration = 4m/s2

What is an 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.

Acceleration = (final velocity - initial velocity ) /time

For object W

Acceleration = ( 29-11)/6 = 3m/s2

For object X

Acceleration = (34-10)/12 = 2m/s2

For object Y

Acceleration = ( 40-12)/7 = 4m/s2

For object Z

Acceleration = (28-20)/8 = 1m/s2

Object Y has the greatest acceleration = 4m/s2

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An Olympic long jumper is capable of jumping 8.0 m. Assuming his horizontal speed is 9.1 m/s as he leaves the ground, how long is he in the air and how high does he go?

Answers

Given data:

* The initial velocity of the jumper is u = 9.1 m/s.

* The horizontal range in the given case is 8 m.

Solution:

(a). By the kinematic equation, the time taken by the jumper in terms of the initial velocity and the horizontal range is,

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

where a is the acceleration of the jumper in the horizontal direction,

As there is no force acting on the jumper in the horizontal direction, thus, the value of acceleration is zero.

Substituting the known values,

[tex]\begin{gathered} 8=9.1\times t \\ t=\frac{8}{9.1} \\ t=0.88\text{ s} \end{gathered}[/tex]

Thus, the time for which the jumper remains in the air is 0.88 seconds.

(b). By the kinematics equation, the initial velocity of the jumper in the upward direction is,

[tex]v_y-u_y=gt^{\prime}_{}\ldots\ldots\ldots(1)[/tex]

where u_y is the initial velocity, v_y is the final velocity of the jumper at the top of vertical displacement, g is the acceleration due to gravity, and t' is the time taken by the jumper to reach the top of vertical displacement,

The jumper will come to rest at the higher position, thus, the final velocity of the jumper at the highest position is zero.

The time taken by the jumper to reach the maximum height is equal to the time taken by the jumper to reach the ground from the maximum height.

As the total time for the complete motion of the jumper is t, thus, the time taken by the jumper to reach the maximum height from the ground is,

[tex]\begin{gathered} t^{\prime}=\frac{t}{2} \\ t^{\prime}=\frac{0.88}{2} \\ t^{\prime}=0.44\text{ s} \end{gathered}[/tex]

Substituting the known values in the equation (1),

[tex]\begin{gathered} 0-u_y=-9.8\times0.44_{} \\ u_y=4.312\text{ m/s} \end{gathered}[/tex]

By the kinematics equation, the maximum height reached by the jumper is,

[tex]h=u_yt^{\prime}+\frac{1}{2}gt^{\prime}^2[/tex]

Substituting the known values,

[tex]\begin{gathered} h=4.312\times0.44+\frac{1}{2}\times(-9.8)\times(0.44)^2 \\ h=1.9-0.95 \\ h=0.95\text{ m} \end{gathered}[/tex]

Thus, the maximum height reached by the jumper is 0.95 meters.

A ray of light travels from air into a liquid, as shown in figure below. The ray is incident upon the liquid at an angle of 30.0°. The angle of refraction is 22.0%,

Answers

If a ray of light travels from air into a liquid, as shown in figure below. The ray is incident upon the liquid at an angle of 30.0°. The angle of refraction is 22°, and the refractive index of the liquid would be 1.334.

What is refraction?

It is the phenomenon of bending of light when it travels from one medium to another medium. The bending towards or away from the normal depends upon the medium of travel as well as the refractive index of the material.

By using Snell's law,

Refractive index of the liquid = sin(i) /sin(r)

                                                =sin(30) /sin(22)

                                                = 1.334

Thus, the refractive index of the liquid would be 1.334.

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The highest frequencies humans can hear isabout 20000 Hz.What is the wavelength of sound in airat this frequency?The speed of sound is310 m/s.

Answers

Given

The highest frequency human can hear is f=20000 Hz

Speed of the sound,v=310 m/s

To find

The wavelength

Explanation

We know,

The wavelength is given by

[tex]\begin{gathered} \lambda=\frac{v}{f} \\ \Rightarrow\lambda=\frac{310}{20000} \\ \Rightarrow\lambda=0.0155\text{ m} \end{gathered}[/tex]

Conclusion

The wavelength is 0.0155 m

A kid is on a stationary sled, onsnowy ground with fls 0.105.It takes 71.2 N of force to setthe sled moving. How muchnormal force is acting?(Unit = N)

Answers

ANSWER

[tex]678.10N[/tex]

EXPLANATION

Parameters given:

Coefficient of static friction, μs = 0.105

Force, F = 71.2 N

The minimum force required to set the sled moving must be equal to the frictional force acting on the sled.

The frictional force is given mathematically as:

[tex]F_f=\mu_sN[/tex]

where N = normal force

Since the force required is equal to the frictional force, we have that:

[tex]F=F_f[/tex]

Therefore:

[tex]F=\mu_sN[/tex]

Solve for N:

[tex]\begin{gathered} N=\frac{F}{\mu_s} \\ N=\frac{71.2}{0.105} \\ N=678.10N \end{gathered}[/tex]

That is the normal force.

How many cubic inches are there in 3.25 yd3?Express the volume in cubic inches to three significant figures.What is the mass in grams of 16.86 mL of acetone?Express your answer to four significant figures and include the appropriate units.What is the volume in milliliters of 7.06 g of acetone?Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

[tex]\text{ 3.25 yd}^3=151,632in^3[/tex]

Explanation: We need to convert cubic-yards into cubic inches, this can be simply done as follows:

[tex]\frac{46656\text{ Cubic inches}}{1\text{ Cubic Yard}}^{}[/tex]

Therefore we have:

[tex]\begin{gathered} 3.25\text{ Cubic yards }\times\text{ }\frac{46656\text{ Cubic Inches}}{1\text{ Cubic Yard}}=151,632\text{ Cubic Inches} \\ \therefore\rightarrow \\ \text{ 3.25 yd}^3=151,632in^3 \end{gathered}[/tex]

nd in Atlanta you decide to drive around the city. You turn a corner and are driving up a steep hill. Suddenly, a small boy runs out on the street chasing a ball. You slam on the brakes and skid to a stop leaving a 50-foot-long skid mark on the street. The boy calmly walks away but a policemen watching from the sidewalk walks over and gives you a speeding ticket. He points out that the speed limit on this street is 25mph. After you recover your wits, you begin to examine the situation. You determine that the street makes an angle of 25◦with the horizontal and that the coefficient of static friction between your tires and the street is0.80. You also find that the coefficient of kinetic friction between your tires and the street is 0.60. Your car’s information book tells you that the mass of your car is 1600 kg. You weigh 140 lbs. Will you fight the ticket

Answers

skid = 50 ft

Speed limit = 25 mph

angle = 25°

friction coefficient = 0.80

mass = 1600 kg

weight = 140 lbs

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

A NASA probe is moving horizontally above the surface of the moon at a constant speed to the right, as depicted in the diagram below. It releases an instrument package when it is directly above Point P. As seen from the lunar surface, which path would the package likely follow after the release and why?
B, because the gravity of the moon will pull the instrument to the ground with constant acceleration and the lack of an atmosphere allows the package to fall straight down
D, because the gravity of the moon will pull the instrument to the ground with constant acceleration so that its vertical velocity increases as it falls, while the horizontal component of the velocity remains constant
C, because the reduced gravity of the moon pulls the package down vertically at a constant speed while the package travels horizontally at a constant speed resulting in a straight-line trajectory to the lunar surface.
A, because the gravity of the moon pulls the package down with constant acceleration, while the atmosphere of the moon creates horizontal drag on the package which reduces the horizontal component of the package’s velocity causing the package to be pulled backward as it falls

Answers

As seen from the lunar surface, the path followed by the package after the release is D, because the gravity of the moon will pull the instrument to the ground with constant a, so that its vertical velocity increases.

A is incorrect because there is no wind pushing the package backwards. B is incorrect because the package has an initial velocity. C is incorrect because vertical velocity is not constant due to the presence of gravity. E and F are incorrect because gravity acts  immediately after the package is dropped.

D is correct because the horizontal component remains constant because there is no horizontal force acting on the package. This is because in outer space there is no atmosphere, so there will be no air resistance. The vertical component increases with respect to time because of constant acceleration due to gravitational pull on the package.

Therefore, as seen from the lunar surface, the path followed by the package after the release is D, because the gravity of the moon will pull the instrument to the ground with constant acceleration so that its vertical velocity increases as it falls, while the horizontal component of the velocity remains constant

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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.

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

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