You’re standing at the highest point on the Moon, 10,786 mm above the level of the Moon’s mean radius. You’ve got a golf club and a golf ball.

Part A

(How fast would you need to hit the ball horizontally so it goes into a circular orbit?)

Express your answer with the appropriate units.

Part B

(If you hit the ball vertically with the same speed, to what height above you would it rise?)

Express your answer with the appropriate units.

Answers

Answer 1

Answer:

A)   v = 1,675 10³ m / s  , B)    r₂ = 11,673 10⁶ m

Explanation:

A) This exercise we must use Newton's second law, where the forces of gravity are the Moon

        F = m a

acceleration is centripetal

        a = v² / r

force is the force of universal attraction

         F = G m M / r²

we substitute

        G m M / r² = m v² / r

        v² = G M / r

distance

        r = R_moon + h

        r = 1.74 10⁶ +1.0786 10⁴

        r = 1,750786 10⁶ m

we calculate

        v = √ (6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶)

        v = √ (2,8052 10⁶)

        v = 1,675 10³ m / s

B) let's use energy conservation

    Starting point. In the mountain

          Em₀ = K + U = ½ m v² + G m M / r

    Final point. Where the speed is zero

          [tex]Em_{f}[/tex] = U = G mM / r₂

           Em₀ = Em_{f}

           ½ m v² + G m M / r = G mM / r₂

           1 / r₂ = (½ v₂ + G M / r) / GM

let's calculate

 1 / r₂ = (½ (1,675 10³)² + 6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶) /(6.67 10⁻¹¹ 7.36 10²²)

           1 / r₂ = (1,4028 10⁶ + 2,805 10⁶) / 49.12 10¹¹

           1 / r₂ = 8.5664 10⁻⁷

            r₂ = 11,673 10⁶ m


Related Questions

A student is conducting an experiment that involves adding hydrochloric acid to various minerals to detect if they have carbonates in them. The student holds a mineral up and adds hydrochloric acid to it. The acid runs down the side and onto the student’s hand causing irritation and a minor burn. If they had done a risk assessment first, how would this situation be different? A. It would be the same, there is no way to predict the random chance of acid dripping off the mineral in a risk assessment. B. The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. C. The student would be safer because he would have been wearing goggles, but his hand still would not have been protected. D. The student would not have picked up the mineral because he would know that some of the minerals have dangerous chemicals in them.

Answers

The answer would be D because it could have been prevented

By  the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "

What is experiment ?

An experiment would be a technique used to confirm or deny a hypothesis, as well as assess the likelihood or effectiveness of something that has never been tried before.

What is hydrochloric acid?

Hydrochloric acid is a kind of compound in which hydrogen and chlorine element is present.

Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.

By  the experiment "By  the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "

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Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)

Answers

Answer:

*If the particles are deflected in opposite directions, it implies that their charges must be opposite

*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

Explanation:

When a charged particle enters a magnetic field, it is subjected to a force given by

        F = q v x B

where bold letters indicate vectors

   

this expression can be written in the form of a module

        F = qv B sin θ

and the direction of the force is given by the right-hand rule.

In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1

If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.

Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

describe the relation among density, temperature, and volume when the pressure is constant, and explain the blackbody radiation curve

Answers

Answer:

in all cases with increasing temperature the density should decrease.

Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body,

Explanation:

Let's start for ya dream gas

        PV = nRT

Since it indicates that the pressure is constant, we see that the volume is directly proportional to the temperature.

The density of is defined by

        ρ = m / V

As we saw that volume increases with temperature, this is also true for solid materials, using linear expansion. Therefore in all cases with increasing temperature the density should decrease.

Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body, since all the radiation that falls on it is absorbed or emitted.

This type of construction has a characteristic curve where the maximum of the curve is dependent on the tempera, but independent of the material with which it is built, to explain the behavior of this curve Planck proposed that the diaconate in the cavity was not continuous but discrete whose energy is given by the relationship

             E = h f

The buoyant force on an object placed in a liquid is (a) always equal to the volume of the liquid displaced. (b) always equal to the weight of the object. (c) always equal to the weight of the liquid displaced. (d) always less than the volume of the liquid displaced.

Answers

Answer:

(c) always equal to the weight of the liquid displaced.

Explanation:

Archimedes principle (also called physical law of buoyancy) states that when an object is completely or partially immersed in a fluid (liquid, e.t.c), it experiences an upthrust (or buoyant force) whose magnitude is equal to the weight of the fluid displaced by that object.

Therefore, from this principle the best option is C - always equal to the weight of the liquid displaced.

Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.0° above the horizontal, and the second arrow is fired straight upward. Assume an isolated system and choose the reference configuration at the initial position of the arrows.
(a) what is the maximum height of each of the arrows?
(b) What is the total mechanical energy of the arrow-Earth system for each of the arrows at their maximum height?

Answers

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

[tex]U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{x,1}[/tex], [tex]K_{x,2}[/tex] - Initial and final horizontal translational kinetic energy, measured in joules.

[tex]K_{y,1}[/tex], [tex]K_{y,2}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

[tex]m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0[/tex]

[tex]g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})[/tex]

[tex]y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}[/tex]

Where:

[tex]y_{1}[/tex]. [tex]y_{2}[/tex] - Initial and final height of the arrow, measured in meters.

[tex]v_{y,1}[/tex], [tex]v_{y,2}[/tex] - Initial and final vertical speed of the arrow, measured in meters.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

The initial vertical speed of the arrow is:

[tex]v_{y,1} = v_{1}\cdot \sin \theta[/tex]

Where:

[tex]v_{1}[/tex] - Magnitude of the initial velocity, measured in meters per second.

[tex]\theta[/tex] - Initial angle, measured in sexagesimal degrees.

If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the initial vertical speed is:

[tex]v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}[/tex]

[tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex]

If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex] and [tex]v_{y,2} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:

[tex]y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]y_{2} - y_{1} = 56.712\,m[/tex]

Second arrow

[tex]U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,3}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{y,1}[/tex], [tex]K_{y,3}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.

[tex]m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0[/tex]

[tex]g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})[/tex]

[tex]y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}[/tex]

If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} = 82\,\frac{m}{s}[/tex] and [tex]v_{y,3} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:

[tex]y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]y_{3} - y_{1} = 342.816\,m[/tex]

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

[tex]E = U + K_{x}[/tex]

The expression is now expanded:

[tex]E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}[/tex]

Where [tex]v_{x}[/tex] is the horizontal speed of the arrow, measured in meters per second.

[tex]v_{x} = v_{1}\cdot \cos \theta[/tex]

If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the horizontal speed is:

[tex]v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}[/tex]

[tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex]

If [tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{max} = 56.712\,m[/tex] and [tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex], the total mechanical energy is:

[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}[/tex]

[tex]E = 201.720\,J[/tex]

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

[tex]E = m\cdot g \cdot y_{max}[/tex]

[tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{max} = 342.816\,m[/tex]

[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)[/tex]

[tex]E = 201.720\,J[/tex]

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.


An ac circuit consist of a pure resistance of 10ohms is connected across an ae supply
230V 50Hz Calculate the:
(i)Current flowing in the circuit.

(ii)Power dissipated

Answers

Plz check attachment for answer.

Hope it's helpful

Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain

Answers

Complete question:

A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.

Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain

Answer:

The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.

Explanation:

Given;

magnitude of applied force, F = 1.5 N

Apply Newton's second law of motion;

F = ma

[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]

The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;

[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]

Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.

The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g.(a) What force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcherâs hand.

Answers

Answer:

Explanation:

F ×1 = 0.5×0.145×47×47

F = 160.15 N

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A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.

Answers

Answer:

Yes, work has been done on the mud.

Explanation:

Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.

A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.

Answers

Answer:

0.4757 mm

Explanation:

Given that:

Load P = 223,000 N

the length of the height of the aluminium column = 1.22 m

the diameter of the aluminum column = 10.2 cm = 0.102 m

The amount that the column has shrunk ΔL can be determined by using the formula:

[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]

where;

A = πr²

2r = D

r = D/2

r = 0.102/2

r = 0.051

A = π(0.051)²

A = 0.00817

Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:

[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]

[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]

[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]

ΔL = 4.757 × 10⁻⁴ m

ΔL =  0.4757 mm

Hence; the amount that the column has shrunk is 0.4757 mm

An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.10 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 37.5 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target

Answers

Answer:

The speed of the arrow after passing through the target is 30.1 meters per second.

Explanation:

The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:

[tex]m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}[/tex]

Where:

[tex]m_{a}[/tex], [tex]m_{t}[/tex] - Masses of arrow and target, measured in kilograms.

[tex]v_{a,o}[/tex], [tex]v_{a,f}[/tex] - Initial and final speeds of the arrow, measured in meters per second.

[tex]v_{t,o}[/tex], [tex]v_{t,f}[/tex] - Initial and final speeds of the target, measured in meters per second.

The final speed of the arrow is now cleared:

[tex]m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})[/tex]

[tex]v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})[/tex]

If [tex]v_{a,o} = 2.1\,\frac{m}{s}[/tex], [tex]m_{t} = 0.3\,kg[/tex], [tex]m_{a} = 0.0225\,kg[/tex], [tex]v_{t,o} = 2.10\,\frac{m}{s}[/tex] and [tex]v_{t,f} = 0\,\frac{m}{s}[/tex], the speed of the arrow after passing through the target is:

[tex]v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )[/tex]

[tex]v_{a,f} = 30.1\,\frac{m}{s}[/tex]

The speed of the arrow after passing through the target is 30.1 meters per second.

When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?

Answers

Answer: If we have equilibrium, the magnitude must be zero.

Explanation:

If the charges are in equilibrium, this means that the total charge is equal to zero.

And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero

If you could see stars during the day, this is what the sky would look like at noon on a given day. The Sun is near the stars of the constellation Gemini. Near which constellation would you expect the Sun to be located at sunset?

Answers

Answer:

The sun will be located near the Gemini constellation at sunset

Which describes any compound that has at least one element from group 17? Halide;noble gas; metalliod; transition metal

Answers

Answer:

Halide

Explanation:

It has at least one element from the halogen group (17)

Halide describes any compound that has at least one element from group 17, therefore the correct option is option A.

What are halides?

When the elements belonging to group 17 of the periodic table form ionic compounds with other electropositive elements, then these compounds are known as halides.

These elements from group 17 are also known as halogens. Generally, these halides have very high electronegativity as they reside on the right side of the periodic table.

Generally, the valency of the halogens element involved in the halide compound is one and they form ionic compounds with the alkali and alkaline earth metals.

Thus, halides are compounds that have at least one element from group 17.

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Stress is a factor that contributes to heart disease risk.true or false

Answers

The answer for this question would be True

The cost of energy delivered to residences by electrical transmission varies from $0.070/kWh to $0.258/kWh throughout the United States; $0.110/kWh is the average value.

Required:
At this average price, calculate the cost of:

a. leaving a 40-W porch light on for two weeks while you are on vacation?
b. making a piece of dark toast in 3.00 min with a 970-W toaster
c. drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer.

Answers

Answer:

Cost = $ 1.48

Cost = $ 0.005

Cost = $ 0.38

Explanation:

given data

electrical transmission varies = $0.070/kWh to $0.258/kWh

average value = $0.110/kWh

solution

when leaving a 40-W porch light on for two weeks while you are on vacation so cost will be

first we get here energy consumed that is express as

E = Pt    .................1

here E is Energy Consumed and Power Delivered is P and t is time

so power is here 0.04 KW and t = 2 week = 336 hour

so

put value in 1 we get

E = 0.04 × 336

E = 13.44 KWh

so cost will be as

Cost = E × Unit Price    .............2

put here value and we get

Cost = 13.44 × 0.11

Cost = $ 1.48

and

when you making a piece of dark toast in 3.00 min with a 970-W toaster

so energy consumed will be by equation 1 we get

E = Pt

power is = 0.97 KW and time = 3 min = 0.05 hour

put value in equation 1 for energy consume

E = 0.97 × 0.05 h

E = 0.0485 KWh

and we get cost by w\put value in equation 2 that will be

cost =  E × Unit Price

cost = 0.0485 × 0.11

Cost = $ 0.005

and

when drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer

from equation 1 we get energy consume

E = Pt

Power Delivered = 5.203 KW and time = 40 min = 0.67 hour

E = 5.203 × 0.67

E = 3.47 KWh

and

cost will by put value in equation 2

Cost = E × Unit Price

Cost = 3.47 × 0.11

Cost = $ 0.38

The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 0.600 g is supported by a needle, the tip of which is a circle 0.240 mm in radius, what pressure is exerted on the record in N/m2?

Answers

Answer:

[tex]P=3.25x10^{4}\frac{N}{m^2}[/tex]

Explanation:

Hello,

In this case, since pressure is defined as the force applied over a surface:

[tex]P=\frac{F}{A}[/tex]

We can associate the force with the weight of the needle computed by using the acceleration of the gravity:

[tex]F=0.600g*\frac{1kg}{1000g}*9.8\frac{m}{s^2} =5.88x10^{-3}N[/tex]

And the area of the the tip (circle) in meters:

[tex]A=\pi r^2=\pi (0.240mm)^2=\pi (0.240mm*\frac{1m}{1000mm} )^2\\\\A=1.81x10^{-7}m^2[/tex]

Thus, the pressure exerted on the record turns out:

[tex]P=\frac{5.88x10^{-3}N}{1.81x10^{-7}m^2} \\\\P=3.25x10^{4}\frac{N}{m^2}[/tex]

Which is truly a large value due to the tiny area on which the pressure is exerted.

Best regards.

When you are told that the wind has a "Small Coriolis force" associated with it, what is that "small force" exactly

Answers

Answer:

Coriolis force is a type of force of inertia that acts on objects that is in motion within a frame of reference that rotates with respect to an inertial frame. Due to the rotation of the earth, circulating air is deflected result of the Coriolis force, instead of the air circulating between the earth poles and the equator in a straight manner. Because of the effect of the Coriolis force,  air movement deflects toward the right in the Northern Hemisphere and toward the left in the Southern Hemisphere, eventually taking a curved path of travel.

Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop

Answers

Answer:

v = 1.7 m/s

Explanation:

By applying conservation of energy principle in this situation, we know that:

Loss in Potential Energy of Car = Gain in Kinetic Energy of Car

mgΔh = (1/2)mv²

2gΔh = v²

v = √(2gΔh)

where,

v = velocity of car at top of the loop = ?

g = 9.8 m/s²

Δh = change in height = 45 cm - Diameter of Loop

Δh = 45 cm - 30 cm = 15 cm = 0.15 m

Therefore,

v = √(2)(9.8 m/s²)(0.15 m)

v = 1.7 m/s

Which scientist's work led to our understanding of how planets move around
the Sun?
A. Albert Einstein
B. Lord Kelvin
C. Johannes Kepler
D. Edwin Hubble

Answers

Answer:

Johannes Kepler

Explanation:

He made rules about planetary motion.The scientist Johannes Kepler was a German astronomer.He found out that the planets evolved around the Sun.He also made the laws of planetary motion.

Hope this helped,

Kavitha

It was Johanne keplers work

A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice​

Answers

Answer:

nvbnncbmkghbbbvvvvvvbvbhgggghhhhb

An alarm clock is plugged into a 120 volt outlet and has a resistance of 15,000 ohms. How much power does it use?

Answers

Answer:

The power used is 0.96 watts.

Explanation:

Recall the formula for electric power (P) as the product of the voltage applied  times the circulating current:

[tex]P=V\,\,I[/tex]

and recall as well that the circulating current can be obtained via Ohm's Law as the quotient of the voltage applied divided the resistance:

[tex]V=I\,\,R\\I=\frac{V}{R}[/tex]

Then we can re-write the power expression as:

[tex]P=V\,\,I=V\,\,\frac{V}{R} =\frac{V^2}{R}[/tex]

which in our case becomes:

[tex]P=\frac{V^2}{R}=\frac{120^2}{15000} =0.96\,\,watts[/tex]

An electric heater is constructed by applying a potential different of 120V across a nichrome wire that has a total resistant of 8 ohm .the current by the wire is

Answers

Answer:

15amps

Explanation:

V=IR

I=V/R

I = 120/8

I = 15 amps

What is the relationship between the magnitudes of the collision forces of two vehicles, if one of them travels at a higher speed?

Answers

Explanation:

The collision forces are equal and opposite.  Therefore, the magnitudes are equal.

A circuit element maintains a constant resistance. If the current through the circuit element is doubled, what is the effect on the power dissipated by the circuit element

Answers

Answer:

This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.

Explanation:

The formula for calculating the power expended in a circuit is P =  I²R... 1

i is the current (in amperes)

R is the resistance (in ohms)

If  a circuit element maintains a constant resistance and the current through the circuit element is doubled, then new current I₂ = 2I

New power dissipated P₂ = (I₂)²R

P₂ = (2I)²R

P₂ = 4I²R ... 2

Dividing equation 2 by 1 will give;

P₂/P = 4I²R/I²R

P₂/P = 4

P₂ = 4P

This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 12.9 rad/s in 2.98 s.
(a) Find the magnitude of the angular acceleration of the wheel.
(b) Find the angle in radians through which it rotates in this time interval.

Answers

Explanation:

(a) Find the magnitude of the angular acceleration of the wheel.

angular acceleration = angular speed /timeangular acceleration = 12.9/2.98 = 4.329rad/s²

(b) Find the angle in radians through which it rotates in this time interval.

angular speed = 2x3.14xf12.9rad = 2 x3.14

rad = 6.28/12.9rad = 0.487

Now we convert rad to angle

1 rad = 57.296°0.487 = unknown angleunknown angle =57.296 x 0.487 = 27.9°

The angle in radians = 27.9°

Bromine, a liquid at room temperature, has a boiling point

Answers

Yes it does !  The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided.  The boiling point is higher than room temperature.

In an undergraduate physics lab, a simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min. What are the period and length of the pendulum

Answers

Explanation:

We have

A simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min.

The frequency of a pendulum is equal to the no of oscillation per unit time. so,

[tex]f=\dfrac{N}{t}\\\\f=\dfrac{71}{1.8\times 60}\\\\f=0.65\ Hz[/tex]

Tim period is reciprocal of frequency. So,

[tex]T=\dfrac{1}{0.65}\\\\T=1.53\ s[/tex]

The time period of a pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is length of pendulum

[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(1.53)^2\times 9.8}{4\pi ^2}\\\\l=0.58\ m[/tex]

So, the period and length of the pendulum are 1.53 s and 0.58 m respectively.

You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern's central maximum

Answers

Answer:

The width is  [tex]Z = 0.0424 \ m[/tex]

Explanation:

From the question we are told that

    The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]

    The wavelength of the light is  [tex]\lambda = 721 \ nm[/tex]

      The position of the screen is  [tex]D = 2.83 \ m[/tex]

Generally angle at which the first minimum  of the interference pattern the  light occurs  is mathematically  represented as

        [tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]

Where m which is the order of the interference is 1

substituting values

       [tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]

      [tex]\theta = 0.5317 ^o[/tex]

 Now the width of first minimum  of the interference pattern is mathematically evaluated as

       [tex]Y = D sin \theta[/tex]

substituting values

       [tex]Y = 2.283 * sin (0.5317)[/tex]

       [tex]Y = 0.02 12 \ m[/tex]

 Now the width of  the  pattern's central maximum is mathematically evaluated as

        [tex]Z = 2 * Y[/tex]

substituting values

      [tex]Z = 2 * 0.0212[/tex]

     [tex]Z = 0.0424 \ m[/tex]

On Apollo missions to the Moon, the command module orbited at an altitude of 160 km above the lunar surface. How long did it take for the command module to complete one orbit?

Answers

Answer:

T = 2.06h

Explanation:

In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:

[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex]         (1)

T: time for a complete orbit = ?

r: radius of the orbit

G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2

Mm: mass of the moon = 7.34*10^22 kg

The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:

[tex]r=R_m+160km\\\\[/tex]

Rm: radius of the moon = 1737.1 km

[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]

Then, you replace all values of the parameters in the equation (1):

[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]

In hours you obtain:

[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]

The time that the Apollo takes to complete an orbit around the moon is 2.06h

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