Suppose you walk 16 m straight east and then 22 m straight south. At what angle, in degrees South of East, is a line connecting your starting point to your final position?

Answers

Answer 1

53.9 angle, in degrees South of East and 27.20 m is a line connecting your starting point to your final position.

What is initial position and final position?

The distance in decimeters between the starting point and the ending position is in a straight line. The distance between an object's original position and its ultimate position is known as displacement.

Briefing:

You walk 16 m straight east and then 22 m straight south. This forms a right angled triangle with a horizontal distance of 16 m, a vertical distance of 22 m and the hypotenuse is the distance between the ending and starting point. Let x represent the distance between the ending and starting point. Using Pythagoras theorem:

x² = 16² + 22²

x² = 256 + 484

x² = 740

Taking square root of both sides:

√x² = √740

x = √740

x = 27.20 m = distance between the ending and starting point.

Now use trigonometry:

sinθ=B/R

sinθ=22/27.20

sinθ= 0.808

θ = 53.9 degree. This is your angle.

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Related Questions

If sound waves travel at 350 m/s on a warm humid day. And you are submerged underwater. Is the sound you hear while below the surface traveling faster, slower or at the same rate as the sound you hear above the water? Explain your answer

Answers

ANSWER:

Sound waves are faster in water

STEP-BY-STEP EXPLANATION:

Sound travels differently in water than it does in air.

Sound waves travel faster in denser substances because neighboring particles will more easily collide with each other.

Water is denser than air, which means that sound waves travel much faster in water than in air.

Although they travel faster in water, we must keep in mind that the human ear evolved to hear sounds in the air and is not as useful when submerged in water.

The plow was an important invention for the Sumerian civilization because it:
A. made it easier for farmers to prepare land for farming.
B. simplified communication with other cultures.
C. allowed important information to be recorded.
D. protected city-states from foreign invaders.





I am in the middle of the test help pls

Answers

The plow was an important invention for the Sumerian civilization because it made it easier for farmers to prepare land for farming.

The plow made farming more efficient for the Sumerians, as people no longer had to waste time planting food by hand. The cart allowed people to carry goods back and forth from different civilizations in a much easier way as people did not have to carry the goods on their backs.

It is estimated that the ancient Sumerians invented the plow. The plow was designed to utilize domestic animals as a means of easing the workload on farmers. Oxen were harnessed to a wooden plow and pulled it through the fields to loosen the soil for easier planting of crops.

Plow, also spelled plough, most important agricultural implement since the beginning of history, used to turn and break up soil, to bury crop residues, and to help control weeds.

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Alex tickles his brother by stroking adjacent ________ spots on his skin.

Answers

Answer:

adjacent flat spots on his skin

Answer:

You didn't provide answer choices but I'd assume it's pressure.

If the slope of a position versus time graph is negative, what does that imply about the motion of an object? A. The object has a negative acceleration. B. The object turned around. C. The object is moving in the negative direction. D. All of the above

Answers

ANSWER

A. The object has a negative velocity

EXPLANATION

We want to describe the slope of a position-time graph that is negative.

If the slope of a position-time graph is negative, it implies that the velocity is also negative. In other words, as time increases, the velocity also decreases.

As velocity decreases, it implies that the acceleration also decreases.

Therefore, we have that the acceleration of the object is negative.

The answer is option A.

Who exhibits more power?A) Bob transfer 2000 J in 3 minutes.B) Carl transfers 2 J in 0.1 s.C) Alan transfers 20 J in 0.5 s.

Answers

Given that:

A) Energy,

[tex]E_B=\text{ 2000 J}[/tex]

The time is

[tex]\begin{gathered} t_B=3\text{ min} \\ =3\times60\text{ s} \\ =180\text{ s} \end{gathered}[/tex]

B) Energy,

[tex]E_C=2\text{ J}[/tex]

and time

[tex]t_C=0.1\text{ s}[/tex]

C) Energy

[tex]E_A=20\text{ J}[/tex]

and time

[tex]t_A=0.5\text{ s}[/tex]

We have to find the power for all three cases.

The formula to calculate power is

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

The power transferred by Bob will be

[tex]\begin{gathered} P_B=\frac{E_B}{t_B} \\ =\frac{2000}{180} \\ =11.11\text{ W} \end{gathered}[/tex]

The power transferred by Carl will be

[tex]\begin{gathered} P_C=\frac{E_C}{t_C} \\ =\frac{2}{0.1} \\ =20\text{ W} \end{gathered}[/tex]

The power transferred by Alan will be

[tex]\begin{gathered} P_A=\frac{E_A}{t_A} \\ =\frac{20}{0.5} \\ =40\text{ W} \end{gathered}[/tex]

Thus, Alan exhibits more power.

how fast is the angle of depression of the telescope changing when the boat is 190 meters from the shore

Answers

ANSWER:

- 0.01943 rad/sec

STEP-BY-STEP EXPLANATION:

The first thing is to make a drawing of what is mentioned in the statement, it would be the following:

Now, we have the following information:

[tex]\begin{gathered} \frac{dy}{dt}=15\text{ m/s} \\ x=50\text{ m} \\ y=190\text{ m} \end{gathered}[/tex]

In this right angle triangle formed by telescope of the boat, e can apply the tangent trigonometric ratio, like this:

[tex]\begin{gathered} \tan \theta=\frac{x}{y} \\ \text{ replacing} \\ \theta=\tan ^{-1}\mleft(\frac{x}{y}\mright) \end{gathered}[/tex]

Now, we implicitly derive with respect to t:

[tex]\begin{gathered} \frac{d}{dt}(\theta)=\frac{d}{dt}(\tan ^{-1}(\frac{x}{y})) \\ \frac{d}{dt}(\theta)=\frac{1}{1+(\frac{x}{y})^2}\cdot\frac{d}{dt}(\frac{x}{y}) \\ \frac{d}{dt}(\theta)=\frac{y^2}{x^2+y^2}\cdot x\cdot(-\frac{1}{y^2}\cdot\frac{dy}{dt}) \\ \frac{d}{dt}(\theta)=\frac{-x}{x^2+y^2}(\frac{dy}{dt}) \\ \text{ replacing} \\ \frac{d}{dt}(\theta)=\frac{-50}{50^2+190^2}\cdot(15) \\ \frac{d}{dt}(\theta)=-0.01943 \end{gathered}[/tex]

The angle of depression is changing at a rate of -0.01943 rad/sec when the boat is 190 m from the shore

Four charges are arranged in a square formation. Take q to be 1 C of charge and a to be 2 cm in length. Four charges are arranged in a square formation. Find the net electric field at the center of the square.

Answers

We have to calculate the electric field in the center, so we need the distance to the center R and the interaction of each charge with that point

[tex]E=\sum_{n\mathop{=}0}^{\infty}E_i[/tex]

To calculate the distance R we have a triangle

[tex]R=\sqrt{\frac{2a^2}{4}}=\frac{a}{\sqrt{2}}=0.014m=1.41cm[/tex][tex]\begin{gathered} \sum_{n\mathop{=}0}^{\infty}Ex=\frac{9\cdot10^9}{2\cdot10^{-4}}\cdot(cos45)\cdot(6C)=1.91\cdot10^{14}N/C \\ \sum_{n\mathop{=}0}^{\infty}Ey=4.5\cdot10^{13}\cdot(cos45\degree)(2C)=0.636\cdot10^{14}N/C \\ Etot=\sqrt{Ex^2+Ey^2}=2.01\cdot10^{14}N/C \end{gathered}[/tex]

Is important to remember that E has direction so you have to calculate each axis, x, and y

A turntable turning at rotational speed 78 rpm stops in 45 s when turned off. The turntable's rotational inertia is 1.2 ×10−2kg⋅m2.

Determine the magnitude of the resistive torque that slows the turntable.

Answers

The magnitude of the torque that slows the turntable is equal to 2.17 × 10⁻³ N.m.

What is torque?

Torque can be described as the measure of the force that can cause the rotation of an object about an axis. The force causes an object to accelerate similarly, torque causes an angular acceleration. Therefore, torque is the rotational equivalent of linear force.

Given, the rotational speed of the turntable, ω = 78 rpm = 8.16 rad/s

The time turntable takes to stop, t = 45 s

The rotational inertia of the turntable, I = 1.2 × 10⁻² Kg.m²

From the rotational kinetics equation, the angular acceleration:

[tex]\omega_f = \omega +\alpha t[/tex]

The final speed of the turntable = 0

[tex]\alpha =\frac{\omega_f-\omega}{t}[/tex]

[tex]\alpha =\frac{0-8.16}{45}[/tex]

[tex]\alpha =-0.1814 \;rad/s^2[/tex]

Calculation of torque by using inertia and angular acceleration:

[tex]\tau = I\alpha[/tex]

[tex]\tau =1.2 \times 10^{-2} \times (-0.1814)[/tex]

[tex]\tau = 2.17\times 10^{-3} \; N.m[/tex]

Therefore, the magnitude of the torque that slows the turntable is 2.17 × 10⁻³ N.m.

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**NEED USEFUL ANSWER ASAP, H.W QUESTION**
Given that hotter blackbodies produce more energy than cooler blackbodies, why do cooler red giants have much higher luminosities than much hotter white dwarfs?

Answers

Given that hotter blackbodies produce more energy than cooler blackbodies,  cooler red giants have much higher luminosities than much hotter white dwarfs.

The hotter blackbodies emits more light as compared to the cooler blackbodies over all the wavelength. A blackbody emits light over all wavelength. the red giant is a star , hydrogen is fused to helium core to stat the fusion reactions. this hydrogen fusion caused the star to expand. the sun becomes red giant it will swell up to size of earth orbit.  This gives a red color and make red giant much hotter. while the white dwarf comes after the red giant . white dwarf is also very hot but it is not hot enough to star fusion reaction in carbon and oxygen.

Thus, Given that hotter blackbodies produce more energy than cooler blackbodies,  cooler red giants have much higher luminosities than much hotter white dwarfs.

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The cross product of two vectors (X and Y) is a negative vector when the angle between them is:
A. 0
B. 90
C. 180
D. 270

Answers

Answer:

C. 180 

Explanation:

Answer:

Explanation:

 The cross product:

[ a × b ] = | a | · | b | · sin α

sin α < 0   if  α ∈ (180°, 360°)

Answer:

D. 270°

A football blocking sled has a mass of roughly 100 kg. If a football player applies a force of 500 N to the sled, and there is a 350 N frictional force acting on the sled, what is the acceleration of the sled?

Answers

The the acceleration of the sled of mass 100 Kg will be 1.5 m/s².

What is Friction?

A drag is force that opposes the motion of an object by acting in the direction opposite to its motion. It is of two types namely - Static friction and kinetic friction. The static friction is given by - F[S] = μ[s]​ x η and the  kinetic friction is given by F[K] = μ[k] x η.

Given is a football blocking sled which has a mass of roughly 100 kg. A football player applies a force of 500 N to the sled, and there is a 350 N frictional force acting on the sled.

Assume that the force applied by the player is - F[P] = 500 N and the force of friction is - F = 350 N.

Now, for the motion of the sled with acceleration [a], we can write -

F[P] - F = ma

a = {F[P] - F}/m

Substituting the values, we get -

a = (500 - 350)/100

a = 150/100

a = 15/10

a = 1.5 m/s²

Therefore, the the acceleration of the sled of mass 100 Kg will be

1.5 m/s².

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assume that the brakes in your car create a constant deceleration if you double your driving speed how does this affect a0 the time required to come to a stop and b) the distance needed to stop?

Answers

For the moving car, if you double your driving speed;

a. the time required to come to a stop will increase by a factor of 2.

b. the distance needed to stop will increase by a factor of 2.

What is the formula relating acceleration, speed, time, and distance traveled?

The formula relating to the acceleration, speed, time, and distance traveled of an object is a formula of the equations of motion.

The equations of motion are given below as follows:

v = u + at

s = ut + ¹/₂at²

v² = u² + 2as

where:

v = final speed

u = initial speed

t = time taken

s = distance traveled

a = acceleration

Considering the given scenario:

a. if the speed is doubled, the equation v = u + at applies for the time required to come to a stop;

t =2 * v - u / t

the time will be increase.

b. if the speed is doubled, the equation v² = u² + 2as applies for the distance needed to stop;

s = 2 * v² - u²

the distance will increase with an increase in speed.

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Which of the following sources is most likely to be free of bias? OA. Textbook OB. Blog post c. Diary D. Video interview

Answers

Answer:

Textbook

Explanation:

Pretty obvious thi :v

Answer:

OA

Explanation:

Text books are often reviewed by experts hence, is likely to be biased.

A cyclist travels a distance S from A to B consisting of three segments: s = S1 + $2 + $3, Time and speed over the segments are t1, t2, t3 and v1= 20(km/h), v2 = 15(km/h), V3= 10(km/h). Calculate the average speed over the distance s. Consider two cases: a) s1=s2=s3 b) t1=t2=t3

Answers

The average speed over the distance s consisting of three segments s1, s2, and s3 in first case a) s1 = s2= s3 is 13.85 km/h and in the second case b) t1 = t2 = t3 is 15 km/h.

In the first case where the distance for three intervals is the same a) s₁ = s₂ = s₃, the average speed for three segments is calculated by the formula:

Average speed = 3 ÷ ( 1/v₁ + 1/v₂ +1/v₃)

Average speed = 3 ÷ ( ¹/₂₀ + ¹/₁₅ + ¹/₁₀)

Average speed = 13.85 km/h

In the second case where the time taken for all three intervals is the same b) t₁ = t₂ = t₃, the average speed for three segments is calculated by the formula:

Average speed = (v₁ + v₂ +v₃) ÷ 3

Average speed = (20 + 15 + 10) ÷ 3

Average speed = 15 km/h

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The cubic expansivity of mercury is 1.8×10^-4K^-1 and the linear expansivity of glass is 8.0×10^-6K^-1, calculate the apparent expansivity of mercury in a glass container?

Answers

[tex]\begin{gathered} \text{cubic expansivity of mercury=1.8}\times10^{-4}K^{-1} \\ \text{ Linear expansivity of glass = 8.0}\times10^{-6}K^{-1} \end{gathered}[/tex]

Therefore,

[tex]Apparent\text{ cubic expansivity=}real\text{ cubic expansivity - cubic expansivity of the container}[/tex][tex]\begin{gathered} \text{ cubic expansivity of the container =3(}8.0\times10^{-6})=24\times10^{-6}=2.4\times10^{-5} \\ \text{note we converted linear expansivity to cubic expansivity} \end{gathered}[/tex]

Finally,

[tex]\text{Apparent cubic expansivity=}1.8\times10^{-4}-2.4\times10^{-5}=1.56\times10^{-4}K^{-1}[/tex]

The outer ear _____ sound waves

Answers

According to the research, the correct term to complete the statement is collect. The outer ear collects sound waves.

What is the outer ear?

It is a structure that captures or collects sound or sound vibrations before channeling them to the middle ear.

In this sense, it is made up of the pinna that projects from the side of the head and the external auditory meatus that automatically amplifies sounds that pass inward to the tympanic membrane.

Therefore, we can conclude that the outer ear captures sound waves and conducts them to the middle ear, thus the correct term is collect.

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How is the direction of the velocity of a satellite differ from the direction of its acceleration?
PLEASE HELP!!!!!

Answers

The direction of the velocity of a satellite differ from the direction of its acceleration by an angle of 90° as they act perpendicular to each other.

In an orbital motion, the acceleration of the object is always directed towards the center of the orbit. This acceleration is called as centripetal acceleration. It is denoted by [tex]a_{c}[/tex].

[tex]a_{c}[/tex] = v² / r

In an orbital motion, the velocity of the object is always tangential to the orbit. It can be calculated in two ways,

v = 2 π r / T

v = r ω

Therefore, the direction of the velocity of a satellite differ from the direction of its acceleration by an angle of 90°.

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A rocket toy moves straight upward, starting from rest with an acceleration of 4 m/s2. It runs out of fuel at the end of 10 s and continues to coast upward, reaching a maximum height before falling back to earth
a. Find the rocket’s velocity and position at the end 10 s
B. Find the velocity before the rocket crashes on the ground
C. Find the time flight of rocke

Answers

I think the answer is C

A particle starts from rest of move with constant acceleration 4 m/s² along straight line then, Calculate velocity when it travel distance of 80 m​ step - by step explaination

Answers

Answer:

25.3 m/s

Explanation:

According to the laws of motion, the final velocity and the distance traveled by a particle moving at constant acceleration a m/s²  is given by the equation

[tex]v^2 = v_0^2 + 2a(x - x_0)[/tex]

where

v= final velocity m/s
v₀ = initial velocity m/s
x = final position m
x₀ = initial position m
a = acceleration in m/s²

x - x₀ is called the displacement, d

We are given
v₀ = 0 since particle is starting at rest
d = 80 m
a = 4 m/s²

Plugging these values into the above equation.

v² = 0 + 2 x 4 x 80 m²/s²

v² = 640 m²/s²

v = √640  = 8√10 ≈ 25.3 m/s

What is the impulse of a car hitting another car if it's initial velocity was 50m/s
and it's velocity after hitting it is 4.4m/s? The car weighs 2168kg.
J=A P
AP=mvf-mvi

Answers

The impulse of the car after hitting is 98860.8 kg.m/s

In physics, the term "impulse" is used to characterize or measure the impact of force operating gradually to alter an object's motion. It is often stated in Newton-seconds or kilograms per second and is denoted by the symbol J.

            J = m ( v-u)

where, m = 2168 kg

           u = 50 m/s

           v = 4.4 m/s

J = 2168 ( 4.4 - 50)

J = 2168(45.6)

J = 98860.8 kg.m/s

since the final velocity is decreasing the impulse is a vector term thus a negative sign will also be included in this.

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A boy walks 4 km east and then turns around and walks 1.5 km west. If east is taken as the positive direction and the west as the negative direction on a number line then what is the distance the boy covers and his displacement?

Answers

ANSWER

[tex]undefined[/tex]

EXPLANATION

First, let us make a sketch of the question:

From the diagram, the blue circle represents his starting position while the black circle represents his final position.

The total distance the boy covers is the sum of his two journeys, to the east and then to the west.

That is:

[tex]\begin{gathered} 4+1.5 \\ 5.5\operatorname{km} \end{gathered}[/tex]

The boy's displacement is the distance between his starting position and his final position.

Since the west is taken as negative direction and the east taken as the positive direction, it means that 4km east means +4km and 1.5

what is the difference between these two formulas [tex]fc = \frac{m(v {}^{2}) }{r} [/tex][tex]ac = \frac{v {}^{2} }{r} [/tex]and how would I apply them to a promblem?

Answers

One is centripetal acceleration (ac) and the other one is centripetal force (fc)

Since = F = m* a

F = force

m= mass

a= acceleration

For centripetal force (Fc), we use centripetal acceleration

Fc= m * ac

ac= centripetal acceleration = v^2 / r

r= radius

v= linear velocity

So, in the end, we have:

Fc = m * v^2 /r

You can apply the to find the Centripetal force of an object with a certain mass, that negotiates a certain radius curve, with a certain speed.

An object is formed by attaching a uniform, thin rod with a mass of mr = 8kg and length L = 6 m to a uniform sphere with mass ms = 36.25 kg and radius R = 1.5m.1) What is the moment of inertia of the object about an axis at the left end of the rod?

Answers

ANSWER:

2167.68 kg*m^2

STEP-BY-STEP EXPLANATION:

Given:

mr = 8 kg

L = 6 m

ms = 36.25 kg

R = 1.5 m

Moment of inertia of sphere about its center is:

[tex]I_{CM}=\frac{2}{5}m_s\cdot R^2[/tex]

Using paraller theorem, moment of inertia of shpere about end of rop is:

[tex]I_{\text{rod}}=m_s\cdot(R+L)^2+\frac{1}{3}m_r\cdot L^2[/tex]

Therefore:

[tex]\begin{gathered} I=I_{cm}+I_{\text{rod}}_{} \\ I=\frac{2}{5}\cdot m_s\cdot R^2+m_s\cdot(R+L)^2+\frac{1}{3}\cdot m_r\cdot L^2 \end{gathered}[/tex]

Replacing:

[tex]\begin{gathered} I=\frac{2}{5}\cdot36.25\cdot1.5^2+36.25\cdot(1.5+6)^2+\frac{1}{3}\cdot8\cdot6^2 \\ I=2167.69\text{ kg}\cdot m^2 \end{gathered}[/tex]

The moment of inertia is 2167.68 kg*m^2

draw each of the following vectors, label an angle that specifies the vectors direction, then find its magnitude and direction.a. B= -4.0 I+ 4.0jb. r= (-2.0i-1.0j) cmc. v= (-10-100j) m/s d. a= (20i+10j) m/s^2the I's and j's have the hat and its the vector simble on the letters

Answers

[tex]\begin{gathered} A) \\ B=-4.0i+4.0j \\ |B|=\sqrt{(-4.0)^2+(4.0)^2} \\ |B|=\sqrt[]{16^{}+16^{}} \\ |B|=\sqrt[]{32^{}} \\ |B|=4\sqrt{2} \\ \text{The magnitude of B is }4\sqrt[]{2} \\ \theta=\tan ^{-1}(\frac{4.0}{-4.0})=135\text{ \degree} \\ The\text{ angle of B is }135\text{ \degree} \\ B) \\ R=(-2.0i-1.0j)cm \\ |R|=\sqrt[]{(-2.0)^2+(-1.0)^2} \\ |R|=\sqrt{4.0+1.0} \\ |R|=\sqrt[]{5.0} \\ \text{The magnitude of R is }\sqrt[]{5.0}cm \\ \theta=\tan ^{-1}(\frac{-1.0}{-2.0})=26.57\text{ \degree, but it is below of negative x-axis, hence} \\ \theta=180+26.57=206.57\text{ \degree} \\ \text{The angle of R is }206.57\text{ \degree} \\ C) \\ V=(-10i-100j)\text{ m/s} \\ |V|=\sqrt{(-10)^2+(-100)^2} \\ |V|=\sqrt[]{100+10000} \\ |V|=\sqrt[]{10100}=10\sqrt{101}\approx100.5\text{ m/s} \\ \text{The magnitude of V is }100.5\text{ m/s} \\ \theta=\tan ^{-1}(\frac{-100}{-10})\approx264.29 \\ \text{The angle of V is }264.29\text{ \degree} \\ \\ D) \\ A=(20i+10j)m/s^2 \\ |A|=\sqrt{20^2+10^2} \\ |A|=\sqrt{400+100} \\ |A|=\sqrt{500}=10\sqrt{5}\approx22.36m/s^2 \\ \text{The magnitud of A is }22.36m/s^2 \\ \theta=\tan ^{-1}(\frac{10}{20})=26.57\text{ \degree} \\ \text{The angle of A is }26.57\text{ \degree} \end{gathered}[/tex]

An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a
mass of 5.8x 104 kg, and the force causes its speed to change from 6800 to 5600m/s. (a) What is the work done by the force? (b) If the
asteroid slows down over a distance of 2x 106 m determine the magnitude of the force.

Answers

The work done on the asteroid by the constant force is 4.06 × 10¹¹ joules and the force experienced by the asteroid is 1.89 × 10⁵ newtons

what is work energy theorem?

work energy theorem is an important phenomenon in physics which states that the change in kinetic energy is equal to the work done by the object.

This law is basic part of conservation of energy.

what is newtons law of motion?

Newton's laws of motion are three basic laws of classical mechanics that describe the relationship between the motion of an object and the forces acting on it. These laws can be paraphrased as follows: A body remains at rest, or in motion at a constant speed in a straight line, unless acted upon by a force.

given:

mass of the asteroid = 5.8 x 10⁴ kg

initial speed of the asteroid v₁ = 6800m/s

final speed of the asteroid v₂ = 5600 m/s

displacement of the asteroid = 2 × 10⁶ meters

a)

using work energy theorem for the asteroid we get,

work done = change in kinetic energy

work done = 1/2 × m × (v₂) ² - 1/2 × m × (v₁) ²

work done by the force on the asteroid is = 1/2 × 5.8 x 10⁴ (1.4 × 10⁷)

work done by the force on the asteroid is = 4.06 × 10¹¹ joules

b)

first, we will have to find the acceleration of the asteroid,

using newtons third law of motion we get,

a = 14/4

a = 3.5 m/s²

therefore, the force can be calculated using relation f = m × a

here,

m is mass

a is acceleration

therefore, the force on the asteroid is = 5.8 x 10⁴ × 3.5

the force on the asteroid is = 1.89 × 10⁵ newtons

therefore, the work done on the asteroid by the constant force is 4.06 × 10¹¹ joules and the force experienced by the asteroid is 1.89 × 10⁵ newtons

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What is the specific heat c of a 0.500 kg metal sample that rises 5.40 C when 305J of heat is added to it?

Answers

ANSWER

[tex]\begin{equation*} 112.96\text{ }J\/kgK \end{equation*}[/tex]

EXPLANATION

Parameters given:

Mass of sample, m = 0.5 kg

Temperature change, ΔT = 5.40 °C = 5.40 K

Heat energy, H = 305 J

To find the specific heat capacity of the sample, we have to apply the formula for heat energy:

[tex]H=mc\Delta T[/tex]

Where c = specific heat capacity

Therefore, solving for c, the specific heat capacity of the metal sample is:

[tex]\begin{gathered} 305=0.5*c*5.4 \\ c=\frac{305}{0.5*5.4} \\ c=112.96\text{ }J\/kgK \end{gathered}[/tex]

That is the answer.

If an astronaut weighed 200 pounds on the surface of the earth how much would the astronaut weigh 4000 miles above the earths surface. Show working equation

Answers

Recall that the gravitational pull is given by

[tex]F=G\frac{m_{}\cdot M_E}{r^2}[/tex]

Where m is the mass of the astronaut, ME is the mass of Earth, r is the distance between them, and G is the gravitational constant.

The above relation states that the force of gravitation is directly proportional to the mass of two objects and is inversely proportional to the square of the distance between them.

If the astronaut weighs 200 pounds and the radius of the Earth is 4000 miles then

[tex]\begin{gathered} 200=G\frac{m\cdot M_E}{4000^2} \\ G\cdot m\cdot M_E=200\cdot4000^2 \end{gathered}[/tex]

When the astronaut moves 4000 miles above the earth's surface then the distance between them is

4000+4000 = 8000 miles

The distance is measured from the center of the earth so the radius must be included.

So, the new force of gravitation is

[tex]\begin{gathered} F=\frac{G\cdot m_{}\cdot M_E}{r^2} \\ F=\frac{200\cdot4000^2}{8000^2} \\ F=50\: lb \end{gathered}[/tex]

Therefore, the astronaut weighs 50 pounds 4000 miles above the earth's surface.

This makes sense because as you go further away from the center of the Earth then your weight becomes less due to less force of gravity.

Simpson s drives his car with an average velocity of 85 km/h eastward how long will it take him to drive 560 km on a perfectly straight highway?

Answers

Simpson will it take 112 / 17 km to drive 560 km on a perfectly straight highway.

What relationship does speed have with distance?

You can replace this computation with d = rt, which stands for "distance equals rate times time." To calculate speed or rate, use the speed formula, s = d/t, which asserts that speed is equal to distance divided by time.

The difference between speed and velocity is the amount of distance an object covers in a given amount of time and in a certain direction. Speed is a scalar quantity while velocity is a vector.

We know that:

v = 85

d = 560

Substitute

v = 85

d = 560 into formula d = v x t  :

560 = 85t

Cross out the common factor: 112 / 17 km

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what is velocity ratio​

Answers

Answer:

Definition of velocity ratio

: the ratio of a distance through which any part of a machine moves to that which the driving part moves during the same time.

pls brainlies me me new pls

Explanation:

Answer:
Le rapport de vitesse, parfois appelé rapport de distance, est une comparaison de la quantité de force qu’un objet, comme une voiture, crée par rapport aux autres forces autour de lui qui agissent contre lui.

Explanation:
Lorsque le premier engrenage (le conducteur ou l'engrenage d'entrée) tourne, le deuxième engrenage (l'engrenage entraîné ou en sortie) se transforme en réponse. La différence entre les vitesses des deux vitesses est appelée rapport de vitesse ou rapport de réduction.

A ball rolls off a table and falls 0.75m to the floor, landing with a speed of 4 m/s. (A) What is the acceleration of the ball just before it strikes the ground? (B) What was the initial speed of the ball? (C) What initial speed must the ball have if it is to land with a speed of 5 m/s?

Answers

Answer:

We had this question yesterday, let me check my book quickly

(A) The acceleration of the ball is 9.81 m/s² just before it strikes the ground.

(B) The initial speed of the ball is equal to 1.14 m/s.

(C) The initial speed must be 3.21 m/s if it is to land with a speed of 5 m/s.

What are the equations of motion?

The equation of motion is the way to represent the relation between the time, acceleration, initial and final velocity, and distance covered by a moving object.

The three equations of motion are:

[tex]v= u+ at\\v^2 = u^2 +2aS\\S = ut +(1/2) at^2[/tex]

The acceleration of the ball just before it strikes the ground is equal to gravitational acceleration, g = 9.81 m/s².

Given, the final velocity of the ball, v = 4m/s

The height of the table, h = 0.75 m

The initial velocity of the ball, [tex]u = \sqrt{v^2-2gh}[/tex]

[tex]u = \sqrt{(4)^2-2\times 9.8\times 0.75}[/tex]

u = 1.14 m/s

When the final velocity of the ball, v = 5m/s

The initial velocity will be :[tex]u = \sqrt{(5)^2-2\times 9.8\times 0.75}[/tex]

u = 3.21 m/s.

Learn more about equations of motion, here:

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