A rocket weighs 1.000. N on the surface of Earth. What is its weight at a distance of two Earth radii above the
surface of Earth?
P.5B
0 111.0 N
250.0 N
333.0 N
5000 N

Answers

Answer 1

Answer:

0 111.0N be cause the rest do not add up


Related Questions

In which direction does the magnetic field in the center of the coil point?

Answers

Each segment of circular loop carrying current produces magnetic field lines in the same direction with in the loop. The direction of magnetic field at the centre of circular coil is perpendicular to the place of the coil. i.e. along the axis of the coil. Please mark brainiest!!!

Answer:

Right

Explanation:

Coil move right yes

We say that the mass comes to rest if, after a certain time, the position of the mass remains within an arbitrary small distance from the equilibrium position. Will the mass ever come to rest

Answers

Answer:

No, the mass will never come to rest

Explanation:

It is so because even at arbitrarily small distance it will experience some amount of force (irrespective of how small the value of force is).

This does not allow the mass to become stationary or in a equilibrium state as it is still subject to some amount of force.

Hence, the the mass will never come to rest

What should stainless steel be considered when using it?

Answers

Strength, ductility, and toughness are the three top mechanical qualities that should be given importance. Stainless steel comes with 10–30% chromium as its alloying element. It is this element that helps it resist corrosion.

Which diagram best represents the gravitational forces, F, be-
tween a satellite, S, and Earth?

Answers

Answer:

Diagram (3).

Explanation:

N3L states that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A ([tex]F_{A} = -F_{B}[/tex]).

The diagram which best represents the gravitational forces, F, between a satellite, S, and Earth is; Choice (3).

The Newton's law of gravitation states that the Force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

The law clearly states a Force of attraction; the two objects come towards each other.

Consequently, Choice (3) best represents the gravitational forces, F, between a satellite, S, and Earth.

Read more:

https://brainly.com/question/11460810

In a movie production, a stunt person must leap from a balcony of one building to a balcony 3.0 m lower on another building. If the buildings are 2.0 m apart, what is the minimum horizontal velocity the stunt person must have to accomplish the jump? Assume no air resistance and that ay = −g = −9.81 m/s2 . (Ans. 2.6m/s) PLS SHOW WORK

Answers

You are looking for Vx0 (initial velocity in X direction) use this formula
remember there is no acceleration in x direction, only Y

X = x0 + Vx0(t) + (1/2)at^2
X= 0 + Vx0 (t) + 0
x= Vx0 (t)
Vx0 = x/t

what we are missing is Time, so u can get this from using the Y formula to find time
Remember there is no velocity in Y direction

Y=y0 + Vy0(t) - (1/2)gt^2
y= 0 +0 - (1/2)gt^2
(3.0 m *2) / 9.80m/s/s = t^2
0.61224 = t^2
t = 0.78 s

now plug this time back into this formula from above
Vx0 = x/t
= 2.0m / 0.78s
=2.6

This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.

The minimum horizontal velocity required is "2.6 m/s".

First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.

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

where,

h = height = 3 m

vi = initial vertical speed = 0 m/s

t = time interval = ?

g = acceleration due to gravity = 9.81 m/s²

therefore,

[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]

t = 0.78 s

Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:

[tex]s = vt\\\\v = \frac{s}{t}[/tex]

where,

s = horizontal distance = 2 m

t =0.78 s

v = minimum horizontal velocity = ?

Therefore,

[tex]v = \frac{2\ m}{0.78\ s}[/tex]

v = 2.6 m/s

Learn more about equations of motion here:  

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

What is the mass of a mallard duck whose speed is 9.1 m/s and whose momentum has a magnitude of 12 kg⋅m/s?

Answers

Answer:

m = 1.31 kg

Explanation:

Given that,

The speed of duck, v = 9.1 m/s

The magnitude of momentum, p = 12 kg-m/s

We need to find the mass of the duck. We know that the momentum of an object is given by :

p = mv

Where

m is the mass of the duck

[tex]m=\dfrac{p}{v}\\\\m=\dfrac{12\ kg-m/s}{9.1\ m/s}\\\\m=1.31 kg[/tex]

So, the mass of the duck is equal to 1.31 kg.

The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.3 m/s . Initially, the car is also traveling at a speed 19.3m/s and its front bumper is a distance 25.0m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.560m/s^2 , then pulls back into the truck's lane when the rear of the car is a distance 26.5m ahead of the front of the truck. The car is of length 4.50m and the truck is of length 20.7m .
Part A) How much time is required for the car to pass the truck?
Part B ) What distance does the car travel during this time?
Part C) What is the final speed of the car?

Answers

Answer:

A)    t = 10.56 s, B)  x = 235 m, C) v = 25.2 m / s

Explanation:

A) We can solve this problem using kinematics expressions.

The distance traveled by the truck is

       x_c = v_c t

Distance traveled by the car.

The car must travel the distance that separates them from the truck x₀=25.0.   Return to the lane at x₁ = 26.5 m. the length of the truck x₂=20.7m and the length of the car x₃ = 2  4.5 = 9 m, therefore the total length traveled by the car is

          x_t = x₁ + x₂ + x₃

          x_t = 26.5 + 20.7 +9 = 56.2 m

the distance traveled by the car when it returns to the lane is

         x_c + x_t = x₀ + v₀ t + ½ a t²

when the car passes the car the distance traveled by the two vehicles is the same, we substitute

         v_c  t + x_t = x₀ + v₀ t + ½ a t²

         ½ a t² + t (v₀ -v_c) + (x₀ - x_t) = 0

we substitute the values

         ½ 0.560 t² + t (19.3 -19.3) + (25.0 - 56.2) =

         0.28 t² -31.2 = 0

         t = [tex]\sqrt{ \frac{31.2}{0.28} }[/tex]

         t = 10.56 s

This is the time it takes for the car to pass the truck and back into the lane.

B) the distance traveled is

        x = v₀ t + ½ a t²

        x = 19.3 10.56 + ½ 0.560 10.56²

        x = 235 m

C) the final velocity is

         v = v₀ + a t

         v = 19.3 + 0.560 10.56

         v = 25.2 m / s

Three 5 Ohm resistors are connected in series to a 10 Volt power supply. What is the current through each resistor?​

Answers

Answer74.3

:

Explanation:

A rifle fires a pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by from its unstrained length. The pellet rises to a maximum height of 6.10 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

Answers

This question is incomplete, the complete question is;

A rifle fires a 2.10 × 10⁻² kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 9.10 × 10⁻² m from its unstrained length. The pellet rises to a maximum height of 6.10 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

Answer:

the spring constant is 303.5 N/m

Explanation:

Given the data in the question;

There is a potential energy associated with the spring;

we know that potential energy = kinetic energy

mgh = [tex]\frac{1}{2}[/tex]kx²

where k is the spring constant and x is the compression

as the pallet is fired, the spring gives kinetic energy which is converted into gravitational potential energy

so

m = 2.10 × 10⁻²

g = 9.81 m/s²

h = 6.10 m

x = 9.10 × 10⁻² m

we substitute

mgh = [tex]\frac{1}{2}[/tex]kx²

2.10 × 10⁻² × 9.81 × 6.10 =  [tex]\frac{1}{2}[/tex] × K × ( 9.10 × 10⁻² )²

1.256661 = 0.0041405 × K

K = 1.256661 / 0.0041405

K = 303.5 N/m

Therefore, the spring constant is 303.5 N/m

The diagram below represents a light ray striking the surface of a flat, shiny object.
Light ray
Object's surface
What most likely happens to the light ray as it strikes the object's surface?
A. It is completely absorbed by the object
B. It is reflected back in the direction it came from
C. It is reflected at an angle equal to the angle of the incidence
D. It is reflected at an angle greater than the angle of incidence

Answers

Answer:

option (C) is right answer

Explanation:

I think it's helps you

Answer:

Option C is correct

Explanation:

hope it helps you

Do Planets with more mass have more gravity than planets with less mass

Answers

Here you go it was too long to type

PLEASE HELPPPPPP!ASAPPPPPPP​

Answers

Answer:

Altitude is the height of landmass calculated from sea level. No it has no effect on weight.

Explanation:

The altitude at sea level is always assumed to be 0°, thus as you move away from altidude is calculated in increments while moving into the sea, the vice versa is true.

Hydrocarbons are by-products of which of the following:
a. Fossil oils, mines and fossil fuels
b. Burning fires, water treatment plants and fossil oils
C. Combustion in fossil, lighting and exhaust fumes
d. Petrol, sea waves and combustion in vehicles
e. Burning fires and combustion in factories and vehicles​

Answers

Answer:

A

Explanation:

Key word part: hydro - water
I believe the answer is A
Please give me Brainliest

the density of gold is 19g/m³ what is the volume of 38g​

Answers

let the volume be x

19 = 38/x

x=38/19

x=2

volume is 2 cm^3

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.3 kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.0 m/s in the opposite direction.

Required:
a. What is the angular speed of the gate just after it is struck by the unfortunate raven?
b. During the collision, why is the angular momentum conserved but not the linear momentum?

Answers

Answer:

its a. and jusing that youl

A car crashes into a wall with an impulse of 40,000 kg.m/s. The collision took 2 seconds. How much force did the wall exert?


Answers

Answer:

i hate physics lol but its 20,000N

Explanation:

Helppppp plsssssssssssssssssssssss thanks 2+2== plzzz help me and 20+67

Answers

Answer:

8,6 is the answer to your question

A ball with a weight of 70 N hangs from a string that is coiled around a 3 kg pulley with a radius of 0.4 m. Both the ball and the pulley are initially at rest. The rotational inertia of the pulley is 0.5mr2 and the ball is released from rest. Calculate the ball's velocity after it falls a distance of 2 meters.

Answers

Answer:

v = 6.195 m / s

Explanation:

For this exercise we can use the conservation of energy, for the system formed by the ball and the pulley

starting point. Higher before releasing the system

        Em₀ = U = M g h

final point. When the ball has lowered h = 2

        Em_f = K = ½ M v² + ½ I w²

the energy is preserved

        Em₀ = Em_f

        M g h = ½ M v² + ½ I w²

angular and linear velocity are related

         v = w r

        w = v / r

         

indicate that the moment of inertia is

         I = ½ m r²

we substitute

          M g h = ½ M v² + ½ (½ m r²)  (v/r) ²

          ½ v² (M + [tex]\frac{1}{2}[/tex] m) = M g h

          v² = [tex]2gh \ \frac{M}{M + \frac{m}{2} }[/tex]

let's calculate

          v = [tex]\sqrt{ 2 \ 9.8 \ 2 \ \frac{70}{70 + 1.5} }[/tex]

          v = 6.195 m / s

Is this right or they wrong definitions which ones are the right ones someone !!!!!

Answers

Answer:

They are right.

Explanation:

Answer:

Mechanical Energy : KE + PE

Conversion : "When energy transfers from one form to another"

Potential Energy: the energy possessed by a body by virtue of its position relative to others , stresses within itself, electric charge , and other factors .'

Kinetic Energy: energy of an object in motion

Law of conservation of energy: KE+PE+friction=KE

Explanation:

First of all mechanical energy is kinetic energy plus potential energy (it is the energy of movement) So:

Mechanical Energy : KE + PE

Conversion is when energy converts or becomes a different form. So:

Conversion : "When energy transfers from one form to another"

Potential energy is stored energy, in Physics I or AP Physics I, it is often due to it being at a height, but batteries, foods, etc. are also example of it, so:

Potential Energy: the energy possessed by a body by virtue of its position relative to others , stresses within itself, electric charge , and other factors .'

Kinetic energy is for objects in motion so you got it right!

Kinetic Energy: energy of an object in motion

The law of conservation of energy means there is the same amount of energy before, as there is after, so when you see an equation with energy on both sides, it is usually this. Also, this is the last question left, so this has to be the answer.

Law of conservation of energy: KE+PE+friction=KE

Plz help, will Mark brainliest. A 20.0 Ohm and 60.0 Ohm resistor are connected in series to a 9.00 V battery. How much current flows out of the battery?
(Unit = A)

Answers

Answer: 0.1125A

Explanation:

We need to know the total resistance which will be:

R = R1+R2

where,

R1 = 20 ohm

R2 = 60 ohm

Therefore, R = 20 + 60 = 80 ohm

We should note that from Ohm's law

v = iR

i = v/R

where,

i = 9/80 = 0.1125 A

Therefore, the current that flows out of the battery is 0.1125A

List two types of current

Answers

Answer:

static and current electricity

As air pressure decreases, what happens
to the density of the atmosphere?
A. increases
B. decreases
C. stays the same
D. not enough information to tell

Answers

Answer:

Increases

Explanation:

This is because if there are more gas particles available within a given area, the greater the air density. Therefore, the greater the air density, there will be more gas particles per unit volume. Which indicate that there are more air particles bombarding with a surface. Therefore, the air pressure increases as the air density increases and visa versa.

3. Provide two examples of static electric charge.

Answers

Answer: 1.  walking across a carpet and touching a metal door handle            2. pulling your hat off and having your hair stand on end.

Explanation

:)

A uniform electric field is present in the region between infinite parallel plane plates A and B and a uniform electric field is present in the region between infinite parallel plane plates B and C. When the plates are vertical, is directed to the right and to the left. The signs of the charges on plates A, B and C may be:_______.
A. -- ,--, --
B. + , -- , --
C. + , --, +

Answers

Answer:

the correct answer is C     +, - , +

Explanation:

The electric field for positive charges is outgoing and for negative charges it is directed towards the charge.

Let's apply this to our case:

On plates A and B the field goes to the right, therefore plate A must be positive

In plates B and C the way it goes to the left, so the field is selected from it, this implies that plate C is positive

therefore plate B must be negative for both cases

when checking the correct answer is C     +, - , +

Which waves can travel through space?
a. Electromagnetic waves only
b. Mechanical waves only
c. Electromagnetic and mechanical waves
d. Longitudinal and electromagnetic waves

Answers

Answer:

electromagnetic waves only

Explanation:

I just took the test, Hope it helps!

Answer:

A: Electromagnetic waves only

Explanation:

Ilus
A bat bounces a sound wave off a
bird 19.4 m away, and hears the
echo after 0.117 s. What is the
speed of sound in the air?
(Hint: It is not necessarily 343 m/s.)
(Unit = m/s)

Answers

Answer:

332m/s

Explanation:

We know the formula for velocity is v=d/t, but this particular question is asking about an echo. The problem tells us the distance to the bird from the bat, which is 19.4. To find the echo, you need to find the distance from the bat to the bird and back.

So

19.4 x 2 = 38.8

And then plug into the equation

v = 38.8 / 0.117

= 332m/s

In the process of fluorescence, a molecule in its ground state will absorb a photon with a certain energy Eex, called the excitation energy, and then emit a photon with energy Eem, the emission energy. Obviously, the molecule cannot emit more energy than was absorbed.
a) You wrote in your notebook that you excited a fluorescent bead at λ = 640 nm and found that it fluoresced (emitted) at λ = 520 nm. Or was it the other way around? Do a calculation and see if those figures should be switched.
b) You shine a laser on your bead at the excitation wavelength. The laser has a power of 1 mW. How many photons are emitted by the laser in one second?
c) You wish to build a microscope that allows you to excite the bead at its excitation wavelength and detect the fluorescence at the emission wavelength. This requires a pair of filters, one that allows only the excitation wavelength to pass through, and the other that allows only the emission wavelength to pass through. What color will these filters appear (i.e. what color light will they transmit?) You may have to look at an electromagnetic spectrum to figure this out…

Answers

Answer:

a) the excitation energy is E₂ λ = 520 nm

the emission energy is E₁,  λ= 640 nm

b)   #_photons = 2.6 10¹⁸ photons,  

c) he excitation wavelength   λ = 520 nm is green, therefore the filter is also green

the emission wavelength is lam = 640 nm is orange

Explanation:

a) the energy of a photo is given by the planck relation

           E = h f

the speed of light is

          c = λ f

          f = c /λ

we substitute

          E = hc /λ

let's calculate the energy for the two photons

λ = 640 nm = 640 10⁻⁰ m

            E₁ = 6.63 10⁻³⁴ 3 10⁸/640 10⁻⁹

            E₁ = 3.1 10⁻¹⁹ J

λ = 520 nm = 520 10⁻⁹ m

             E₂ = 6.63 10⁻³⁴ 3 10⁸/520 10⁻⁹

            E₂ = 3.825 10⁻¹⁹  J

therefore the excitation energy is E₂ λ = 520 nm

the emission energy is E₁,  λ= 640 nm

b) For this part let's use a direct proportion rule (rule of three). If a photon (lam = 520 nm) has an energy of 3.825 10⁻¹⁹ J, how many photons have an energy of E = 1 10-3 J. Remember that the power is the energy per unit of time

        #_photons = 1 10⁻³ J (1 photon / 3.825 10⁻¹⁹ J)

        #_photons = 2.6 10¹⁸ photons

c) the excitation wavelength   λ = 520 nm is green, therefore the filter is also green

the emission wavelength is lam = 640 nm is orange

The diagram shows two bowling balls, A and B, each having a
mass of 7 kilograms, placed 2 meters apart.
А
B
7.00 kg
(7.00 kg
-2.00 m
What is the magnitude of the gravitational force exerted by ball
A on ball B?
(A) 8.17x10-9 N
(B) 1.63x10-9N
(C) 8.17x10-10 N
(D) 1.17 x 10-10 N

Answers

Answer:

b) 1.63x10-9N

Explanation:

i dont think this is right answer

Answer:

C or 8.17x10^-10 N

Explanation:

I was looking for the explanation but the answer is correct.

Relative velocity: An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination of the flight. The plane flies with an air speed of 120 m/s. If a constant wind blows at 10.0 m/s due west during the flight, what direction must the plane fly relative to north to arrive at the destination

Answers

Answer:

4.78 degrees east of north

Explanation:

The computation is shown below:

The Speed of the plane is 120 m/s

The Speed of the wind is 10 m/s

Sin [tex]\theta[/tex] = 10 ÷ 120

= 1 ÷ 12

= 0.0833

Now

\theta = 4.78 degrees

It should be 4.78 degrees east of north

Since the wind blows due west plane and it should fly east of north to  reach at the final destination

a student starts his lawnmower by applying a constant tangential force of 150 N to the 0.3 kg disk-shaped flywheel. the radius of the flywheel is 18 cm. what is the flywheels angular acceleration? b. what is the angular speed of the wheel after it has turned through one revolution,( neglect friction and motor compression.) c. what is the tangent speed of a point on the rim of the flywheel? ​

Answers

Answer:

okay

Explanation:

please I don't know

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