an observer sees two spaceships flying apart with speed .99c. What is the speed of one spaceship as viewed by the other

A.

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

Substituting the values of the variables into the  equation, we have

Q = mcΔT

Q = ρVcΔT

Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C

Q = 482630400 J

So, the energy of the hot water is 482630400 J

B.

The elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

Substituting the values of the variables into the equation, we have

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.

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Answer:

a

Explanation:

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Answer:

Albert Einstein WAS a very well known Theoretical physicist

The USA claims he did not ever get his hands directly on an atomic bomb but in fact, other country textbooks like in Germany say he did.

Explanation:

A fun fact is that he was hired by the United States to make the Atomic bomb. Albert Einstein was a german yet many believed him to be a Smart American, they were wrong.

Answer:

Explanation:

Δx means a change in the magnitude of the x variable, often used in reference to a number line on the horizontal axis of a plot.

Answer:

Light can travel in three ways from a source to another location: (1) directly from the source through empty space; (2) through various media; (3) after being reflected from a mirror.

Explanation:

Answer:

the answer is weight=10N

Answer:

[tex]\boxed {\boxed {\sf 9.8 \ Newtons}}[/tex]

Explanation:

Weight is also called the force of gravity. This force acts on all objects at all times, pulling them down toward the center of the Earth.

It is calculated by multiplying the mass by the acceleration due to gravity.

[tex]F_g=mg[/tex]

The mass of the object is 1 kilogram. This scenario is occurring on Earth, so the acceleration due to gravity is 9.8 meters per second squared.

Substitute the values into the formula.

[tex]F_g= 1 \ kg *9.8 \ m/s^2[/tex]

Multiply.

[tex]F_g= 9.8 \ kg*m/s^2[/tex]

Convert the units. 1 kilogram meter per second squared is equal to 1 Newton, so our answer of 9.8 kilogram meters per second squared is equal to 9.8 Newtons.

[tex]F_g= 9.8 \ N[/tex]

A 1 kilogram mass at Earth's surface weighs 9.8 Newtons.

Answer:

Explanation:

conservation of momentum during the collision

0.035(214) + 0.15(0) = 0.185v

v = 40.486 m/s

The kinetic energy after impact will convert to gravity potential energy

(ignoring air resistance)

mgh = ½mv²

     h = v²/2g

     h = 40.486² / (2(9.8))

     h = 83.6303...

     h = 84 m

Answer:

Explanation:

a) The spring force will equal the weight.

b) If up is positive

kx - mg = 0

mg = kx              kx = 25 N

c) m = kx/g = 25/10 = 2.5 kg

Answer:

jsbdhdndmlsusgsbkaksudgnslsosufhbf ffb

The magnitude of the load L on the crane before the collapse is 3211.81 N

To determine the magnitude of the load on the crane (L), we will need to make use of the equilibrium conditions of the torque.

It is always an ideal process to list out all the parameters given as this will let you understand how you can determine the answer to the question from the given parameters.

From the given information;

From the question;

the angle at which the crane is positioned can be determined by taking the tangent of the angle θ. i.e.

[tex]\mathbf{tan \ \theta = \dfrac{h}{d-s}}[/tex]

[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{h}{d-s} \Big )}[/tex]

[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{2.070 }{5.580 - 0.522} \Big )}[/tex]

[tex]\mathbf{\theta =22.26^0}[/tex]

Consider the equilibrium conditions of the torques with respect to the magnitude of the load at point P.

∴

[tex]\mathbf{Tsin \theta (d-s) - W_L (d-x) -(Mg) (\dfrac{d}{2}) = 0}[/tex]

By making the magnitude of the load [tex]\mathbf{W_L}[/tex] the subject of the formula, we have:

[tex]\mathbf{W_L = \dfrac{Tsin \theta (d-x) -(Mg) (\dfrac{d}{2})}{ (d-s) } }[/tex]

[tex]\mathbf{W_L = \dfrac{(11650 )sin (22.26) (5.580-1.350) -(88.50\times 9.81) (\dfrac{5.580}{2})}{ (5.580-0.522) } }[/tex]

[tex]\mathbf{W_L = 3211.81 \ N }[/tex]

Therefore, we can conclude that the magnitude of the load is 3211.81 N

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Hi there!

Part A:

The only quantities explicitly given to us are:

3. mass (m)

4. Maximum velocity (vmax)

5. Amplitude (A)

8. Position x from equilibrium

Part B:

To solve, we must begin by calculating the force constant, 'k'.

We can use the following relationship:

[tex]v = \sqrt{\frac{k}{m}(A^2-x^2)[/tex]

We are given the max velocity which occurs at a displacement of 0 m, because the mass is the fastest at the equilibrium point. We can rearrange the equation for k/m:

[tex]\frac{v^2}{(A^2-x^2)} = \frac{k}{m}[/tex]

[tex]\frac{3.2^2}{(0.06^2-0)} = \frac{k}{m} = 2844.44[/tex]

Now, we can find the velocity at 4.7cm (0.047m) using the equation:

[tex]v = \sqrt{(2844.44)(0.06^2-0.047^2)} = 1.989 m/s[/tex]

Plug this value into the equation for kinetic energy:

[tex]KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(0.05)(1.989^2) = \boxed{0.0989 J}[/tex]

Part C:

The potential energy of a spring is given as:

[tex]U = \frac{1}{2}kx^2[/tex]

Find 'k' using the derived quantity above:

[tex]\frac{k}{m} = 2844.44\\\\k = 2844.44m = 142.22 N/m[/tex]

Now, calculate potential energy:

[tex]U = \frac{1}{2}(142.22)(0.047^2) = \boxed{0.157 J}[/tex]

Answer:

heat

Explanation:

a. Assuming all energy involved is conserved, at the lowest point of the swing (which includes the moment the student grabs the rope), the student only has kinetic energy,

K = 1/2 m (3.5 m/s)²

and at the highest point of the swing, the student only has potential energy

P = mgh

The energies at the bottom and top of the swing must be equal, so

1/2 m (3.5 m/s)² = mgh

h = (3.5 m/s)² / (2g)

h = 0.625 m ≈ 63 cm

b. In part (a), we found the relationship

h = v²/(2g)

If we cut the speed v in half, we replace v in the equation above with v/2 :

h = (v/2)²/(2g)

and simplifying this gives

h = (v²/4)/(2g) = 1/4 • v²/(2g)

The factor of 1/4 tells you that reducing the speed by a factor of 1/2 reduces the height by a factor of 1/4. So he can swing as high as

1/4 (3.5 m/s)²/(2g) = 0.15625 m ≈ 16 cm

Answer:

440 m

Explanation:

S=(u+v) t / 2

S = (11+33) × 20/2

S= 44× 20/2

S=440 m

Answer:

circuit

Explanation:

Answer:

it's ahfdfhhh hhgfdjjjjuyggffdddcff

I think static is the correct answer

Answer:

Remember, NORTH ^, EAST >, SOUTH v, WEST <

Explanation:

It doesn't have to be a super complex answer. All you have to do is look to the left (west) of the North American plate. What are the 2 plates that you see? The Pacific and the Juan de Fuca, yeah? To the South, there is the Cocos amongst a few others.

I am not going to share the answer for sure as I haven't completed the test yet but that's how I'm solving it. You should write the answer in your own words anyways. Hope this helps! Have a good day :)

Answer:

The Juan de Fuca Plate and the Pacific Plate both border the west side of the North American Plate.

Explanation:

Edmentum

Hi there!

We can begin by using the work-energy theorem in regards to an oscillating spring system.

Total Mechanical Energy = Kinetic Energy + Potential Energy

For a spring:

[tex]\text{Total ME} = \frac{1}{2}kA^2\\\\\text{KE} = \frac{1}{2}mv^2\\\\PE = \frac{1}{2}kx^2[/tex]

A = amplitude (m)

k = Spring constant (N/m)

x = displacement from equilibrium (m)

m = mass (kg)

We aren't given the mass, so we can solve for kinetic energy by rearranging the equation:

ME = KE + PE

ME - PE = KE

Thus:

[tex]KE = \frac{1}{2}kA^2 - \frac{1}{2}kx^2\\\\[/tex]

Plug in the given values:

[tex]KE = \frac{1}{2}(20)(0.3^2) - \frac{1}{2}(20)(0.3^2) = \boxed{0 \text{ J}}[/tex]

We can also justify this because when the mass is at the amplitude, the acceleration is at its maximum, but its instantaneous velocity is 0 m/s.

Thus, the object would have no kinetic energy since KE = 1/2mv².

Answer:D: the velocity is zero

Explanation:

Answer:

Explanation:

1 Object C has an acceleration that is greater than the acceleration for D.

2 B

3 17M

4 The velocity is zero.

5 a straight line with negative slope

just took it

Answer:

The objects outside the reference frame aren't moving. It appears this way since the vehicle you are inside is moving, but unless the objects are people, animals, or other vehicles, the objects aren't moving.

Answer:

[tex]1.74\times10^3 kg; 100m[/tex]

Explanation:

Step a: mass of the car. Let's grab the definition of kinetic energy: [tex]K= \frac12 mv^2[/tex]. We have K, we have v (which we should convert in meters per second, dividing by 3.6) to get:[tex]4.22\times10^5 = \frac12m(22.03)^2 \rightarrow m= 2\times4.22 / 495.22 \times 10^5 = 1.74 \times 10^3 kg[/tex]

Point a is done.

Now for the (b)reaking part. (I'm sorry, it's an horrible joke, but I couldn't resist)

In theory we have the mass, we have the force, so we could find the acceleration, find how long it takes to slow down, and then find the distance traveled. Too long. Let's do things more easily: when the car slows down to 56 km/h it will have a different kinetic energy. The difference in kinetic energy is the work done by the breaking force ofer the slowing distance.

[tex]K_f-K_i=W[/tex] A quick note on signs: if you look carefully the final kinetic energy will be less than the initial value, thus the work will be negative: it means it's correct, since the work is against the motion, slowing it down. Let's get calculating, first by converting 56 kmh in m/s (15,56 m/s), then finding the final kinetic energy:

[tex]K_f =\frac12 (1.74\times10^3) (15.56)^2 =2.11 \times 10^5 J[/tex]

The difference will be the work done by the force, or

[tex](2.11 - 4.22) \times 10^5 = \vec F\cdot \vec x=Fx[/tex] where  we are assuming that force and displacement have the same line of actions to simplify the dot product.

[tex]2.11\times 10^5 = 2100x = 1.00\times 10^2 m[/tex]

Answer:

In 1924 Louis de Broglie introduced the idea that particles, such as electrons, could be described not only as particles but also as waves. This was substantiated by the way streams of electrons were reflected against crystals and spread through thin metal foils.

Explanation:

I know I probably didn't answer your question, I just used all of my knowledge that I learned about Louie De Broglie. Hope it helps!

Answer:

200 kilocalories

Explanation:

Answer:

.192 kg x m^2

Explanation:

I= mass of a times radius of a squared + mass of b times radius of b squared +...

I= .6 kg x .4m^2 + .6 kg x .4m^2

= .192 kg x m^2

Hope this helps :)

Answer:

Explanation:

F = ma

F = 2.60(14.0)

F = 36.4 N

Answer:

Explanation:

Without friction, a roller coaster continuously converts potential energy to kinetic energy and back again. Total energy will be constant.

Let m be the mass of the car and ground level is the origin.

on the 5.5 m hill, total energy is

E = PE + KE

E = mgh + ½mv²  

E = m(9.8)(5.5) + ½m(9.3)² = 97m J

a) The maximum height will occur when the total energy is all potential energy.

E = mgh

h = E/mg

h = 97m/m(9.8) = 9.9 m  

As this value is greater than the height of the third hill at 5.5 + 4.0 = 9.5 m The car will cross the last hill with some remaining velocity in kinetic energy.

b) As 9.5 m is greater than 9.3 m, the 9.5 m hill will have more of the total energy of the system as potential energy, This mean there is less kinetic energy and therefore less velocity (and speed) on top of the 9.5 m hill.

c) KE = E - PE

KE = 97m - m(9.8)(9.5 - 1.0)

KE = 97m = 83.3m

KE = 13.7m = ½mv²

v² = √(2(13.7)

v = 5.2345...

v = 5.2 m/s

Answer:it’s C

Explanation:

by using graph of velocity vs. time for two objects, Item 4 and Item 5 statement makes an accurate comparison of the motions for objects. thus option C is correct.

velocity is the  rate of change of the position of the object with respect to  reference and it is complicated but velocity is basically speeding a particular object in a specific direction.

Velocity is a vector quantity which means both magnitude (speed) and direction are combinedly define define velocity. The SI unit of velocity is meter per second (ms-1) and the magnitude or the direction of velocity of a body changes leads to acceleration.

Speed and velocity are the two closest term but the major difference between speed and velocity is that speed gives us an idea that the object with the faster rate of movement r where as  velocity  speed up as well as  tells us the direction of the body

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