I was reading an old thermodynamics textbook and came across this equation describing change in internal energy. What does the (dU/dT)V and (dU/dV)T mean? I think it’s partial derivatives, but I’ve never seen a derivative sub (something). Please help

Answer:

B. 15 miles an hour going west

Answer:

1210 ohm

Explanation:

Given :

P=40 W

V=220 V

Now,

[tex]P=\frac{V^{2} }{R} \\40=\frac{(220)^{2} }{R} \\40R=48400\\R=\frac{48400}{40} \\R=1210 ohm[/tex]

Therefore, resistance of bulb will be 1210 ohm

Answer: 90

Explanation: a=Fnet/M

=27/0.30

=90

the acceleration is 90

Answer:

The acceleration of the ball is 83.333ms2 [forward].

Explanation:

i hope it helps :)

Answer:

D I think I’m not for sure

Explanation:

Answer:

The Escape Velocity Of Earth is

11.19 km/s

Explanation:

Hope it Helps!

Answer:

The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.

Explanation:

Given that,

The total mass of the wheelbarrow and the road is 80 kg.

The weight of an object is given by :

W = mg

where

g is acceleration due to gravity

So,

W = 80 × 9.8

= 784 N

So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.

Answer:

arge number of electrons free to move between the charged ions in the lattice.

Explanation:

The metallic bond occurs when an atom with few electrons is united in its last level, therefore the best way to decrease the total energy of the system is to lose all its electrons to remain with the configuration of a noble gas. The electrons that it loses cannot be acquired by other atoms since they all have few electrons, thus leaving a large number of electrons free to move between the charged ions in the lattice.

Some important characteristics emerge from this description of the metallic bond:

* It has many free electrons therefore its electrical conductivity is high

* As the charged ions are fixed, the material can be malleable, bent without breaking since the free electrons create the bond that keeps the system stable.

* As the electrons are free when heating a part of the material, these electrons acquire energy and rapidly propagate it to the other side, giving a high thermal conductivity

* As the temperature increases, the electrons acquire more kinetic energy, which is why there are more collisions between them and consequently the resistivity of the material increases.

Answer:

From Earth, the Sun looks like it moves across the sky in the daytime and appears to disappear at night. This is because the Earth is spinning towards the east. The Earth spins about its axis, an imaginary line that runs through the middle of the Earth between the North and South poles.

Answer:

∴ [T]=[WF−1V−1]

Hope this answer is right!!

Answer :

[T] = [W(F)^-1(V)^-1]

Answer:

[tex]V_2=3.3m/s[/tex]

Explanation:

From the question we are told that:

Distance [tex]d_1=1.4m[/tex]

Tangential speed [tex]V=2.2m/s[/tex]

Distance 2 [tex]d_2=2.1m[/tex]

Generally the equation for Angular velocity is mathematically given by

 [tex]w=\frac{v}{r}[/tex]

Therefore

 [tex]\frac{v_1}{r_1}=\frac{v_2}{r_2}[/tex]

 [tex]V_2=\frac{2.2*2.1}{1.4}[/tex]

 [tex]V_2=3.3m/s[/tex]

Answer:

We need, mass, gravity and height.

Explanation:

When a body falls freely from a height, its initial velocity is zero, but due to the height it has some potential energy at the top and the kinetic energy is zero.

As it falls, the potential energy is gradually converted in to the kinetic energy so that the total energy of the falling body is conserved.

At the time as the body strikes the ground, the entire potential energy is converted into the kinetic energy.

Potential energy is given by

U = m g h

where, m is the mass, g is the gravity and h is the height,

So, to get the kinetic energy we require mass, gravity and height of the body.

Answer:

1. 5.825 × 10¹⁷ J

2. i. $ 29.125 billion ii. I would not invest in the company

3. A nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.

4. a i. The momentum will not be relativistic

ii. This is because objects with large masses do not move at relativistic speeds

b i. 155 m/s

ii. This speed wouldn't be a problem for the ship.

Explanation:

1.  A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?

The kinetic energy of the payload is K = (γ - 1)mc² where m = mass of payload = 1 kg, c = speed of light = 3 × 10⁸ m/s and γ = 1/√(1 - β²) where β = 0.9 (since the payload moves at 90 % speed of light)

So, K = (γ - 1)mc²

= (1/√(1 - β²) - 1)mc²

= (1/√(1 - (0.9)²) - 1) × 1 kg × (3 × 10⁸ m/s)²

= (1/√(1 - 0.81) - 1) × 1 kg × (3 × 10⁸ m/s)²

= (1/√0.19 - 1) × 1 kg × (3 × 10⁸ m/s)²

= (1/0.436 - 1) × 1 kg × (3 × 10⁸ m/s)²

= (2.294 - 1) × 1 kg × (3 × 10⁸ m/s)²

= 1.294 × 1 kg × 9 × 10¹⁶ m²/s²

= 11.65 × 10¹⁶ kgm²/s²

= 1.165 × 10¹⁷ J

Let E be the total electrical energy of the rail gun. Since 20 % of this energy is converted to kinetic energy of the payload, we have

20 % of E = K

0.2E = K

E = K/0.2

= 1.165 × 10¹⁷ J/0.2

= 5.825 × 10¹⁷ J

2 Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?

i. how much would this launch cost?

Since the total energy required is E = 5.825 × 10¹⁷ J = 5.825 × 10¹¹ MJ and it costs 5 cent/MJ. So the total cost of energy will be total energy rate = 5.825 × 10¹¹ MJ × 5 cent/MJ = 29.125 × 10¹¹ = 2.9125 × 10¹² cents. Converting this to dollars, we have 2.9125 × 10¹² cents/100 cents/dollar = 2.9125 × 10¹⁰ dollars = 29.125 × 10⁹ dollars = 29.125 billion dollars = $ 29.125 billion

ii. Would you invest in this company?

I would not invest in the company

3. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.

Since the kinetic energy of the payload is 1.165 × 10¹⁷ J and a nuclear explosion generates about 10¹⁵ J of energy,  then a nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.

4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the rail gun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).

a. Will the recoil momentum of the ship be relativistic? Justify your argument.

i. Will the recoil momentum of the ship be relativistic?

The momentum will not be relativistic.

ii. Justify your argument.

This is because objects with large masses do not move at relativistic speeds. Since the speed cannot be relativistic, its momentum which is the product of mass and speed is non-relativistic

b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?

i. At what speed will the ship recoil after it launches a probe?

Since the total energy of the payload E' = K + mc² = 1.165 × 10¹⁷ J + 1 kg × (3 × 10⁸ m/s)² = 1.165 × 10¹⁷ J + 1 kg × 9 × 10¹⁶ m²/s² = 11.65 × 10¹⁶ J + 9 × 10¹⁶ J = 20.65 × 10¹⁶ J

Also, E'² = (pc)² + (mc²)² where p = momentum of payload

So, making p subject of the formula, we have

(pc)² = E'² - (mc²)²

pc = √[E'² - (mc²)²]

p = √[E'² - (mc²)²]/c

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

p = √[E'² - (mc²)²]/c

p = √[(20.65 × 10¹⁶ J)² - 1kg × (3 × 10⁸ m/s²)²]/3 × 10⁸ m/s

p = √[(20.65 × 10¹⁶ J)² - (1kg × 9 × 10⁸ m²/s²)²]/3 × 10⁸ m/s

p = √[426.4225 × 10³² J² - 81 × 10³² J²]/3 × 10⁸ m/s

p = √[345.4225 × 10³² J²]/3 × 10⁸ m/s

p = 18.59 × 10¹⁶/3 × 10⁸ m/s

p = 6.20 × 10⁸ kgm/s

From the law of conservation, this momentum of the payload equals the momentum of recoil of the ship.

So, p = m'v where m' = mass of navy ship = 4,000 metric tons = 4,000 × 1000 kg = 4 × 10⁶ kg and v = speed of navy ship

So, v = p/m'

= 6.20 × 10⁸ kgm/s ÷ 4 × 10⁶ kg

= 1.55 × 10² m/s

= 155 m/s

ii. Do you think that this is a problem for the ship?

Since the ship's speed is 155 m/s, which is small for an object with such a large mass, this speed wouldn't be a problem for the ship.

Answer:

Localized convective lifting

Answer:

Marine geologists learn about the rocks and geologic processes of the ocean basins.

Answer:

parasites are creatures the gain benefit off of other animals usually harming them eg:ticks on dogs

Answer:

What is the answer bro idont now

Answer:

a) the distance between her and the wall is 13 m

b) the period of her up-and-down motion is 6.5 s

Explanation:

Given the data in the question;

wavelength λ = 26 m

velocity v = 4.0 m/s

a) How far from the wall is she?

Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;

x = λ/2

we substitute

x = 26 m / 2

x = 13 m

Therefore, the distance between her and the wall is 13 m

b) What is the period of her up-and-down motion?

we know that the relationship between frequency, wavelength and wave speed is;

v = fλ

hence, f = v/λ

we also know that frequency is expressed as the reciprocal of the time period;

f = 1/T

Hence

1/T = v/λ

solve for T

Tv = λ

T = λ/v

we substitute

T = 26 m / 4 m/s

T = 6.5 s

Therefore, the period of her up-and-down motion is 6.5 s

Let g be the acceleration due to gravity on the surface of the star. By Newton's second law, the gravitational force felt by the object has a magnitude of

F = GMm/r ² = mg

where

• G = 6.67 × 10⁻¹¹ Nm²/kg² is the gravitational constant,

• M = 2.08 × 10³⁰ kg is the mass of the star,

• m is the unknown mass of the object, and

• r = 6.73 × 10³ m is the radius of the star

Solving for g gives

g = GM/r ²

g = (6.67 × 10⁻¹¹ Nm²/kg²) (2.08 × 10³⁰ kg) / (6.73 × 10³ m)²

g ≈ 3.06 × 10¹² m/s²

The object is in free fall with uniform acceleration and starting from rest, so its speed after falling 0.0093 m is v such that

v ² = 2g (0.0093 m)

v = √(2g (0.0093 m))

v ≈ 240,000 m/s ≈ 240 km/s

Answer:

The second statement.

Answer:

The water usually rushes back too enthusiastically, causing a splash – and the bigger the rock, the bigger the splash. The splash then creates even more ripples that tend to move away from where the rock went into the water.

Answer:

The moon Phobos orbits Mars (m = 6.42 x 1023 kg) at a distance of 9.38 x 106 m.

When we exhale, 90% waste material is Carbon Dioxide ( CO2 ) , so, it gets exhaled out in the form of CO2 rich air and it gets removed from the body, therefore our internal body becomes more pure and helps in making our internal temperature constant at a suitable level.

Answer:

θ = 33.54 rad

Explanation:

This is a circular motion exercise

         θ= θ₀ + w₀ t + ½ α t²

suppose that for t = 0 the body is at its initial point θ₀ = 0 and from the graph we see that the initial angular velocity w₀ = -9.0 rad / s

we look for the angular acceleration,

         α = [tex]\frac{\Delta w}{ \Delta t}[/tex]

from the graph taken two points

         α = [tex]\frac{0 - (-9.0)}{3.0 - 0}[/tex]

         α = 3 rad / s²

we substitute in the first equation

            θ = 0 -9 t + ½ 3 t²

the displacement is requested for t = 8.6 s

           θ = = -9  8.6 + 3/2  8.6²

           θ = 33.54 rad

The angular displacement of the wheel from 0 to 8.6s is 33 radians.

Given to us

t = 8.6s

We know acceleration is the ratio of velocity and time, therefore, the slope of the velocity-time graph will give us acceleration, therefore,

At point t=3, ω =  0

At point t = 5, ω = 6

Acceleration = slope of the Velocity-time graph = 3 rad/sec²

Using the equation,

[tex]\theta = \omega_0t+\dfrac{1}{2}\alpha t^2[/tex]

SUbstitute values,

[tex]\theta = (-9.0\times 8.6)+\dfrac{1}{2}(3\times 8.6^2)\\\theta = 33\rm\ radians[/tex]

Hence, the angular displacement of the wheel from 0 to 8.6s is 33 radians.

Learn more about Angular displacement:

https://brainly.com/question/14613604

Answer:

Explanation:

This is a work problem...energy is created and used in the form of work.

W = FΔx where W is work, F is the force needed to move the object Δx in meters.

W = 110(140) ∴

W = 15000 J

Answer:

c. Only the linear acceleration is zero.

Explanation:

The linear acceleration is defined as the rate of change of linear velocity. Since the bicycle is moving in the same direction, with the same speed, without speeding up or slowing down. Therefore, there will be no change in linear velocity and as a result, linear acceleration will be zero.

The angular acceleration is the rate of change of angular velocity. Since the angular velocity is changing its direction constantly. Therefore, it has a certain component of acceleration at all times called centripetal acceleration.

Therefore, the correct option is:

c. Only the linear acceleration is zero.

Answer:

a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.

Explanation:

Since the tension in the rope, t is greater than the kinetic friction fk, the box is moving forward because there is a net force on it. That is, t - fk = f = ma.

Since there is a net force, there is an acceleration and thus an increasing velocity.

When the rope breaks, the tension, t = 0. So, t - fk = 0 - fk = -fk = ma'.

Now, the net force acting on the box is friction in the opposite direction. This force tends to slow the box down from its initial velocity at acceleration, 'a' until its velocity is zero, where it stops. Since the frictional force is constant, the acceleration, a' on the box is thus constant and the box undergoes uniform deceleration until its velocity is zero.

So, the box keeps its speed for a short while, then slows and stops. slows steadily until it stops.

So, the answer is a.