If you were capable of converting mass to energy with 100%, efficiency, how much mass would you need to produce 3.5x10^12 Joules of energy? ​

Answer:

The blue one.

Explanation:

Im pretty sure its because the blue is hotter.

Answer:

271 K

Explanation:

Answer:

The temperature of Nitrogen is 270.85 K.

Explanation:

A gas that follows Boyle's law , Charle's law and law of pressure , is called perfect gas.

Given data;

[tex]P =101.3 KPa , V = 0.08 m^{3} , \\ n = 3.6 mol , R = 8.31\\T = ?[/tex]

By the formula,

P.V = n .R.T

On putting values

[tex]1.013[/tex] x [tex]10^{5}[/tex] x [tex]0.08[/tex] = [tex]3.6[/tex] x [tex]8.31[/tex] x [tex]T[/tex]

So [tex]T =\frac{1.013*10^{5} * 0.08}{3.6 * 8.31}[/tex]

 [tex]T =\frac{8104}{29.92}[/tex]

[tex]T = 270.85 K[/tex]

Hence the temperature of Nitrogen is 270.85 K

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Answer:

Option D

Explanation:

The options for the question are

a) How do galaxies rotate?

b) What is the weather on Neptune?

c) What is the core of Saturn made of?

d) What other solar systems have planets?

Solution

The Hubble space telescope was designed and integrated into the extraterrestrial system in order to capture information about the surrounding universe. If the  Keck Observatory has a better observation capacity than the Hubble space telescope then the scientist would be interested to know the surrounding planets in the solar system.

Hence, option D is correct

Answer:

I got 2508 J

Explanation:

q=mct

125-95

= 30 * 4.18 * 20

= 2508

The energy principle states that:

[tex]\sf\purple{Energy \:cannot \:be \:created \:or\: destroyed.✅}[/tex]

[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{.}}}}}[/tex]

Answer:

Electromotive force or EMF is equal to the terminal potential difference when no current flows. EMF (ϵ) is the amount of energy (E) provided by the battery to each coulomb of charge (Q) passing through.

Answer:

This plasma mostly consists of electrons

Answer:

yes because there is also a theory about this I studied in biology

theory name -Lamarck's theory

Answer :light

Explanation:  

A television set is another device that operates by the transformation of energy. An electrical beam from the back of the television tube strikes a thin layer of chemicals on the television screen, causing them to glow. In this case, electrical energy is converted into light.

Answer:

t = 1.22 s

Explanation:

Given that,

The initial upward velocity component of a football = 12 m/s

The horizontal velocity component is 20 m/s

We need to find the time required for the football to reach the highest point of the trajectory. Let the time is t.

Using first equation of motion to solve it such that,

[tex]v=u+at[/tex]

u is initial velocity

v is final velocity

a = -g

so,

[tex]t=\dfrac{u}{g}\\\\t=\dfrac{12}{9.8}\\\\t=1.22\ s[/tex]

So, the required time taken by the football to reach the highest point is 1.22 seconds.

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)

Where:

[tex]I[/tex] - Impulse, in kilogram-meters per second.

[tex]m[/tex] - Mass, in kilograms.

[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.

[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.

If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:

[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]

[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Answer:

Explanation:

PE = [tex]\frac{1}{2}k[/tex](Δ[tex]x^2[/tex])

where k is the spring constant and Δx is the displacement of the spring when an object is hung from it. Plugging in:

[tex]PE=\frac{1}{2}(2.5)(.2^2)[/tex] which gives us

PE = .05 J

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

we substitute

          F = k 4q² / 9 r²

          F = k q² r² 4/9

          F = ⅔ F₀

Answer:

A

Explanation:

A. The Moon orbits the Earth, and the Earth-Moon system orbits the Sun

B. The Earth and Moon both orbit the Sun separately

C. The Sun orbits the Earth-Moon system

D. The Moon revolves around Earth but not the Sun

Answer:

a)  3.0  10⁴ m / s, b) 3.7 10¹ m, c) 0.750 kg, d) 4 10¹² m²,  e)  75 m, f) 60 m²

g) 5.207 10³ m², e) 4.847 10⁷ s

Explanation:

The international system (SI) of measurements has as fundamental units the meter for length, the second for time and kilogram for mass.

Let's reduce the different magnitudes to the SI system

a) 30 km / h (1000m / 1 km) (1 h / 3600 s) = 3.0 10⁴ m / s

b) 37 Dm (10 m / 1 Dm) = 3.7 10¹ m

c) 750 g (1 kg / 10,000 g) = 0.750 kg

d) 4 10⁶ km² (1000 m / 1km) ² = 4 10¹² m²

e) 7500 cm (1 m / 100 cm) = 75 m

f) 600000 cm² (1m / 10² cm) ² = 60 m²

g) 520700000 mm³ (1 m / 10³ mm) ³ = 5.20700000 109/10 ^ 6

  = 5.207 10³ m²

e) 3.4 years (l65 days / 1 yr) (24 h / 1 day) (3600 s / 1h) = 4.847 10⁷ s

Answer:

The given potential energy of the spring is expressed as follows;

ΔUsp = -(WB + WW)

Where;

WB = Th work done by the spring on the block to which it is attached

WW = The work done by the spring on the wall

We recall that work done, W = Force applied × Distance moved in the direction of the force

The work done by the spring on the block, WB = The spring force × The distance the block moves

The work done by the spring on the wall, WW = The spring force × The distance the wall moves

However, given that the wall does not move, we have;

The distance the wall moves = 0

∴ The work done by the spring on the wall, WW = The spring force × 0 = 0 J

Therefore, WW = 0 J, and the spring does not do work on the wall, and WW can be ignored in the next subsequent) steps

Explanation:

Answer:

7. (D) uniformly accelerated vertical motion

8. (A) zero

9. (A) zero

10. (C) parabolic

Answer:

7.Uniformly accelerated vertical motion

8.0m/s²

9.9.8m/s

10.parabolic

11.vertical component.

Answer:

I honestly don't know but that thing looks cool lol

Answer:

gravity

Explanation:

Answer:

I am not sure if this is the answer

(B) what is the height of the building if it hits the ground after those 5 seconds.

Answer:

Divide the total measurement of the stack/ream of paper by the number of sheets. For example: 2 inches/100 pages = 0.01-inches sheets.

In case you measured the stack, and it gave you less than an inch, then you need to do the same. For example: 0.5-inches/100 sheets = 0.005-inches per sheet.

Answer:

Divide the total measurement of the stack/ream of paper by the number of sheets. For example: 2 inches/100 pages = 0.01-inches sheets.

In case you measured the stack, and it gave you less than an inch, then you need to do the same. For example: 0.5-inches/100 sheets = 0.005-inches per sheet.

Explanation:

Answer:

Explanation:

Gravity pulls everything down at the same rate of 9.8 m/s/s. If you're looking for the normal force, which is the same as the weight of the object, we'll find that, just in case.

w = mg which says that the normal force/weight of an object is equal to its mass times the pull of gravity:

w = 4.0(9.8) so

w = 39N

Answer:

Explanation:

We will use the KE equation you wrote here and fill in what we are given:

[tex]36=\frac{1}{2}m(12)^2[/tex] and isolating the m:

[tex]m=\frac{2(36)}{12^2}[/tex] which gives us

m = .50 kg

Answer:

Explanation:

Magnitude of frictional force = μ mg

μ is either static or kinetic friction.

To start the crate moving , static friction is calculated .

a ) To start crate moving , force required = μ mg where μ is coefficient of static friction .

force required =.517 x 56.6  x 9.8 = 286.76  N .

b ) to  slide the crate across the dock at a constant speed , force required

= μ mg where μ is coefficient of kinetic  friction , where μ is kinetic friction

= .26 x 56.6  x 9.8 = 144.21 N .

Answer:

a)   a = - 0.0833 m / s²,  b)   t = 4.4 s

Explanation:

a) this is a kinematics exercise where the acceleration is along the inclined plane

         v = v₀ - a t

         a = v₀ - v / t

         a = 3 - 8/60

         a = - 0.0833 m / s²

b) in this case the final velocity is zero

         v = v₀ - a t

         0 = v₀ - at

         t = v₀ / a

         t = 28 / 6.4

         t = 4.375 s

         t = 4.4 s

Answer:

The second one.

Explanation:

It caused both to change speed because they have both the same mass.

Answer:

The gravitational force between earth and any object is known as gravity.

...

Answer:

67.5

Explanation:

The plane accelerates at 2.7m/s,^2

Time is 25 seconds

The velocity can be calculated as follows

= 25×2.7

= 67.5

Hence the speed f the plane is 67.5

Answer:

0.893 rad/s in the clockwise direction

Explanation:

From the law of conservation of angular momentum,

angular momentum before impact = angular momentum after impact

L₁ = L₂

L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)

L₂ = angular momentum of cylinder and angular momentum of bullet after collision.

L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision

So,

L₁ = L₂

L₁ = (I₁ + I₂)ω

ω = L₁/(I₁ + I₂)

ω = L₁/(1/2MR² + mR²)

ω = L₁/(1/2M + m)R²

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

ω = L₁/(1/2M + m)R²

ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²

ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)

ω = + 9 kgm²/s/(2.52 kg)(4 m²)

ω = +9 kgm²/s/10.08 kgm²

ω = + 0.893 rad/s

The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.

Answer:

[tex]W=17085KJ[/tex]

Explanation:

From the question we are told that:

Height [tex]H=16m[/tex]

Radius [tex]R=3[/tex]

Height of water [tex]H_w=9m[/tex]

Gravity [tex]g=9.8m/s[/tex]

Density of water [tex]\rho=1000kg/m^3[/tex]

Generally the equation for Volume of water is mathematically given by

 [tex]dv=\pi*r^2dy[/tex]

 [tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 [tex]dw=(pdv)g (H-y)[/tex]

Substituting dv

 [tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]

 [tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]

Therefore

 [tex]W=\int dw[/tex]

 [tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]

 [tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]

 [tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]

 [tex]W=3420.84*0.25[2401-65536][/tex]

 [tex]W=17084965.5J[/tex]

 [tex]W=17085KJ[/tex]

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