Which computational model can be applied to the total internal energy of a system with two point masses?

{Energy Flows Quick Check}

Answer:

Yes, geothermal energy is a renewable energy resource because the water can be heated by pumping it through the rocks.

Answer:

2

Explanation:

there are 2 significant figures in there

Answer:

1. A. 8.22. m/s

Explanation:

Answer:

A) 140 k

b ) 5.22 *10^3 J

c) 2910 Pa

Explanation:

Volume of Monatomic ideal gas = 1.20 m^3

heat added ( Q ) = 5.22*10^3 J

number of moles  (n)  = 3

A ) calculate the change in temp of the gas

since the volume of gas is constant no work is said to be done

heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT

make ΔT subject of the equation

ΔT = Q / n.(3/2).R

    = (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )

    = 140 K

B) Calculate the change in its internal energy

ΔU = Q  this is because no work is done

therefore the change in internal energy = 5.22 * 10^3 J

C ) calculate the change in pressure

applying ideal gas equation

P = nRT/V

therefore ; Δ P = ( n*R*ΔT/V )

                        = ( 3 * 8.3144 * 140 ) / 1.20

                        = 2910 Pa

A) The change in the temperature of the gas is; ΔT = 139.5 K

B) The change in internal energy of the gas is; ΔU = 5.22 × 10³ J

C) The change in pressure of the gas is; ΔP = 2899.5 Pa

We are given;

Volume of Monatomic ideal gas; V = 1.2 m³

Amount of heat added; Q = 5.22 × 10³ J

number of moles; n = 3

A) To calculate the change in temperature of the monatomic idea gas, we will use formula;

Q = ³/₂nRΔT

Where R is a constant = 8.314 J/mol.K

ΔT is the change in temperature

Making ΔT the subject of the formula;

ΔT = ²/₃(Q/(nR))

ΔT = ²/₃(5.22 × 10³)/(3 × 8.314)

ΔT = 139.5 K

B) Due to the fact that no work was done, then from first law of thermodynamics, we can say that;  

ΔU = Q

Thus;

change in internal energy; ΔU = 5.22 × 10³ J

C) The change in pressure will be calculated from the formula;

ΔP = (n*R*ΔT)/V

ΔP = (3 * 8.314 * 139.5)/1.2

ΔP = 2899.5 Pa

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Answer:

ur answer is d

Explanation:

it has to be coiled to give off magnetic energy

Answer:

The final velocity of the car is 16.893 m/s

The final velocity of the truck is 4.393 m/s

Explanation:

Given;

mass of the car, m₁ = 715 kg

mass of the truck, m₂ = 1490 kg

initial velocity of the car, u₁ = 0

initial velocity of the truck, u₂ = 12.5 m/s

let the final velocity of the car, = v₁

let the final velocity of the truck, = v₂

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(715 x 0) + (1490 x 12.5) = 715v₁ + 1490v₂

18625 = 715v₁ + 1490v₂ -----equation (1)

Apply one-directional velocity formula;

u₁ + v₁ = u₂ + v₂

0 + v₁ = 12.5 + v₂

v₁ = 12.5 + v₂

Substitute v₁ into equation (1)

18625 = 715(12.5 + v₂) + 1490v₂

18625 =8937.5 + 715v₂ + 1490v₂

18625 - 8937.5 = 715v₂ + 1490v₂

9687.5 = 2205v₂

v₂ = 9687.5 / 2205

v₂ = 4.393 m/s

solve for v₁

v₁ = 12.5 + v₂

v₁ =  12.5 + 4.393

v₁ = 16.893 m/s

Answer:

See explanation

Explanation:

Given that the density of the unknown substance is 0.79 g/cm3 and is soluble in water, the possible substances it could be are;

i) t-butanol

ii) ethanol

iii) 2-propanol

iv) acetone

However, the actual identity of the unknown substance can be obtained by carrying out a boiling point test. The four substances listed above have different boiling points. Hence the boiling point of the unknown substance ultimately discloses its identity.

[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]

➡ 15.85/5

➡ 3.17 ans.

Since, no external force is acting , so the system is in equilibrium .

Initial total energy = Final total energy

[tex]mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s[/tex] ( Here , g = acceleration due to gravity = 9.8 m/s² )

Therefore, option 1) is correct.

Hence, this is the required solution.

Answer:

$4.00

Explanation:

The cost per pound is $4.00, $11 divided by 2.75 is 4.

Answer:

Th cost per pound is $4.00.

Explanation:

11/2.75= 4

Answer:

option c is correct

Explanation:

we know that

2as=vf^2-vi^2

vf=24 m/s

vi= 0 m/s

a=g= 9.8 m/s^2

s=vf^2-vi^2/2a

s=(24)²-(0)²/2*9.8

s=576/19.6

s=29.4 m

therefore option c is correct

Voltage:

It is basically the difference between the charges of the materials on the ends of the Wire

also known as potential difference

It is very similar to the movement of air, it moves from higher density to lower density. in this case, the change in density is the potential difference

So, since voltage is the difference between the charge available on the ends of a wire. Even if the wire splits in parallel circuit, the difference of the charges remains the same

the more the potential difference, the faster electrons will move to the material with lower charge

Current:

Current is the amount of electrons moving through a cross-section of a wire in a period of time

So basically, it is the amount of electrons that move across a given point on a wire in a period of time

If the wire splits, we will have the same amount of electrons moving through as they would if the wire was not split but now, the electrons passing are divided and hence, if we measure the current after the split, we will find that we have a lower current

that's because we have less charge moving through the cross-section of the wire since some of those electrons are moving through a different wire

That's why the current splits in a parallel circuit

Answer:

Explanation:

m1v1=m2v2

m1=70 kg

m2=10 g=0.01 kg

v2=500 m/s

m1v1=m2v2

v1=m2v2/m1

v1=0.01*500/70

v1=0.07

Answer:half-life (T1/2) of this isotope =

Explanation:

The number of nuclei of any radioactive substance at a given time is expressed by

Nt = N0e⁻kt

Nt=decay of material  at a time t, =3110 decays per minute

N=decays at t=0, 8255 decays per minute

k=constant

Nt=N0e−kt

3110= 8255 e⁻k(4.50)

3110/ 8255=e−k(4.50)

0.3767 =  e−k(4.50)

In 0.3767  =   -k (4.50)

0.976=-4.5k

k=0.976/4.5

=0.2159

Also we know that t 1/2= time that it takes half the original material to decay.it is  related to the rate constant by

T₁/₂=ln  2 / k

Therefore half-life (T1/2) of this isotope

T₁/₂=ln  2/0.2159

T₁/₂=3.12 days

Answer:

ou have I=200mA, E=40J, t=30s, and you want to find the voltage drop.

First, you should know that  P=V⋅I , so V=PI

Second, you have the amount of energy converted in a certain amount of time, so E=P⋅t

So, find the power and use it to find the voltage drop.

this works , but i thought energy was defined by W = P * t whitch would then be P = W/t

[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the Kinematics in real world.

So, as given here, we have to find the Mass of the bus from the given momentum, so we get as,

P = m * V

momentum = mass * velocity

here, P= 152625 kgm/s and v= 11.1 m/s

so substituting we get as,

m = 152625 ÷ 11.1 => 13,750 kg

hence,the mass of the bus is 13,750 kg.

Answer:

sorry I wish I could it help you

Answer:

Car A would have a better average speed

Explanation:

added weight to a object that is self propelled will be slower than a identical object with no added weight

Let w, n, p, and f denote the magnitudes of the 4 forces acting on the box.

• w = weight = 319 N

• n = normal force

• p = pushing force = 485 N

• f = friction = µ n, where µ is the coefficient of static friction

The net force on the box points in the direction that the box moves, which is to the right. In particular, this means the box is vertically in equilibrium. Split up the vectors into their vertical and horizontal components, and apply Newton's second law. (I take up and right to be the positive vertical and horizontal directions, respectively.)

• vertical:

p sin(-35°) + n - w = 0

and solving for n,

- (485 N) sin(35°) + n - 319 N = 0

n ≈ 597 N

• horizontal:

p cos(-35°) - f = m a

where a is the magnitude of the net acceleration on the box. Solve for a. Since f = µ n and m = w / g (where g = 9.80 m/s² is the mag. of the acc. due to gravity) we get

p cos(35°) - µ n = (w / g) a

(485 N) cos(35°) - µ (597 N) = (319 N) / (9.80 m/s²) a

a ≈ (12.2 - 18.3 µ) m/s²

(a) If µ = 0.57, then the net acceleration on the box is

a ≈ (12.2 - 18.3 • 0.57) m/s² ≈ 1.75 m/s²

so that the time t required to move the box 4 m is

4 m = 1/2 a t ²

t ≈ √((8 m) / (1.75 m/s²))

t ≈ 2.14 s

(b) The box does not move.

If µ = 0.75, then

a = (12.2 - 18.3 • 0.75) m/s² ≈ -1.55 m/s²

but a negative acc. here means the applied acc. points *opposite* the direction of movement, thus making the box move backward which doesn't make sense. The coefficient of friction is too large for the given applied force to get the box moving. With µ = 0.75, the frictional force to overcome has mag. f ≈ 448 N. But the given push contributes a horizontal force of (485 N) cos(-35°) ≈ 397 N. This mag. needs to be increased in order to get the box moving.

(a) The time taken to move the box 4 meters is 2.14 s.

(b) The box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.

The given parameters;

The mass of the books is calculated as;

[tex]m = \frac{W}{g} \\\\m = \frac{319}{9.8} \\\\m = 32.55 \ kg[/tex]

The normal force on the box is calculated as follows;

[tex]F_n = -W - Fsin\theta\\\\F_n = -319 - (485\times sin35)\\\\F_n = -597.18 \ N[/tex]

The frictional force when the coefficient of friction is 0.57;

[tex]F_f = \mu F_n\\\\F_f = 0.57 \times -597.18\\\\F_f = -340.39 \ N[/tex]

The acceleration of the box is calculated as follows;

[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ - 340.39 = 32.55 a\\\\56.899 = 32.55a\\\\a = \frac{56.899}{32.55} \\\\a = 1.75 \ m/s^2[/tex]

The time taken to move the box 4 meters is calculated as;

[tex]s = v_0t + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2\times 4}{1.75} } \\\\t = 2.14 \ s[/tex]

(b) The frictional force when the coefficient of friction is 0.75;

[tex]F_f = \mu F_n\\\\F_f = 0.75 \times -597.18\\\\F_f = -447.885 \ N[/tex]

The acceleration of the box is calculated as follows;

[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ -447.885 = 32.55 a\\\\-50.596 = 32.55a\\\\a = \frac{-50.596}{32.55} \\\\a = -1.55\ m/s^2[/tex]

Thus, the box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.

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Answer:

The head

Explanation:

Answer:

option 3

Explanation:

can i get brainliest

Answer:

wavelength =  0.01 m

distance = 162.8 m

Explanation:

Given that;

Speed of sound in water = 1,480 meters per second

Frequency of ultrasound = 125KHZ

From=

v=λf

v= speed of sound

λ= wavelength of sound

f= frequency of sound

λ= 1,480 ms-1/125 * 10^3 Hz

λ= 0.01 m

From

v = 2x/t

where;

v= velocity of sound in water

x= distance traveled

t = time taken

x = vt/2

x = 1,480 ms-1 *  0.220 s/2

x= 162.8 m

Answer:

Option 1

Explanation:

I always see animals do that

Answer:

Approximately [tex]35.2\; \rm m[/tex].

Explanation:

Given:

Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].

Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)

Implied:

Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)

Required:

Displacement, [tex]x[/tex].

Not required:

Time taken, [tex]t[/tex].

Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:

[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].

In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.

Answer:

1) k = 10 [N/m]

2) a-) x = 0.4 [m]

b)  x = 0.075 [m]

Explanation:

To be able to solve this type of problems that include springs we must use Hooke's law, which relates the force to the deformed length of the spring and in the same way to the spring coefficient.

F = k*x

where:

F = force [N] (units of Newtons]

k = spring constant  [N/m]

x = distance = 10 [cm] = 0.1 [m]

Now, the weight is equal to the product of the mass by the gravity

W = m*g = F

where:

m = mass = 100 [g] = 0.1 [kg]

g = gravity acceleration = 10 [m/s²]

F = 0.1*10 = 1 [N]

Now clearing k

k = F/x

k = 1/0.1

k = 10 [N/m]

2)

a ) if the force is 4 [N]

clearing x

x = F/k

x = 4/10

x = 0.4 [m]

m = 75 [g] = 0.075 [kg]

W = m*g = F

F = 0.075*10 = 0.75 [N]

x = .75/10

x = 0.075 [m]

Answer:

A

Explanation:

just did it

As the mass of an object increases its momentum increases, and it takes more force to change its motion. So, option A.

Mass in motion is quantified by momentum, which is the measure of the amount of mass in motion.

Here,

Momentum of an object, which is under motion can be defined as the product of the mass and velocity of the object.

Momentum, P = mv

According to Newton's second law, the force is defined as the rate of change of momentum, or the momentum per unit time.

F = dP/dt

So, force is proportional to the amount of momentum imparted on the object.

Therefore, if the mass or velocity of the object increases, it will eventually cause the momentum to be increased and as a result, the force required to exert on the object will increase.

Hence,

As the mass of an object increases its momentum increases, and it takes more force to change its motion. So, option A.

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Answer:

A plane mirror is placed to the right of an object. The image formed by the mirror will be a virtual image that appears to be on the left of the mirror.

Explanation:

Answer:

0.25 m.

Explanation:

mass of the block = 7.40 kg, height = 0.83 m, force constant of the spring = 9.50 x [tex]10^{2}[/tex] N/m.

The maximum compression on the spring can be determined by;

Potential energy stored in the spring = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]

But, potential energy = mgh

So that,

mgh = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]

7.4 x 9.8 x 0.83 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]

60.1916 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]

[tex]x^{2}[/tex]= [tex]\frac{60.1916}{9.50*10^{2} }[/tex]

  = 0.06336

x = 0.2517

x = 0.25 m

The maximum compression of the spring is 0.25 m.