Two balloons become equally charged once they are rubbed against each other. If the force between the balloons is 6.2 * 10^23 N, what would happen to the force if the charge were to triple on one of the balloons?

A) the force would triple

B) the force would become one-nineth

Answer:

Aluminum board will move with a higher velocity

Explanation:

The velocity of the block will be higher when the impulse imparted by the bullet is higher.

In case of bullet bouncing off, the impulse imparted on the aluminum board is high and hence, it will move with a high velocity as compared to that of the clay board.

Answer:

Displacement and Velocity

Answer:

Work done gets doubled.

Explanation:

The work done by a force is given by :

W = Fd

Where

F is force and d is distance move

If the force is doubled and the distance moved remain the same, it would mean that the work done becomes double of the initial work done.

At the time when force is doubled and distance moved remain the same so Work done gets doubled.

The following information should be considered:

W = Fd

Where

F is force and d is distance move

In the case when the force is doubled and the distance moved remain the same, it would mean that the work done becomes double of the initial work done.

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Answer:

Positive, Negative

Explanation:

The image I've attached shows that vectors point into the negative source and vectors point away from the positive source.

Answer:

8.08 J/g °C

Explanation:

Q=m*Cp*ΔT-->

Cp=Q/(m*ΔT) -->

Cp=1600/[18*(31-20)]-->

Cp=8.08 J/g °C

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   [tex]\frac{m}{m+ \frac{I}{r^2} }[/tex]

          v² = 2gh    [tex]\frac{1}{1 + \frac{I}{m r^2} }[/tex]

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = [tex]g \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }[/tex]

let's use trigonometry

         sin θ = h / L

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    [tex]\frac{1}{1 + \frac{mr^2 }{m r^2 } }[/tex]1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  [tex]\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }[/tex] = g sin θ   [tex]\frac{1}{1+ \frac{1}{2} }[/tex]

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   [tex]\frac{1}{1 + \frac{2}{3} }[/tex]

     a₃ = g sin θ [tex]\frac{3}{5}[/tex]

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  [tex]\frac{1 }{1 + \frac{2}{5} }[/tex]

     a₄ = g sin θ  [tex]\frac{5}{7}[/tex]

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      [tex]\frac{105}{210}[/tex]

b) a₂ = g sinθ ⅔ = g sin θ     [tex]\frac{140}{210}[/tex]

c) a₃ = g sin θ [tex]\frac{3}{5}[/tex]= g sin θ       [tex]\frac{126}{210}[/tex]

d) a₄ = g sin θ [tex]\frac{5}{7}[/tex] = g sin θ      [tex]\frac{150}{210}[/tex]

the order of acceleration from lower to higher is

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Answer:

60mph=26.8224meters per second

Explanation:

Answer:

atoms

Explanation:

Hormones are derived from amino acids or lipids. Amine hormones originate from the amino acids tryptophan or tyrosine. Larger amino acid hormones include peptides and protein hormones. Steroid hormones are derived from cholesterol.

The amount of heat absorbed by the water will be 8368 J.

Heat gain is defined as the amount of heat required to increase the temperature of a substance by some degree of Celsius. While heat loss is inverse to heat gain.

It is given by the formula as ;

[tex]\rm Q= mcdt[/tex]

The given data in the problem is;

Equilibrium temperature = 30°C.

mass of water  = 50 g  ,

Temperature change = 70ºC

The specific heat of water =4.184 J//g °C

The amount of heat absorbed by the water is;

[tex]\rm Q= mcdt \\\\ Q=50 \times 4.184 \times (70^0 -30^0)C\\\\ Q= 8368 J[/tex]

Hence, the amount of heat absorbed by the water will be 8368 J.

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Solve the differential equation:

dV/dt = k V   →   1/V dV/dt = k

→   d/dt [ln(V)] = k

→   ln(V) = k t + C

→   V (t )= exp(k t + C ) = C exp(k t ) = C e ᵏᵗ

At t = 0, the glacier has volume 400 km³ of ice, so

V (0) = 400   →   C e⁰ = C = 400

Find when the glacier's volume is 300 km³:

V (t ) = 400 e ᵏᵗ = 300   →   e ᵏᵗ = 3/4

→   k t = ln(3/4)

→   t = 1/k ln(3/4)

At this time, the volume is decreasing at a rate of 15 km³/yr, so

V ' (t ) = C k e ᵏᵗ   →   V ' (1/k ln(3/4)) = 400 k exp(k × 1/k ln(3/4)) = -15

→   3/4 k = -3/80

→   k = -1/20

Then the volume V (t ) of the glacier at time t is

V (t ) = 400 exp(-1/20 t )

The volume in terms of time will be "V(t) = 400 exp (-[tex]\frac{1}{20}[/tex] t)".

According to the question,

Glacier's volume, V(0) = 400 km³

→     [tex]\frac{dV}{dt}[/tex] = kV

   [tex]\frac{1}{V}[/tex] [tex]\frac{dV}{dt}[/tex] = k

   ln(V) = kt + C

Now,

   V (t) = exp (kt + C)

           = C exp (kt)

           = C[tex]e^{kt}[/tex]

When volume of glacier be "300 km³",

V(t) = 400 [tex]e^{kt}[/tex] = 300

[tex]e^{kt}[/tex] = [tex]\frac{3}{4}[/tex]

By taking log,

 kt = ln([tex]\frac{3}{4}[/tex])

   t = [tex]\frac{1}{k}[/tex] ln([tex]\frac{3}{4}[/tex])

When volume decrease at 15 km³/yr, then

→ V'(t) = Ck[tex]e^{kt}[/tex]

          = 400 k exp (k × [tex]\frac{1}{k}[/tex] kn ([tex]\frac{3}{4}[/tex]))

          = -15

Now,

[tex]\frac{3}{4}[/tex] k = - [tex]\frac{3}{80}[/tex]

By applying cross-multiplication,

  k = - [tex]\frac{1}{20}[/tex]

Thus the response above is correct.

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