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Marysya12 [62]
2 years ago
9

A toy car of mass 8 kg initially moves with a speed of 5 m/s. How much work must be done on the car to increase its speed to 10

m/s?
Physics
1 answer:
sukhopar [10]2 years ago
3 0

ANSWER:

300 J

STEP-BY-STEP EXPLANATION:

To know the work required, we must calculate the work in both cases, the difference would be the amount of work necessary for the speed to increase. The work done is the same as the amount of energy increase. The formula for kinetic energy is:

\frac{1}{2}m\cdot v^2

We calculate in each case:

\begin{gathered} \frac{1}{2}\cdot8\cdot10^2=400\text{ J} \\ \frac{1}{2}\cdot8\cdot5^2=100\text{ J} \end{gathered}

We calculate the difference between the two to find out the work required:

400-100=300\text{ J}

The work required is 300 J

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Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
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Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

   Pressure at sea level = 1 atm = 101300 Pa

   we know

   P = ρ g h

   h = \dfrac{P}{\rho\ g}

   h = \dfrac{101300}{1.3\times 9.8}

          h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

  at x = 0  air density = 0

  at x= h   ρ_l = ρ_sl

 assuming density is zero at x - distance

 \rho_x = \dfrac{\rho_{sl}}{h}\times x

now, Pressure at depth x

dP = \rho_x g dx

dP = \dfrac{\rho_{sl}}{h}\times x g dx

integrating both side

P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx

P =\dfrac{\rho_{sl}\times g h}{2}

 now,

h=\dfrac{2P}{\rho_{sl}\times g}

h=\dfrac{2\times 101300}{1.3\times 9.8}

  h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

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Use the dropdown menus to complete the paragraph about how the troposphere is heated.
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A runner generates 1260 W of thermal energy. If this heat has to be removed only by evaporation, how much water does this runner
sergey [27]

Answer:0.502kg

Explanation:

F4om the relation

Power x time = mass x latent heat of vapourization

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A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much tim
Free_Kalibri [48]

Answer:

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\   where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]

3 0
3 years ago
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