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3 years ago

A car travels with an average speed of 22 m/s.what is this speed in km/s

Physics

2 answers:

Debora [2.8K]3 years ago

4 0

It is .022 kilometers.

maksim [4K]3 years ago

4 0

Answer:

22 m/s = 79.2 km/h

Explanation:

Given that,

The average speed of the car is 22 m/s. Here, we need to convert the speed to km/h. Firstly, we will see the following conversions as :

1 kilometres = 1000 meters

1 hour = 3600 seconds

To convert m/s to km/h we must use these coversions.

$22\dfrac{m}{s}=22\times \dfrac{1/1000\ km}{1/3600\ s}$

After solving the above expression, we get

22 m/s = 79.2 km/h

So, the car travel with a speed of 79.2 km/h. Hence, this is the required solution.

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A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo

Dima020 [189]

Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

E₁ = mechanical energy at initial state [J]

$E_{1}=E_{pot}+E_{kin}+E_{elas}\\$

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

E₂ = mechanical energy at final state [J]

$E_{2}=E_{kin}+E_{pot}$

Now we can use the first statement to get the first equation:

$E_{1}+W_{1-2}=E_{2}$

where:

W₁₋₂ = work from the state 1 to 2.

$E_{k}=\frac{1}{2} *m*v^{2} \\$

$E_{pot}=m*g*h$

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

$70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5$

$58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]$

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3 years ago

(a) What is the escape speed on a spherical asteroid whose radius is 545 km and whose gravitational acceleration at the surface

Keith_Richards [23]

Answer:

1777.92 m/s

Explanation:

R = Radius of asteroid = 545 km

M = Mass of planet

g = Acceleration due to gravity = 2.9 m/s²

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Acceleration due to gravity is given by

$g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}$

The expression of escape velocity is given by

$v=\sqrt{\dfrac{2GM}{R}}\\\Rightarrow v=\sqrt{\dfrac{2G}{R}\dfrac{gR^2}{G}}\\\Rightarrow v=\sqrt{2gR}\\\Rightarrow v=\sqrt{2\times 2.9\times 545000}\\\Rightarrow v=1777.92\ m/s$

The escape speed is 1777.92 m/s

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