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3 years ago

1200 kg car is being driven down a road. If it has 101 kJ of kinetic energy, what is its speed?

Physics

1 answer:

vova2212 [387]3 years ago

5 0

Kinetic Energy = (1/2) (mass) (speed)²

101,000 J = (1/2) (1200 kg) (speed)²

Speed² = 101,000 J / 600 kg

Speed² = 168 and 1/3 m²/s²

Speed = √(168 and 1/3 m²/s²)

Speed = 12.97 m/s

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Light waves have some similarities with water and sound waves, but they are not exactly the same. Describe all the differences y

makkiz [27]

Answer:

<h2>All the waves are pertubations that propagate (transport) energy.</h2><h2></h2>

Nevertheless, they have some differences:

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3 0

3 years ago

Help I need a answer to this

natta225 [31]

Answer:

1700 Joules

Explanation:

Work=force x distance

Force = 170 kg

Distance= 10 Meters

170 x 10 = 1700 Joules of work

3 0

3 years ago

The moon is smaller and more dense than the Earth, and has less extreme temperature changes.

Lerok [7]

The moon is smaller and more dense than the Earth, and has less extreme temperature changes. The statement presented is True. In terms of temperature, since there is no atmosphere on the moon, then it has less extreme temperature changes. The moon can reach 253 Fahrenheit in the day and -387 Fahrenheit at night.

8 0

3 years ago

Read 2 more answers

If the kinetic energy of an electron is 4.1e-18 j, what is the speed of the electron? (you can use the approximate (nonrelativis

arlik [135]

The kinetic energy of the electron is
$K= \frac{1}{2}mv^2$
where $m=9.1 \cdot 10^{-31} kg$ is the mass of the electron and v its speed. Since we know the value of the kinetic energy, $K=4.1 \cdot 10^{-18} J$ , we can find the value of the speed v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 4.1 \cdot 10^{-18}J}{9.1 \cdot 10^{-31}kg} } = 3\cdot 10^6 m/s$

3 0

3 years ago

Suppose that the velocity (in meters per second) of a sky diver falling near the Earth's surface is given by the following expon

Mice21 [21]

Answer:

4.7 s

Explanation:

The complete question is presented in the attached image to this solution.

v(t) = 61 - 61e⁻⁰•²⁶ᵗ

At what time will v(t) = 43 m/s?

We just substitute 43 m/s into the equation for the velocity of the diver and solve for t.

43 = 61 - 61e⁻⁰•²⁶ᵗ

- 61e⁻⁰•²⁶ᵗ = 43 - 61 = -18

e⁻⁰•²⁶ᵗ = (18/61) = 0.2951

In e⁻⁰•²⁶ᵗ = In 0.2951 = -1.2205

-0.26t = -1.2205

t = (1.2205/0.26) = 4.694 s = 4.7 s to the nearest tenth.

Hope this Helps!!!

6 0

2 years ago