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KengaRu [80]
3 years ago
13

How fast is a car going if he travels 2,100 km in 21 hr?

Physics
1 answer:
Tcecarenko [31]3 years ago
3 0

Answer:

about 62mph or 100 kph

Explanation:

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24 POINTS!!!!!!!!!!!!
Karo-lina-s [1.5K]

Answer: Option (c) is the correct answer.

Explanation:

When the child is tossed up into the air then she gains kinetic energy as the child has moved from its initial position.

It is given that mass is 20 kg, velocity is 5 m/s^{2}, and height is 2 m.

Calculate the kinetic energy of child as follows.

             kinetic energy = \frac{1}{2}mv^{2}

                                      = \frac{1}{2}20 kg \times (5 m/s)^{2}

                                      = 10 kg \times 25 m^{2}/s^{2}

                                      = 250 kg m^{2}/s^{2}

Also, when child falls off the ground then she will have gravitational potential energy.

Calculate gravitational potential energy of child as follows.

            Potential energy = m × g × h

                                         = 20 kg \times 9.8 m/s^{2} \times 2 m

                                         = 392 kg m^{2}/s^{2}

3 0
3 years ago
Read 2 more answers
You venture out on a cold winter morning to warm up your vehicle. You have layers of cotton/polyester blend clothes on and from
xxMikexx [17]

Answer:

There is a localization of negative charge near the door handle.

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3 years ago
Please help me I need this for a test
Marysya12 [62]

Answer:

Explanation:

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8 0
3 years ago
Which type of wave forms at the boundary between air and water when you
azamat

D. Transverse wave

Hope it's correct and helps uh.

6 0
3 years ago
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The rate constant of a reaction is 7.8 × 10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter yo
Alexus [3.1K]

Answer:

k_2 = 7.815 * 10^-3 s^-1

Explanation:

Given:

- rate constant of reaction k_1 = 7.8 * 10^-3 s^-1    @  T_1 = 25 C

- rate constant of reaction k_2 = ?    @  T_2 = 75 C

- The activation energy E_a = 33.6 KJ/mol

- Gas constant R = 8.314472 KJ / mol . K

Find:

- rate of reaction k_2 @ T_2 = 75 C

Solution:

- we will use a combined form of Arrhenius equations that relates rate constants k as function of E_a and temperatures as follows:

                              k_2 = k_1 * e ^ [(E_a / R) * ( 1 / T_1 - 1 / T_2 )

- Evaluate             k_2 = 7.8 * 10^-3* e^[(33.6 / 8.314472)*(1/298 -1/348)

- Hence,                k_2 = 7.815 * 10^-3 s^-1

4 0
4 years ago
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