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svetlana [45]
3 years ago
10

Cómo es útil el ejercicio Squat Side Kick para practicar deportes?? Dame dos ejemplos

Physics
1 answer:
svetlana [45]3 years ago
4 0

Answer:

do you speak english

Explanation:

You might be interested in
Help please! Will give brainly! and thanks.
kipiarov [429]
<h2>Answer:</h2>

All the energy sources are correctly matched with their category.

Explanation:

Renewable energy sources:

These are energy sources which can be replenished as they don't in involves the irreversible phase change.

These resources can never be ended as they can be used again and again.

Wind, geothermal, biomass, bio gas are example of renewable energy sources.

Non renewable energy sources:

These are the energy source which can never be replenished after one time use. They undergo the chemical irreversible change.

These sources are lacking with the passage of time because they can never be reused.

Oil, gas, coal and natural gas are examples.

7 0
3 years ago
While traveling along a highway a driver slows from 24 m/sec to 15 m/sec in 12 seconds. What is the
Vladimir [108]
24-15=9 m/s slower in 12 seconds. So 9/12 m/s² slower. Therefore the acceleration is -0,75 m/s²
4 0
3 years ago
Which of the following groups contain three elements with stable electron configurations?
Wewaii [24]

Answer:

B. helium, xenon, neon

Explanation:

6 0
3 years ago
Read 2 more answers
An athlete in a gym applies a constant force of 50 N to the pedals of a bicycle to keep the rotation rate of the wheel at 10 rev
marishachu [46]

Answer:

Explanation:

Given that,

Force applied to pedal F = 50N

Angular velocity ω = 10rev/s

We know that, 1rev = 2πrad

Then, ω = 10rev/s = 10×2π rad/s

ω = 20π rad/s

Length of pedal r = 30cm = 0.3m

Power?

Power is given as

P = τ×ω

We need to find the torque τ

τ = r × F

Since r is perpendicular to F

Then, τ = 0.3 × 50

τ = 15 Nm

Then,

P = τ×ω

P = 15 × 20π

P = 942.48 Watts

power delivered to the bicycle by the athlete is 942.48 W

6 0
3 years ago
A) An automobile light has a 1.0-A current when it is connected to a 12-V battery. Determine the resistance of the light.
kirill [66]

Answer:

The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

Explanation:

Given:

(A)

Current I = 1 A

Voltage V = 12 V

For finding the resistance,

  V = IR

  R = \frac{V}{I}

  R = \frac{12}{1}

  R = 12Ω

(B)

For finding power delivered,

  P = I^{2} R

  P = (1) ^{2} \times 12

  P = 12 Watt

(C)

For finding the potential difference,

   V = IR

   V = 5 \times 10^{-3} \times 2

   V = 10 \times 10^{-3}

   V = 0.01 V

Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

4 0
3 years ago
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