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3 years ago

Choose all the answers that apply.

Physics

2 answers:

kozerog [31]3 years ago

8 0

Answer:

Skipping a cool-down can cause them.

They take rest to heal.

A broken bone is one type.

Tendonitis is one type.

Explanation:

Olenka [21]3 years ago

5 0

Answer:

<h3>Skipping a cool-down can cause them.</h3><h3>They take rest to heal.</h3><h3>A broken bone is one type.</h3><h3>Tendonitis is one type.</h3>

Explanation:

<em>Overuse injuries represent any type of muscle or bone injury caused by repetitive trauma as tendinitis or stress fractures.</em> <u>The most common overuse injuries are from muscle and bone overuse,</u> when the athlete don't rest properly, the excess of activity push muscles and bones out of the limit. Also, training errors conduct to these injuries.

Overuse injuries can be caused by skipping cool-down periods, or rest periods, that's what get the athlete in a overuse state. After these injuries occur, to rest is mandatory, it's the only way to heal, because muscles and bone need to stop to recover from the wide time use.

In addition, as we said before, tendinitis and stress fractures are the most common overuse injuries, due to the overuse of tendons and bones, at that point, they stress and become vulnerable.

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1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t

tiny-mole [99]

Explanation:

Charges, $q_1=8\ \mu C=8\times 10^{-6}\ C$

$q_2=-5\ \mu C=-5\times 10^{-6}\ C$

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

$F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N$

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

6 0

3 years ago

A circuit has a 10 Ω resistor connected to a 1.5 V dry cell. What is the current that can flow in the circuit?

ludmilkaskok [199]

Here is the highly detailed, arcane, complex, technical form of Ohm's Law that is needed in order to answer this question ===> I = V / R .

Current = (voltage) / (resistance)

Current = (1.5 V) / (10 Ω)

<em>Current = 0.15 Ampere</em>

8 0

3 years ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

yaroslaw [1]

Answer:

K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

p₀ = 0

final moment

p_f = m_a v_a + m_b v_b

p₀ = p_f

0 = m_a v_a + m_b v_b

tells us that

m_a = 8 m_b

0 = 8 m_b v_a + m_b v_b

v_b = - 8 v_a (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

Em₀ = K₀ = 73 J

final point. After separating the body

Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

K₀ = K_f

73 = ½ m_a (v_a² + v_b² / 8)

we substitute equation 1

73 = ½ m_a (v_a² + 8² v_a² / 8)

73 = ½ m_a (9 v_a²)

73/9 = ½ m_a (v_a²) = K_a

K_a = 8,111 J

3 0

3 years ago

1 point

e-lub [12.9K]

The engineer built a device called a generator

4 0

3 years ago

F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$