What is the answer to 895cm a Km

An object is pulled with 20 net of force to the right against a friction force of 10 net.what is the net force.in what direction

Alenkasestr [34]

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6 0

3 years ago

What is the speed of sound in solid? <br>

sasho [114]

The speed of sound in a solid would be 6000 metres per second

7 0

3 years ago

Read 2 more answers

A 2.0 g particle moving at 5.2 m/s makes a perfectly elastic head-on collision with a resting 1.0 g object.

sesenic [268]

Answer:

(a) The speed of the first particle is 1.75 m/s. The speed of the second particle is 6.9 m/s after the collision.

(b) The speed of the first particle is 3.45 m/s in the negative direction. The speed of the second particle is 1.73 m/s.

(c) The final kinetic energy of the incident particle in part (a) and part(b) is 0.0031 J and 0.011 J, respectively.

Explanation:

(a)

In an elastic collision, both momentum and energy is conserved.

$\vec{P}_{initial} = \vec{P}_{final}\\m_1v_1 = m_1v_1' + m_2v_2'\\K_{initial} = K_{final}\\\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2$

Combining these equations will give the speed of the second particle.

$v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2*2}{2+1}(5.2) = 6.9~m/s$

We can use this to find the speed of the first particle.

$m_1v_1 = m_1v_1' + m_2v_2'\\2(5.2) = 2v_1' + (1)(6.9)\\v_1' = 1.75~m/s$

(b)

If m_2 = 10g.

$v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2*2}{2+10}(5.2) = 1.73~m/s$

$m_1v_1 = m_1v_1' + m_2v_2'\\2(5.2) = 2v_1' + (10)(1.73)\\v_1' = -3.45~m/s$

The minus sign indicates that the first particle turns back after the collision.

(c)

The final kinetic energy of the particle in part (a) and part (b) is

$K_a = \frac{1}{2}m_1v_1'^2 = \frac{1}{2}(2\times10^{-3})(1.75)^2 = 0.0031 ~J\\K_b = \frac{1}{2}m_2v_1'^2 = \frac{1}{2}(2\times10^{-3})(3.45)^2 = 0.011~J$

8 0

3 years ago

Two electrons, each with mass mmm and charge qqq, are released from positions very far from each other. With respect to a certai

lana66690 [7]

The answer & explanation for this question is given in the attachment below.

3 0

3 years ago

A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

mamaluj [8]

The vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

<h3>Tension in the cable</h3>

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ = ¹/₂WL

T sinθ = ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

<h3>Vertical component of the force</h3>

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Learn more about tension here: brainly.com/question/24994188

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6 0

2 years ago