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Karolina [17]
3 years ago
12

I need huge help understanding this type of questions :( Please and thank you

Physics
1 answer:
Mrrafil [7]3 years ago
5 0

Answer:

67.28 N

Explanation:

Given that,

The mass of an box, m = 15.2 kg

The coefficients of static and kinetic frictions for plastic on wood are 0.55 and 0.36, respectively.

The force of static friction,

F_s=\mu_smg\\\\F_s=0.55\times 15.2\times 9.8\\\\F_s=81.92\ N

The force of kinetic friction,

F_k=\mu_kmg\\\\F_k=0.36\times 15.2\times 9.8\\\\F_k=53.62\ N

Net force acting on the object is :

F = 120.9 -53.62

= 67.28 N

Hence, this is the required solution.

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awhite billiard ball with mass mw = 1.47 kg is moving directly to the right with a speed of v = 3.01 m/s and collides elasticall
pochemuha

Answer:

speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

initial kinetic energy = final kinetic energy

KE = 6.66 J

Explanation:

Since there is no external force on the system of two balls so here total momentum of two balls initially must be equal to the total momentum of two balls after collision

So we will have

momentum conservation along x direction

m_1v_{1i} + m_2v_{2i} = m_1v_{1x} + m_2v_{2x}

now plug in all values in it

1.47 \times 3.01 + 0 = 1.47 v_1cos68 + 1.47 v_2cos22

so we have

3.01 = 0.375v_1 + 0.927v_2

similarly in Y direction we have

m_1v_{1i} + m_2v_{2i} = m_1v_{1y} + m_2v_{2y}

now plug in all values in it

0 + 0 = 1.47 v_1sin68 - 1.47 v_2sin22

so we have

0 = 0.927v_1 - 0.375v_2

v_2 = 2.47 v_1

now from 1st equation we have

3.01 = 0.375 v_1 + 0.927(2.47 v_1)

v_1 = 1.13 m/s

v_2 = 2.78 m/s

so speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

Also we know that since this is an elastic collision so here kinetic energy is always conserved to

initial kinetic energy = final kinetic energy

KE = \frac{1}{2}(1.47)(3.01^2)

KE = 6.66 J

5 0
4 years ago
how much kinetic energy is produced when an object having mass of 50gm throwing with velocity 80m/s?​
Jlenok [28]

Answer:

KE = 160 J

Explanation:

KE = 1/2mv²

mass= 50gm = 0.05kg

velocity = 80

KE = 1/2 x 0.05 x 80²

= 1/2 x 0.05 x 6400

= 160

4 0
3 years ago
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v^{2} =  u^{2}  +  2ar
0 = u^2 + 2*(-2)*9600
u^2 = 38400
u = 195.96 m/s
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An 800-N man stands at rest on two bathroom scales so that his weight is distributed evenly over both scales. The reading on eac
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400 N

The weight of the man is evenly distributed on each scale

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