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Svetllana [295]
3 years ago
11

A modified Atwood machine is at rest. The hanging block has a mass of 3kg. The black on wheels has unknown mass M1. When release

d the block m2 falls at 5.88m/s^2. If frictional forces are considered so small they are negligible, the block M1 must have a mass of _______ kg.

Physics
1 answer:
fgiga [73]3 years ago
7 0

Answer:

From the figure,

The free-body diagrams for  and  are shown in the figures below. The only forces on the blocks are the upward tension  and the downward gravitational forces  and  . Applying Newton’s second law, we obtain:   

which can be solved to yield 

Substituting the result back, we have 

(a) With  and , the acceleration becomes   

(b) Similarly, the tension in the cord is   

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Answer:

mass of the ball will be 1 kg

Explanation:

We have given force F = 0.5 N

This force must be equal to the centripetal force for the ball to revolve in the circle

So F=F_C=0.5N

Centripetal acceleration is given as a_c=0.5m/sec^2

We have to calculate the mass of the ball

We know that force is given by

F_C=ma_c

So 0.5=m\times 0.5

m = 1 kg

So mass of the ball will be 1 kg

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Which circuit component is usually a device that uses electrical energy to to perform a certain function? A. wire B. power sourc
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A

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3 years ago
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Salmon often jump waterfalls to reach their
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The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

  • height of the waterfall, h = 0.432 m
  • distance of the Salmon from the waterfall, s = 3.17 m
  • angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s

The minimum velocity of the Salmon jumping at the given angle is calculated as;

X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

Learn more here: brainly.com/question/20064545

8 0
2 years ago
A hot iron ball of mass 200 g is cooled to a temperature of 22°C from 100°C. How much heat was
Nookie1986 [14]

Answer:

Q= -6900 J

Explanation:

use the formula Q=mC(T_2 - T_1) and sub in givens

Q=mC(T_2 - T_1)

Q= (200 g)(0.444 J/g°C)(22-100)

Q= -6900 J

The negative sign means heat is lost, which agrees with the decrease in temperature

6 0
3 years ago
wo lacrosse players collide in midair. Jeremy has a mass of 120 kg and is moving at a speed of 3 m/s. Hans has a mass of 140 kg
Julli [10]

2.71 m/s fast Hans is moving after the collision.

<u>Explanation</u>:

Given that,

Mass of Jeremy is 120 kg (M_J)

Speed of Jeremy is 3 m/s (V_J)

Speed of Jeremy after collision is (V_{JA}) -2.5 m/s

Mass of Hans is 140 kg (M_H)

Speed of Hans is -2 m/s (V_H)

Speed of Hans after collision is (V_{HA})

Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is  

= =\mathrm{M}_{1} \times \mathrm{V}_{\mathrm{J}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{H}}

Substitute the given values,

= 120 × 3 + 140 × (-2)

= 360 + (-280)

= 80 kg m/s

Linear momentum after the collision of Jeremy and Hans is  

= =\mathrm{M}_{\mathrm{J}} \times \mathrm{V}_{\mathrm{JA}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{HA}}

= 120 × (-2.5) + 140 × V_{HA}

= -300 + 140 × V_{HA}

We know that conservation of liner momentum,

Linear momentum before the collision = Linear momentum after the collision

80 = -300 + 140 × V_{HA}

80 + 300 = 140 × V_{HA}

380 = 140 × V_{HA}

380/140= V_{HA}

V_{HA} = 2.71 m/s

2.71 m/s fast Hans is moving after the collision.

4 0
3 years ago
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