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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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A space probe lands on a newly discovered planet. A small canister is released from the probe and falls a distance of 3 m in 0.5
ivanzaharov [21]

Answer:

24m/s²

Explanation:

Given

Distance S = 3m

Time of fall = 0.5sec

Required

Acceleration due to gravity

Using the equation of motion

S = ut+1/2gt²

Substitute the given values

3 = 0+1/2g(0.5)²

3 = 1/2(0.25)g

3 = 0.125g

g = 3/0.125

g = 24

Hence the value for the acceleration of gravity on this new planet is 24m/s²

3 0
2 years ago
Describe a ball's motion as it rolls up a slanted
emmasim [6.3K]

The ball will decelerate as it moves upwards.

The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

The given parameters;

  • initial velocity of the ball, u = 1.25 m/s
  • time of motion of the ball, t = 4.22 s

As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.

Thus, the ball decelerate as it moves upwards.

The acceleration of the ball is calculate as;

a = \frac{v_f -v_0}{t} \\\\

<em>at the highest point on the incline plane, the final velocity </em>v_f<em> is zero</em>

a = \frac{0-1.25}{4.22} \\\\a = -0.3 \ m/s^2

Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

Learn more here:brainly.com/question/23860763

4 0
3 years ago
Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal grav
DochEvi [55]

Answer:

a) 3673469.39 seconds

b) 6.61×10¹⁴ m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0.12×3×10⁸ m/s

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

Equation of motion

v=u+at\\\Rightarrow 0.12\times 3\times 10^8=0+9.8t\\\Rightarrow t=\frac{0.12\times 3\times 10^8}{9.8}=3673469.39\ s

Time taken to reach 12% of light speed is 3673469.39 seconds

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{(0.12\times 3\times 10^8)^2-0^2}{2\times 9.8}\\\Rightarrow s=6.61 \times 10^{14}\ m

The distance it would have to travel is 6.61×10¹⁴ m

7 0
3 years ago
When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink
Dovator [93]

Answer:

 y <8 10⁻⁶ m

Explanation:

For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.

 Therefore the diffraction equation for slits with m = 1 remains

             a sin θ = λ

in general these experiments occur for oblique angles so

             sin θ = θ

             θ = λ / a

in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant

           θ = 1.22 λ / a

The angles in these measurements are taken in radians, therefore

          θ = s / R

as the angle is small the arc approaches the distance s = y

          y / R = 1.22 λ / s

          y = 1.22 λ R / a

let's calculate

            y = 1.22 500 10⁻⁹ 0.42 / 0.032

            y = 8 10⁻⁶ m

with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)

                 y <8 10⁻⁶ m

4 0
3 years ago
Starting at the same time, an arrow and a ball are shot horizontally with a speeds of 50 m/s and 44 m/s respectively from the to
Yuri [45]

Answer:

option D

Explanation:

given.

horizontal velocity of arrow and a ball given as  50 m/s and 44 m/s respectively from the top of a building over flat ground.

In vertical direction, they are both identical

In vertical direction the initial velocity of arrow and a ball  is 0  m/s

Their acceleration due to gravity  is same for both arrow and a ball  9.8 m/s²

they will react bottom at the same time

 time of flight is same for both

now,

In horizontal direction,

distance = speed × time

Since speed is more for arrow, it will travel more horizontal distance  at the same time.

the correct answer is option D

5 0
3 years ago
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