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aleksandr82 [10.1K]
2 years ago
9

PLEASE HELP!!

Physics
1 answer:
leva [86]2 years ago
8 0
I think the answer is B
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A 5.0 kg block hangs from the ceiling by a mass-less rope. A Second block with a mass of 10.0 kg is attached to the first block
gayaneshka [121]

The tension in the first and second rope are; 147 Newton and 98 Newton respectively.

Given the data in the question

  • Mass of first block; m_1 = 5.0kg
  • Mass of second block, m_2 =10kg
  • Tension on first rope; T_1 =\ ?
  • Tension on second rope; T_2 =\ ?

To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

F = m\ *\ a

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity  g = 9.8m/s^2 )

For the First Rope

Total mass hanging on it; m_T = m_1 + m_2 = 5.0kg + 10.0kg = 15.0kg

So Tension of the rope;

F = m\ * \ g\\\\F = 15.0kg \ * 9.8m/s^2\\\\F = 147 kg.m/s^2\\\\F = 147N

Therefore, the tension in the first rope is 147 Newton

For the Second Rope

Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

F = m\ * \ g\\\\F = 10.0kg \ * 9.8m/s^2\\\\F = 98 kg.m/s^2\\\\F = 98N

Therefore, the tension in the second rope is 98 Newton

Learn More, brainly.com/question/18288215

4 0
2 years ago
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Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou
Gennadij [26K]

Answer:

a. The station is rotating at 1.496 \frac{rev}{min}

b. the rotation needed is 2.8502 \frac{rev}{min}

Explanation:

We know that the centripetal acceleration is

a_{c}= \omega ^2 r

where \omega is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed  is :

2.7 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.02454 \frac{rad^2}{s^2} }

\omega  = 0.1567 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.1567 \frac{rad}{s}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 1.496 \frac{rev}{min}

<h3>b. </h3>

The rotational speed needed is :

9.8 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.08909 \frac{rad^2}{s^2} }

\omega  = 0.2985 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.2985 \frac{rev}{min}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 2.8502 \frac{rev}{min}

3 0
3 years ago
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When sunlight shines on a leaf the leaf looks green why does the leaf look green
KonstantinChe [14]
The correct answer is the third, It reflects the green light waves and absorbs most of the rest.
7 0
3 years ago
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If you measure the amount of work accomplished in a particular time interval, u have measured-
serious [3.7K]

Answer:

The power

Explanation:

We know that the work definition is given by the following expression:

W = F * d

where:

F = force [Newtons] [N]

d = distance [meters] [m]

W = work [Joules]

And the expression that defines the work done by unit of time is called - <u>Power</u>, therefore:

P = W/t

where:

P = power [watts] [w]

W = work [Joules] [J]

t = time [seconds] [s]

6 0
3 years ago
skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If
scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

7 0
3 years ago
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