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aliina [53]
3 years ago
8

Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T

is applied. The system moves at a constant speed across the ground. Block B experiences a friction force of f.T = 50N f = 10N
What is friction force acting on block A?
Physics
1 answer:
Oksanka [162]3 years ago
3 0

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

T-f-T'=m_Ba

Where m_B=Mass of block B

Substitute the values

50-10-T'=m_B\times 0=0

T'==40 N

For block A

T'-f'=m_Aa

Where m_A=Mass of block A

Substitute the values

40-f'=m_A\times 0=0

f'=40 N

Hence, the friction force acting on block A=40 N

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The thermal transfer of energy by electromagnetic waves through the vacuum of space is called
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So, in the outter space is vacuum, this means the energy cannot be transmitted by convection, nor conduction. It must be transmitted by electromagnetic waves that are able to travel with or without a medium, and this is called radiation.

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The Burj Khalifa in Dubai is the world's tallest building. The structure is 828 m (2,716.5 feet) and has more than 160 stories.
Elena L [17]

Answer:

 h = 599.5 m

Explanation:

Given,

height of structure = 828 m

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Potential energy = 187000 J

PE = m gd

d = \frac{PE}{mg}

d = \frac{187000}{83.43\times 9.81}

h = 228.5 m

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Height of the floor above ground is equal to 599.5 m.

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When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache
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Answer:

b) True.    the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

     fr - w = ma

Where the friction force has some form of type.

     fr = G v + H v²

Let's replace

     (G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

      fr - w = 0

      fr = mg

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For low speeds the quadratic depended is not important, so we can reduce the equation to

     G v = mg

     v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

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3 0
3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

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\Delta d=0.8 cm=0.008 m

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We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

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\frac{Q\Delta d}{\epsilon_0 S}=100

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Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
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