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MrRissso [65]
3 years ago
15

A force of 150N at an angle of 60 degree to the horizontal to pull a box through a distance of 50m calculate the work done

Physics
1 answer:
Alexxandr [17]3 years ago
6 0
  • Force=150N
  • Angle=60°
  • Displacement=50m

\boxed{\sf W=Fscos\Theta}

\\ \sf\longmapsto W=150(50)cos 60

\\ \sf\longmapsto W=7500\times \dfrac{1}{2}

\\ \sf\longmapsto W=3750J

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A runner went from 6 m/s and two seconds what was his acceleration
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Answer:

is it 3?

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40ml of Liquid A are poured into a beaker, and 40.0ml of Liquid B are poured into an identical beaker. Stirrers in each beaker a
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d

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2 years ago
A 60 W light bulb is powered by a connection to a wall outlet with 120 V across the plug terminals. a) What is the current passi
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Answer:

The current through the resistor is 0.5 A

Explanation:

Given;

power of the light bulb = 60 W

voltage in the wall outlet across the plug terminals = 120 V

power of the light bulb is the product of voltage in the wall outlet across the plug terminals and the current passing through the resistor.

power = voltage x current

Current = \frac{power}{voltage} = \frac{60}{120} = 0.5 A

Therefore, for a  60 W light bulb powered by a connection to a wall outlet with 120 V across the plug terminals, the current passing through the resistor is 0.5 A

7 0
3 years ago
Read 2 more answers
Explain whether the unit of work is a fundamental or derived unit
klasskru [66]

Answer:

answer here

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An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
Andreas93 [3]

Answer:

v=1.08\times 10^7\ m/s

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, \dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2

Separation between the plates, d=1.75\times 10^{-2}\ m

An electron is released from rest, u = 0

Using equation of kinematics,

v^2-u^2=2ad..........(1)

Acceleration of the electron in electric field, a=\dfrac{qE}{m}............(2)

Electric field, E=\dfrac{\sigma}{\epsilon_o}............(3)

From equation (1), (2) and (3) :

v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}

v = 10840393.1799 m/s

or

v=1.08\times 10^7\ m/s

So, the electron is moving with a speed of 1.08\times 10^7\ m/s before it reaches the positive plate. Hence, this is the required solution.

3 0
3 years ago
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