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beks73 [17]
3 years ago
10

Calculate the work done by a person who exerts a force of 200.0 N in the horizontal direction on a box that moves 10.0 meters al

ong a ramp that makes an angle of 30.0o with the horizontal, moving the box at a constant speed. Note: Do not include the work done on the person, only on the box.
Physics
1 answer:
Westkost [7]3 years ago
7 0

Answer:

the work done by a person is 1,732 J

Explanation:

The computation of the work done by a person is as follows;

The following information is given

Force = 200.N

Distance = 10.0 meters

Angle = 30 degrees

based on the above information

Work done is

\overset{F}{\rightarrow}. \overset{D}{\rightarrow}\\\\fd cos \theta

Work done is

= (200) (cos 30 degrees)  (10.0 meters)

= 1,732 J

hence, the work done by a person is 1,732 J

The same is considered and relevant

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How can you degauss a magnet temporarily?
EleoNora [17]

Answer:

Demagnetization processes include heating past the Curie point, applying a strong magnetic field, applying alternating current, or hammering the metal.

Explanation:

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3 years ago
Levi and Clara are trying to move a very heavy box. Levi is pushing the box with a force of 30 N, and Clara is pulling the box w
musickatia [10]
The net force on the box is:
50 + 30 - 65
= 15 Newtons
This can be used in conjunction with
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3 0
4 years ago
A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.
mafiozo [28]

Answer:

a. f=1.22*10^{-15} N

b. f=53.6*10^{-17} N

Explanation:

The force existing between two charges is given as

f=\frac{kq_{1}q_{2}}{r^{2}}

where q= charge,

k=constant

r= distance between the two charges

Note: this force can either be repulsive or attractive force depending on the charges involve. it is repulsive if they are similar charge and it is attractive if it is opposite charges.

Also the charge of an electron is

-1.602*10^{-19}

A. we first determine the magnitude force between the -4nC and the electron

f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis

for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\  f=1.22*10^{-15} N

b. at  distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni

this this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

The total net force on the electron is thus

f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\  f=53.6*10^{-17} N

8 0
3 years ago
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6 0
3 years ago
Read 2 more answers
1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup
azamat

Answer:

\dot{W} = 339.84 W

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

\dot{W} =\dot{m}\frac{V^{2}}{2}

here \dot{m}is mass flow rate and given as

\dot{m} = \rho*Q

\dot{W} =\rho*Q\frac{V^{2}}{2}

Putting all value to get minimum power

\dot{W} =1.18*9*\frac{8^{2}}{2}

\dot{W} = 339.84 W

7 0
3 years ago
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