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Goshia [24]
3 years ago
15

How do I solve this​

Physics
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

where:

q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

W = work [J]

W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

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From measurements on a uniformly sized material from a dryer, it was found that the surface area of the material is 1200 m2. If
amm1812

Answer:

D = 2.07\times 10^{-4} m

Explanation:

We know that the volume of the material

= mass/density

mass= 360 Kg

density = 1450 kg/m^3

Volume= 360/1450= 0.248 m^3

and Volume = Area× equivalent Diameter

Area= 1200 m^2

⇒Equivalent Diameter= Volume/Area= 0.248/1200=

D = 2.07\times 10^{-4} m

3 0
3 years ago
How does the momentum of a 75 kilogram stationary football player compare with that of a 70 kilogram player moving at a velocity
pashok25 [27]
Momentum of an object = (its mass) x (its speed) .

Since speed is a factor of momentum, an object that
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6 0
3 years ago
Read 2 more answers
What is the relationship between electric field lines and equipotential lines that you observed in doing the lab
lidiya [134]

Answer:

Explained below

Explanation:

Generally speaking, we know in physics that Electric field lines are lines which usually start at positive charges and deflect away from them to terminate at the negative charges. Meanwhile Equipotential lines are lines that are used to connect points located on the same electric potential.

Finally, in conclusion, electric field lines are usually lines that go through in a perpendicular manner across every equipotential lines.

8 0
3 years ago
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Mariana [72]

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5 0
3 years ago
A ball is thrown horizontally from the top of a building 54 m high. The ball strikes the ground at a point 35 m horizontally awa
Ivanshal [37]

Answer:

V=34.2 m/s

Explanation:

Given that

Height , h= 54 m

Horizontal distance , x = 35 m

Given that , the ball is thrown horizontally , therefore the initial vertical velocity will be zero.

In vertical direction :

We know that

V^2_y=U^2_y+2 g h

Now by putting the values in the above equation we got

V^2_y=U^2_y+2 g h

V^2_y=0^2+2\times 9.81 \times 54

Assume g= 9.81 m/s^2

Thus

V^2_y=1059.48

V_y=\sqrt{1059.48}\ m/s

V_y=32.54 m/s

We also know that

V_y=U_y+ g\times t

32.54=9.81\times t

t=\dfrac{32.54}{9.81}=3.31\ s

In horizontal direction :

x=U_x\times t

U_x=\dfrac{35}{3.31}=10.54\ m/s

Thus the resultant velocity

V=\sqrt{V^2_y+U^2_x}

V=\sqrt{32.54^2+10.54^2}=34.2\ m/s

V=34.2 m/s

Therefore the velocity will be 34.2 m/s.

5 0
4 years ago
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