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Marrrta [24]
3 years ago
8

11. A 3.8 kg object is lifted 12 meters. Approximately how much work is performed during the lifting?

Physics
1 answer:
Marianna [84]3 years ago
3 0
I found the answer for you if u need any help ask anytime!

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This is an example of using a hammer as a(n)
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, to hit nails into a piece of wood or a wall, or to break things into pieces.

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momentum A proton interacts electrically with a neutral HCl molecule located at the origin. At a certain time t, the proton’s po
arlik [135]

Answer:

\ m/s

Explanation:

F = Force =

m = Mass of proton = 1.7\times 10^{-27\ kg

t = Time taken = 2\times 10^{-14}\ s

Acceleration is given by

a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{}{1.7\times 10^{-27}}\\\Rightarrow a=\ m/s^2

v=u+at\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\\\Rightarrow v=\ m/s

The velocity of the proton is \ m/s

6 0
3 years ago
Two parallel wires carry a current I in the same direction. Midway between these wires is a third wire, also parallel to the oth
Luda [366]

Answer:

Force is repulsive hence direction of force is away from wire

Explanation:

The first thing will be to draw a figure showing the condition,

Lets takeI attractive force as +ve and repulsive force as - ve and thereafter calculating net force on outer left wire due to other wires, net force comes out to be - ve which tells us that force is repulsive, hence direction of force is away from wire as shown in figure in the attachment.

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3 years ago
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ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the
antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = M_A=5 kg

Velocity of the ball A before collision:U_A=20 m/s

Velocity of ball A after collision=V_A=10 m/s

Mass of ball B= M_B

Velocity of the ball B before collision:U_B=10 m/s

Velocity of ball B after collision=V_B=15 m/s

M_AV_A+M_BV_B=M_AU_A+M_BU_B

5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s

M_B=10kg

The mass of ball B is 10 kg.

8 0
3 years ago
Three-fourths of the area of a rectangular lawn 30 feet wide by 40 feet long is to be enclosed by a rectangular fence. If the en
satela [25.4K]

Answer:

The fence is 5feet less.

Explanation:

We need to determine

The less amount of fence required, if the enclosure has full width and reduced length, compared to full length and reduced width.

Approach & WorkingArea of lawn = 30 × 403/4th of the area of lawn = ¾(30 × 40) = 30 * 30

 When full width will be fenced, and reduced length will be fenced.

Width = 30 feet30 * L = 30 * 30Hence, length = 30 feetLength of fence needed = 2(30 + 30) = 120 feet

When full length will be fenced, and reduced width will be fenced

Length = 40 feet40 * W = 30 * 30W = 22.5 feetLength of fence needed = 2(40 + 22.5) = 125 feet

Difference in length of fence needed = 125 – 120 = 5 feet.

4 0
3 years ago
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