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3 years ago

An electrical connection between an electrical circuit or equipment and the earth is called a

1 answer:

3 0

I'd say ground. <span>Voltage in the US is 110-120 volts while in other countries it is 220-240 volts. if you have a 110-120 volt appliance, you need a step-up transformer (110 to 220)before you could use it in a 220 volt power supply.</span>

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The electrical force on a 2-c charge is 60 n. the electric field where the charge is located is

Kazeer [188]

The electrical force acting on a charge q immersed in an electric field is equal to
$F=qE$
where
q is the charge
E is the strength of the electric field

In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
$E= \frac{F}{q}= \frac{60 N}{2 C}=30 N/C$

5 0

3 years ago

I need help with part D

natka813 [3]

1). The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.

2). The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.

3). The block is also affected by friction with the air (air resistance) as it
falls to the ground. Friction robs some kinetic energy.

8 0

3 years ago

A pingpong ball has 2 kg m/s of momentum when thrown 8 m/s find the mass of the ball

avanturin [10]

Answer:

0.25 kg

Explanation:

p = mv

2 = m(8)

2/8 = m(8)/8 *cancels

m = 1/4 OR 0.25 kg

4 0

2 years ago

What causes the earths plates to move

Norma-Jean [14]

The intense heat in the earth's core that causes molten rock in the mantle layer to move.

8 0

3 years ago

Read 2 more answers

A student team is to design a human powered submarine for a design competition. The overall length of the prototype submarine is

Allushta [10]

Answer:

a) The speed is 61.42 m/s

b) The drag force is 10.32 N

Explanation:

a) The Reynold´s number for the model and prototype is:

$Re_{m} =\frac{p_{m}V_{m}L_{m} }{u_{m} }$

$Re_{p} =\frac{p_{p}V_{p}L_{p} }{u_{p} }$

Equaling both Reynold's number:

$\frac{p_{p}V_{p}L_{p} }{u_{p} }=\frac{p_{m}V_{m}L_{m} }{u_{m} }$

Clearing Vm:

$V_{m} =\frac{p_{p}V_{p}L_{p} u_{m} }{u_{p} p_{m} L_{m} }=\frac{999.1*0.56*8*1.849x10^{-5} }{1.138x10^{-3}*1.184*1 } =61.42m/s$

b) The drag force is:

$\frac{F_{Dm} }{p_{m}V_{m}^{2}L_{m}^{2} } =\frac{F_{Dp} }{p_{p}V_{p}^{2}L_{p}^{2} } \\F_{Dp} =\frac{F_{Dp}p_{p}V_{p}^{2}L_{p}^{2} }{p_{m}V_{m}^{2}L_{m}^{2} } \\F_{Dp}=\frac{2.3*999.1*0.56^{2} *8^{2} }{1.184*61.42^{2}*1^{2} } =10.32N$

6 0

3 years ago

Read 2 more answers