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1 year ago

a simple machine that combines two inclined planes and is used to split objects is called a . group of answer choices

Physics

1 answer:

Semmy [17]1 year ago

5 0

Answer:

The answer is a wedge.

Explanation:

The wedge is a combination of two inclined planes. It is used to separate bodies which are held together by large forces, e.g, splitting timber

M.A.= slant height of wedge/thickness of wedge.

Hence a long thin wedge has a higher mechanical advantage than a short thick one; or the smaller the angle theta between the slant heights, the greater the mechanical advantage.

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What is required for an electric charge to flow through a wire?

max2010maxim [7]

In order to persuade the electrons in the wire to flow, you need
a potential difference between the ends of the wire. Then the
electrons will want to get away from the more-negative end and
go to the more-positive end. If both ends of the wire are at the
same potential, then the electrons have no reason to go anywhere,
and they just stay where they are.

Choice-d says this.

5 0

3 years ago

Read 2 more answers

The total consumption of electrical energy in the United States is about 1.0 1019 joules per year.

dexar [7]

Answer:

a) 316880878140.2895 Joules

b) 1056.26 Joules

c) 372.80103310622 km²

Explanation:

Energy consumption

$E=\frac{P}{t}\\\Rightarrow E=\frac{1\times 10^{19}}{365.25\times 24\times 60\times 60}\\\Rightarrow E=316880878140.2895\ W$

Average rate of electrical energy consumption is 316880878140.2895 Joules

Total population = 300 million

$E=\frac{316880878140.2895}{300\times 10^6}\\\Rightarrow E=1056.26\ J$

Each person consumes 1056.26 Joules.

Efficiency = 0.85

Power from Sun

$0.85\times 1000=850\ W/m^2$

Area

$\frac{316880878140.2895}{850}=372801033.10622\ m^2$

Converting to km²

$372801033.10622=\frac{372801033.10622}{1000\times 1000}=372.80103310622\ km^2$

Area required to collect the electrical energy used in the United States is 372.80103310622 km²

4 0

2 years ago

Read 2 more answers

The blank tells how many and the blank tells of what

motikmotik

Answer: and a theme

Explanation:

3 0

3 years ago

Read 2 more answers

What is the magnitude of the electric field at a point midway between a −5.6μC and a +5.8μC charge 9.0cm apart? Assume no other

Rashid [163]

Answer:

Magnitude of electric field at mid way is $5.07\times 10^{7} N/C$

Solution:

As per the question:

Q = $- 5.6\mu m = - 5.6\times 10^{- 6} C$

Q' = $5.8\mu m = 5.8\times 10^{- 6} C$

Separation distance, d = 9.0 cm = 0.09 m

The distance between charges at mid-way, O is $\farc{d}{2} = 0.045 m$

Now, the electric at point O due to charge Q is:

$E = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}$

$E = \frac{1}{4\pi \epsilon_{o}}.\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}$

$E = (9\times 10^{9})\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}$

$E = - 2.49\times 10^{7} N/C$

Here, negative sign is indicative of the direction of electric field which is towards the point O

Now, the electric at point O due to charge Q' is:

$E' = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}$

$E' = (9\times 10^{9})\frac{5.8\times 10^{- 6}}{(0.045^{2}}$

$E' = 2.58\times 10^{7} N/C$

Refer to Fig 1.

Since, both the fields are in the same direction:

$E_{net} = E + E' = 2.49\times 10^{7} + 2.58\times 10^{7} = 5.07\times 10^{7} N/C$

3 0

3 years ago

A farmhand pushes a 26-kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 88 N on the hay, how

Tamiku [17]

Answer:

W = 343.2 J

Explanation:

Given that,

Mass of bale of hay = 26 kg

Horizontal force exerted = 88 N

Distance moved, d = 3.9 m

Work done, W = Fd

Put all the values,

W = 88 N × 3.9 m

= 343.2 J

So, the work done is 343.2 J.

7 0

3 years ago