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4 years ago

In the thomson model of the atom, where was the positive charge located?

1 answer:

7 0

I don't know if you still need this, but here's an answer anyways.

It was a giant cloud of matter, with negatively charged electrons inside.

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What part of the gun position equipment supports all of the elevating parts of the gun?

BARSIC [14]

The gunstock it’s also called stock or shoulder stock

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3 years ago

Derive the value of electric field due to a uniform sphere of charge.

Deffense [45]

<h2>Hey there!</h2>

The Force "F" applied on the unit electric charge "q" at a point describes the electric field.

<h3>☆ Formula to find electric charge:</h3>

E = F/q

<h2>Hope it helps </h2>

4 0

2 years ago

It is easier to climb up a slanted slope than a vertical slope

V125BC [204]

IT IS EASIER TO CLIMB A SLANTED SLOPE

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3 years ago

Read 2 more answers

A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55

SpyIntel [72]

Answer:

The answer is 2,416 m/s. Let's jump in.

Explanation:

We do work with the amount of energy we can transfer to objects. According to energy theory:

W = ΔE

Also as we know W = F.x

We choose our reference point as a horizontal line at the block's rest point.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>

Under the force block gains;

W = F.x → $W=2,55.0,71=1,8105\frac{N}{m}$

In the second position block has both kinetic and potential energy. Following the law of conservation of energy;

W = ΔE = Kinetic energy + Potantial Energy

W = ΔE = $\frac{1}{2} mV^{2} + mgh$

Here we can find h in the triangle i draw in the picture using sine theorem;

In a triangle $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

In our situation

$\frac{0,71}{sin90} =\frac{h}{sin32}$ → $h=0,376$

Therefore

$1,8105=\frac{1}{2} 0,274V^{2} +0,274.9,81.0,376$

→ $V=2,416$

7 0

3 years ago

In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1850 J of work is done on the gas

hammer [34]

Answer:

The value of change in internal energy of the gas = + 1850 J

Explanation:

Work done on the gas (W) = - 1850 J

Negative sign is due to work done on the system.

From the first law we know that Q = Δ U + W ------------- (1)

Where Q = Heat transfer to the gas

Δ U = Change in internal energy of the gas

W = work done on the gas

Since it is adiabatic compression of the gas so heat transfer to the gas is zero.

⇒ Q = 0

So from equation (1)

⇒ Δ U = - W ----------------- (2)

⇒ W = - 1850 J (Given)

⇒ Δ U = - (- 1850)

⇒ Δ U = + 1850 J

This is the value of change in internal energy of the gas.

7 0

3 years ago