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Lostsunrise [7]

3 years ago

8

What happens when an ionic compound is melted?

1 answer:

Dahasolnce [82]3 years ago

7 0

It conducts electricity. This is because the electrons are able to move around more for it has gained energy, after being heated.

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A diamond with a mass of 45 g hangs motionless from a chain. what is the upward force of the chain on the diamond?

Oksi-84 [34.3K]

The upward force of the chain on the diamond would be the tension in the chain, and this tension would have to support the weight of the 45g that hangs from the chain.

mass = 45 g = 45/1000 kg = 0.045kg

Weight = mg = 0.045 * 10 ≈ 0.45N, g ≈ 10 m/s²

<span>So the upward force is ≈ </span><span>0.45N. </span>

6 0

3 years ago

Which term refers to the upper part of a glacier that receives the most snowfall?

liq [111]

The answer is A) accumulation zone

6 0

3 years ago

An object is moving in a circle. If the radius of the object is doubled, and the period remains constant, the magnitude of the v

Anton [14]

Answer:

Twice

Explanation:

From the formula for velocity in a circle

V= 2πr/T

Where V is velocity

r is raduis

T is period

We see that as r increases V increases so if r is doubled V becomes doubled

4 0

3 years ago

How much force is required to accelerate a 2kg mass at 3 m/s2? answer key?

mr Goodwill [35]

F = ma so u can plug in the given numbers and solve:

F = (2)(3)

4 0

3 years ago

Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig

HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

$y=-\frac{1}{2} gt^2+v_0t+y_0$

With the given values in the question the equation has one unknown v₀:

$v_0=\frac{y-y_0}{t}+\frac{1}{2}gt$

Solving for t=1:

1) $v_0=y-y_0+\frac{g}{2}$

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) $E=\frac{1}{2}m{v_0}^2+mgy_0$

If h is the height of the tower, the energy on top of the tower:

3) $E=mgh$

Combining equation 2 and 3 and solving for h:

4) $h=\frac{{v_0}^2}{2g}+y_0$

Combining equation 1 and 4:

$h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0$

4 0

3 years ago