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3 years ago

In what ways does energy transform or convert from one form to another?

Physics

1 answer:

JulsSmile [24]3 years ago

5 0

Many ways such as freezing, melting, condensing, evaporating, sublimation, and desublimation.

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If you increase the resistance in a series circuit, ________________

GrogVix [38]

B) the current will decrease

6 0

3 years ago

Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

4 years ago

What is the volume of a cone with a height of 27 cm

JulijaS [17]

Explanation:

→ Volume of cone = πr² × h/3

Here,

Radius (r) = 13 cm
Height (h) = 27 cm

→ Volume of cone = π(13)² × 27/3 cm³

→ Volume of cone = 169π × 9 cm³

→ Volume of cone = 1521π cm³

→ Volume of cone = 1521 × 22/7 cm³

→ Volume of cone = 33462/7 cm³

→ <u>Volume of cone = 4780.28 cm³</u>

4 0

3 years ago

Help

Blizzard [7]

We can approach this in another way.
We know that sin(∅) = height / hypotenuse.

Thus, for x, height is 1 and hypotenuse is 3. Using Pythagoras theorem,
3² = 1² + b²
b = √8
cos(x) = b/hypotenuse
cos(x) = √8 / 3

Now, lets consider y:
sec(y) = 1 / cos(y) = 1 / base / hypotenuse = hypotenuse / base
The hypotenuse is 25 and the base is 24. We again apply Pythagoras theorem to find the third side, which works out to be:
height = 7
sin(y) = height / hypotenuse
sin(y) = 7/25

Now, sin(x + y) =
sin(x)cos(y) + sin(y)cos(x)
= (1/3)(24/25) + (√8 / 3)(7/25)
= 8/25 + 7√8/75
= (24 + 14√2) / 75

6 0

3 years ago

Read 2 more answers

A student pulls a 50-newton sled with a force having a magnitude of 15 newtons. What is the magnitude of the force that the sled

Simora [160]

Answer:

Force = 35 N

Explanation:

From Newton's third law of motion, the boy must apply a force greater than the weight of the sled to lift it.

weight of sled = mg

where m is its mass and g the force of gravity on it.

weight of sled = 50 N

Force applied by the boy on the sled = 15 N

Since the force applied on the sled by the boy is lesser than the weight of the sled, then;

Force that the sled exerts on the student = 50 - 15

= 35 N

The force exerted by the sled on the student is 35 N.

5 0

3 years ago