When gravitational field lines get farther apart, the gravitational field ________

An object has a velocity of 8 m/s and a kinetic energy of 480 J. What is the mass of the object? (Formula: )

Furkat [3]

KE = 1/2 * m * v2

We have to rearrange this so the subject is mass. (because the question asks for the mass of the object) :

Mass = (2 * KE) / v

Now input the values into this equation to get your answer :

Mass = (2 * KE) / v
Mass = (2 * 480) / 8
Mass = 960 / 8
Mass = 120

7 0

2 years ago

Read 2 more answers

The figure shows a horizontal block that is suspended by two wires, a and b, which are identical except for their original lengt

nlexa [21]

It show how people work and what's they should for there living

4 0

3 years ago

A fan is designed to last for a certain time before it will have to be replaced (planned obsolescence). The fan only has one spe

otez555 [7]

Answer:

1355 days

Explanation:

Find the given attachment

3 0

3 years ago

Why is it incorrect to say iron is heavier than wood?

Tomtit [17]

Answer: It is correct to say iron is heavier than the wood.

Explanation:

It's because the density of iron is more than wood. What is this mean is iron the space between iron molecules in iron much less than as that of wood. So in iron there is more molecules than wood in same volume. More molecules means more mass, that's why iron is much heavier than wood for same volume.

8 0

3 years ago

Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106V/m ) if

xxMikexx [17]

(a) The electric field strength between two parallel conducting plates does not exceed the breakdown strength for air ( $3 \times 10^{6} V / m$ )

(b) The plates can be close together to 1.7 mm with this applied voltage

<u>Explanation:</u>

Given data:

Dielectric strength of air = $3 \times 10^{6} V / m$

Distance between the plates = 2.00 mm = $2.00 \times 10^{-3} \mathrm{m}$

Potential difference, V = $5.0 \times 10^{3} V$

We need to find

a) whether the electric field strength between two parallel conducting plates exceed the breakdown strength for air or not

b) the minimum distance at which the plates can be close together with this applied voltage.

The voltage difference (V) between two points would be equal to the product of electric field (E) and distance separation (d). The equation form is and apply all given value,

$E=\frac{V}{d}=\frac{5.0 \times 10^{3}}{2.00 \times 10^{-3}}=2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$

From the above, concluding that The electric field strength between two parallel conducting plates ( $2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$ ) does not exceed the breakdown strength for air ( $3 \times 10^{6} V / m$ )

b) To find how close together can the plates be with this applied voltage:

The formula would be,

$d_{\min }=\frac{V}{E_{\max }}$

Apply all known values, we get

$d_{\min }=\frac{5.0 \times 10^{3}}{3 \times 10^{6}}=1.7 \times 10^{-3} \mathrm{m}=1.7 \mathrm{mm}$

3 0

3 years ago

When gravitational field lines get farther apart, the gravitational field _________.