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VashaNatasha [74]
3 years ago
6

What will be the final velocity of a 5.0 g bullet starting from rest, if a net force of 45 N is applied over a time of 0.02s?

Physics
1 answer:
Wewaii [24]3 years ago
3 0

We know there’s a change in momentum due to a force applied over a time interval. Ft= m[v(final)-v(initial)]. Now simply plug in know values: (45)(0.02)=.005[v(final)-0]. Remember converting grams to kilograms. Solve for v final

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D. a person pulling a sled.

Explanation:

contact force only occurs when something directly comes in contact with another object.

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What is a magnet?<br>just the basic definition ​
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Answer:

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2 years ago
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The nucleus of a hydrogen atom is a single proton, which has a radius of about 1.2 × 10-15 m. The single electron in a hydrogen
Nutka1998 [239]

Answer: 0.86 × 10^14

Explanation:

Given the following :

Radius of proton = 1.2 × 10-15 m

Radius of hydrogen atom = 5.3 × 10-11 m

Density of proton could be calculated thus:

Mass of proton = 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (1.2 × 10^-15)^3 = 7.24 × 10^-45

Density = mass / volume

Density = (1.67 × 10^-27) / ( 7.24 × 10^-45)

= 0.2306 × 10^18

Density of hydrogen atom:

Mass of hydrogen atom= 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (5.3 × 10^-11)^3 = 6.24 × 10^-31

Density = mass / volume

Density = (1.67 × 10^-27) / ( 6.24 × 10^-31)

= 0.2676 × 10^4

Ratio is thus:

Density of proton / density of hydrogen atom

0.2306 × 10^18 / 0.2676 × 10^4 = 0.8617 × 10^14

6 0
3 years ago
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3 0
2 years ago
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When two point charges are a distance d apart, the magnitude of the electrostatic force between them is F. If the distance betwe
EastWind [94]

Answer:

The magnitude of the electrostatic force decreases by a factor 9

Explanation:

The electrostatic force between two charges is given by:

F=k\frac{q_1 q_2}{d^2}

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k is the Coulomb's constant

q1 and q2 are the two charges

d is the distance between the two charges

We see that the magnitude of the force is inversely proportional to the square of the distance. If the distance is increased to 3d: d' = 3d, the new electrostatic force would be:

F'=k\frac{q_1 q_2}{(d')^2}=k\frac{q_1 q_2}{(3d)^2}=\frac{1}{9} k\frac{q_1 q_2}{d^2}=\frac{F}{9}

So, the electrostatic force decreases by a factor 9.

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3 years ago
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