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3 years ago

How is impact force and impact time related when describing the same impulse?

Physics

1 answer:

Radda [10]3 years ago

7 0

Answer:

The correct option is (a)

Explanation:

Impulse is given by the product of force and time. It is denoted by J. So,

$J=F\times t$

$F=\dfrac{J}{t}$

It can be seen that the relation between the force and the impact time is inverse. Hence, the impact force and impact time possess inverse relation.

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Compared to the weak nuclear force the electromagnetic force

Crazy boy [7]

Explanation:

Weak nuclear force:

The interaction between the subatomic particles is called weak nuclear force.

The weak nuclear force is one of the four fundamental forces.

The weak nuclear force is effective at very short distance.

The range and relative strength of weak nuclear force is 10⁻¹⁸ m and 10²⁵ with respect to gravitational force respectively

Deuterium is formed due to the fusion of protons and neutrons under the action the weak force.

Example : Beta decay

Electromagnetic force:

The interaction between the charged particles is called electromagnetic force.

The electromagnetic force is one of the four fundamental forces.

The electromagnetic force is effective at long range distance.

The range and relative strength of electromagnetic force is infinity and 10³⁶ with respect to gravitational force respectively

Example : light

8 0

3 years ago

Read 2 more answers

Why hurricane is dangerous?

Olin [163]

Because you can die from the hurricane that’s why it is dangerous

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3 years ago

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What are three good things plate tectonics provide for humans?

lara31 [8.8K]

1. Fertile soil.
2. Ore deposits.
3. Fossil fuels

7 0

4 years ago

A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?

GalinKa [24]

Answer:

The electric flux through the sphere is $8.58 *10^{5} \frac{Nm^2}{C}$

Explanation:

Given

$Radius,\ r = 0.03m\\Charge,\ q =7.6\µC$

Required

Find the electric flux

Electric flux is calculated using the following formula;

Ф = q/ε

Where ε is the electric constant permitivitty

ε $= 8.8542 * 10^{-12}$

Substitute ε $= 8.8542 * 10^{-12}$ and $q =7.6\µC$ ; The formula becomes

Ф = $\frac{7.6\µC}{8.8542 * 10^{-12}}$

Ф = $\frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}$

Ф = $\frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}$

Ф = $\frac{7.6}{8.8542} *10^{12-6}}$

Ф = $0.85834970974 *10^{12-6}}$

Ф = $0.85834970974 *10^{6}}$

Ф = $8.5834970974 *10^{5}}$

Ф = $8.58 *10^{5} \frac{Nm^2}{C}$

Hence, the electric flux through the sphere is $8.58 *10^{5} \frac{Nm^2}{C}$

7 0

3 years ago

Capacitors C1 = 5.85 µF and C2 = 2.80 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon

posledela

Answer:

Q1_new = 515.68 µC

Q2_new = 246.82 µC

Explanation:

Since the capacitors are charged in parallel and not in series, then both are at 250 V when they are disconnected from the battery.

Then it is only necessary to calculate the charge on each capacitor:

Q1 = 5.85 µF * 250 V = 1462.5 µC

Q2 = 2.8 µF * 250 V = 700 µC

Now, we will look at 1462.5 µC as excess negative charges on one plate, and 1462.5 µC as excess positive charges on the other plate. Now, we will use this same logic for the smaller capacitor.

When there is a connection of positive plate of C1 to the negative plate of C2, and also a connection of the negative plate of C1 to the positive plate of C2, some of these excess opposite charges will combine and cancel each other. The result is that of a net charge:

1462.5 µC - 700 µC = 762.5 µC

Thus,762.5 µC of net charge will remain in the 'new' positive and negative plates of the resulting capacitor system.

This 762.5 µC will be divided proportionately between the two capacitors.

Q1_new = 762.5 µC * (5.85/(5.85 + 2.8)) = 515.68 µC

Q2_new = 762.5 µC * (2.8/(5.85 + 2.8) = 246.82 µC

4 0

4 years ago