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3 years ago

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A 1200-kg car initially at rest undergoes constant acceleration for 9.4 s, reaching a speed of 11 m/s. It then collides with a s

tationary car that has a perfectly elastic spring bumper. What is the final kinetic energy of the two car system?

1 answer:

Natalka [10]3 years ago

8 0

To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,

$KE_i = KE_f$

$KE_f = \frac{1}{2} mv^2$

Here,

m = Mass

V = Velocity

Replacing,

$KE_f = \frac{1}{2} (12000)(11)^2$

$KE_f = 72600J$

Therefore the final kinetic energy of the two car system is 72.6kJ

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A force of 10 N causes a spring to extend by 20 mm. Find a) the spring constant of the spring in N/m b) the extension of the spr

suter [353]

(a) The spring constant is 500 N/m.

(b) The extension of the spring when 25 N force is applied is 0.05 m.

(c) The applied force to cause an extension of 5 mm is 2.5 N.

The given parameters:

Applied force, F = 10 N
Extension of the spring, x = 20 mm

The spring constant is calculated as follows;

$F = kx\\\\k = \frac{F}{x} \\\\k = \frac{10}{20 \times 10^{-3}} \\\\k = 500 \ N/m$

The extension of the spring when 25 N force is applied is calculated as follows;

$F = kx\\\\x = \frac{F}{k} \\\\x = \frac{25}{500} \\\\x = 0.05 \ m$

The applied force to cause an extension of 5 mm is calculated as follows;

$F = kx\\\\F = 500 \times 5 \times 10^{-3}\\\\F = 2.5 \ N$

Learn more about Hook's law here: brainly.com/question/12253978

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2 years ago

Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio

Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

Explanation:

Given that,

First charge $q_{1}= 20\mu C$

second charge $q_{2}= 8\mu C$

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

$E_{1}= \dfrac{kq_{1}}{r^2}$

Put the valueinto the formula

$E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}$

$E_{1}=18\times10^{5}\ N/C$

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

$E_{2}= \dfrac{kq_{2}}{r^2}$

Put the valueinto the formula

$E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}$

$E_{2}=7.2\times10^{5}\ N/C$

Direction of electric field along AO

We need to calculate the net electric field at midpoint

$E_{net}=E_{1}-E_{2}$

$E_{net}=(18-7.2)\times10^{5}\ N/C$

$E_{net}=10.8\times10^{5}\ N/C$

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

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2 years ago

The photons of different light waves:

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Answer: contain different amounts of energy

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As the energy of the photon depends on its frequency:

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2 years ago

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Answer:

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Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

$F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}$ .....(2)

Dividing equation (1) and (2), we get :

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