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g100num [7]
3 years ago
5

Which of the following physical laws or principles can best be used to analyze the collision between the object and the pendulum

bob? Which can best be used to analyze the resulting swing? 1. Newton's first law 2. Newton's second law 3. Newton's third law 4. Conservation of mechanical energy 5. Conservation of momentum
Physics
1 answer:
posledela3 years ago
4 0

3. Newton's third law

5. Conservation of momentum

<u>Explanation:</u>

Conservation of momentum is mostly used for describing collisions between objects. Here, the type of collision is inelastic collision in which the object when collides with the pendulum bob sticks to it and moves as a combined object. In this process the momentum is conserved.

Let the mass of the pendulum be m1 moving with a velocity v1.

Let the mass of the object be m2 moving with a velocity v2.

Since the momentum is conserved during collision, the equation will be

m1 v1 + m2 v2 = (m1 + m2) v

Where, v is the velocity of the combined system.

Conservation of momentum is actually a direct consequence of Newton's third law.

Consider a collision between two objects, object A and object B. When the two objects collide, there is a force on A due to B. However, because of Newton's third law, there is an equal force in the opposite direction, on B due to A

FAB = -FBA

The mechanical energy is not conserved due to the fact that the kinetic energy is not the same before and after the collision.

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An elf pushes a sleigh with force of 9N over a distance of 11m. How much work did the elf. Do on the sleigh
Whitepunk [10]

Answer:

<h2>99 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 9 × 11

We have the final answer as

<h3>99 J</h3>

Hope this helps you

5 0
3 years ago
A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume th
Katyanochek1 [597]

Answer:

9.88 milivolt

Explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf =  - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV

3 0
3 years ago
How long does it take long a light from the sun to reach the earth if it travels a distance of 1.5 × 10^11m? (velocity of light
mojhsa [17]

Answer:

Therefore, light travelling at 3.0x10^8 meters per second takes 500 seconds (8 minutes, 20 seconds) to reach the Earth, which is 1.5x10^11 meters away from the sun

Explanation:

7 0
3 years ago
Across what potential difference does an electron have to be accelerated in order to reach the speed v=3e7 m/s?
zmey [24]

When an electron is accelerated through potential difference then the speed that it attain will be explained by energy conservation

here by energy conservation we can say that

change in kinetic energy of electron = electrostatic potential energy gained through given potential difference

kinetic energy is given as

KE = \frac{1}{2}mv^2

electrostatic potential energy is given as

PE = qV

now by energy conservation

qV = \frac{1}{2}mv^2

given that for electron

m = 9.1 * 10^{-31} kg

v = 3 * 10^7 m/s

q = 1.6 * 10^{-19} C

now by plug in values

1.6 * 10^{-19} * V = \frac{1}{2}*9.1* 10^{-31} *(3*10^7)^2

V = \frac{4.095 * 10^{-16}}{1.6 * 10^{-19}}

V = 2559.4 Volts

So here it is accelerated through potential difference of 2559.4 Volts

4 0
3 years ago
A puck sliding on ice approaches a 20 degrees ramp at 10 m/s with coefficient of friction 0.2. How high will it go up the ramp a
timama [110]

Answer:

it will go up along the inclined plane by d = 9.62 m

Explanation:

As we know that puck is moving upward along the slide

then the net force opposite to its speed is given as

F_{net} = - mgsin\theta - \mu mgcos\theta

so here deceleration is given as

a = -g(sin\theta + \mu cos\theta)

now plug in all values in it

a = -9.81(sin20 + 0.2cos20)

a = -5.2 m/s^2

now the distance covered by the puck along the plane is given as

v_f^2 - v_i^2 = 2 a d

0 - 10^2 = 2(-5.2)d

d = 9.62 m

3 0
3 years ago
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