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1 year ago

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In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for

F⃗ ? Let U=0 when x=0.

1 answer:

Blizzard [7]1 year ago

7 0

Answer:

The potential energy is $-4x^3$

Explanation:

Given that,

Force $F=-\alpha x^2 i$

We need to calculate the potential energy

Using formula of work done

$\Delta U=F(x) dx$

Put the value of F

$\Delta U=-\alpha x^2\ dx$

On integration

$U=-\alpha \dfrac{x^3}{3}+C$

$U=-\dfrac{\alpha x^3}{3}+C$ ...(I)

U = 0, x = 0 so C = 0

Put the value of c and α in equation (I)

$U=-\dfrac{12x^3}{3}+0$

$U=-4x^3$

Hence, The potential energy is $-4x^3$

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A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t

Reil [10]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

7 0

2 years ago

A cylinder is 0.10 m in radius and 0.20 in length. Its rotational inertia, about the cylinder axis on which it is mounted, is 0.

Bumek [7]

Answer: $5 rad/s^2$

Explanation:

Given

Radius of cylinder r=0.1 m

Length L=0.2 in.

Moment of inertia I=0.020 kg-m^2

Force F=1 N

We Know Torque is given by

$Torque =I\alpha =F\cdot r$

where $\alpha =angular\ acceleration$

$I\alpha =F\cdot r$

$0.02\cdot \alpha =1\cdot 0.1$

$\alpha =5 rad/s^2$

5 0

2 years ago

1. Three confused sled dogs are trying to pull a sled across the Alaskan snow. Tim pulls east with a force of 42 N; Sam also pul

Bumek [7]

Answer:

Explanation:

According to the statement, three confused sleigh dogs are trying to pull a sled across the Alaskan snow.

Forces in same direction gets added , so 35N + 42N=77N and the Net Force is 77N -53N as it is acting in opposite direction.

Net force is 25N in east to the maximum without any hassle.

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thankyou : )

6 0

2 years ago

A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harm

Alisiya [41]

Answer:

The moment of inertia is 0.7500 kg-m².

Explanation:

Given that,

Mass = 2.2 kg

Distance = 0.49 m

If the length is 1.1 m

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=\dfrac{1}{12}ml^2+mx^2$

Where, m = mass of rod

l = length of rod

x = distance from its center

Put the value into the formula

$I=\dfrac{1}{12}\times2.2\times(1.1)^2+2.2\times(0.49)^2$

$I=0.7500\ kg-m^2$

Hence, The moment of inertia is 0.7500 kg-m².

5 0

2 years ago

An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect

melisa1 [442]

Answer:

The plate's surface charge density is $-8.056\times10^{-9}\ C/m^2$

Explanation:

Given that,

Speed = 9800 km/s

Distance d= 75 cm

Distance d' =15 cm

Suppose we determine the plate's surface charge density?

We need to calculate the surface charge density

Using work energy theorem

$W=\Delta K.E$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2$

Here, final velocity is zero

$W=0-\dfrac{1}{2}mv_{i}^2$ ...(I)

We know that,

$W=-Fd$

$W=-E\times e\times d$

$W=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$ ...(II)

From equation (I) and (II)

$-\dfrac{1}{2}mv_{i}^2=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$

Charge is negative for electron

$\lambda=\dfrac{mv^2\epsilon_{0}}{(-e)d}$

Put the value into the formula

$\lambda=-\dfrac{9.1\times10^{-31}\times(9800\times10^{3})^2\times8.85\times10^{-12}}{1.6\times10^{-19}\times(75-15)\times10^{-2}}$

$\lambda=-8.056\times10^{-9}\ C/m^2$

Hence, The plate's surface charge density is $-8.056\times10^{-9}\ C/m^2$

3 0

2 years ago