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3 years ago

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A proton is released from rest at the origin in a uniform electric field that is directed in the positive x direction with magni

tude 950 \text{ N/C}950 N/C. What is the change in the electric potential energy of the proton-field system when the proton travels to x = 2.5 \text{ m}x=2.5 m

1 answer:

Vika [28.1K]3 years ago

4 0

Explanation:

It is given that,

Electric field, $E=950\ N/C$

We need to find the change in the electric potential energy of the proton-field system when the proton travels to x = 2.5 m

From the conservation of energy, the loss in potential energy is equal to the gain in kinetic energy and kinetic energy is is equal to the work done.

$W=F\times x$

$W=q\times E\times x$

$W=1.6\times 10^{-19}\times 950\times 2.5$

$W=3.8\times 10^{-16}\ J$

So, the change in electric potential energy of the proton field system is $3.8\times 10^{-16}\ J$ . Hence, this is the required solution.

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P14.003A spherical gas-storage tank with an inside diameter of 8.1 m is being constructed to store gas under an internal pressur

Artist 52 [7]

Answer:

The minimum wall thickness required for the spherical tank is 0.0189 m

Explanation:

Given data:

d = inside diameter = 8.1 m

P = internal pressure = 1.26 MPa

σ = 270 MPa

factor of safety = 2

Question: Determine the minimum wall thickness required for the spherical tank, tmin = ?

The allow factor of safety:

$\sigma _{a} =\frac{\sigma }{factor-of-safety} =\frac{270}{2} =135MPa$

The minimun wall thickness:

$t=\frac{Pd}{4\sigma _{a} } =\frac{1.26*8.1}{4*135} =0.0189m$

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3 years ago

A gun is fired with angle of elevation 30°. What is the muzzle speed if the maximum height of the shell is 544 m? (Round your an

Artemon [7]

Answer:

206.62313 m/s

Explanation:

u = Muzzle speed

g = Acceleration due to gravity = 9.8 m/s²

$\theta$ = Angle at which the bullet is fired = 30°

h = Maximum height = 544 m

Maximum height is given by

$h=\dfrac{u^2sin^2\theta}{2g}\\\Rightarrow u=\sqrt{\dfrac{2gh}{sin^2\theta}}\\\Rightarrow u=\sqrt{\dfrac{2\times 9.81\times 544}{sin^2(30)}}\\\Rightarrow u=206.62313\ m/s$

The muzzle speed is 206.62313 m/s

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4 years ago

A 1250-kg car moves at 20.0 m/s. How much work must be done on the car to increase its speed to 30.0 m/s.

bulgar [2K]

Answer:

312,497.5Joules

Explanation:

Work done = force × distance

W = FS

Get the force

F = ma

F = 1250×9.8

F = 12250N

Get the distances using the equation of motion

v² = u² +2gS

30² =20²+2(9.8)S

900 =400+19.6S

900-400 =19.6S

500 = 19.6S

S = 500/19.6

S = 25.51m

Get the work done

Work done = 12250×25.51

Workdone = 312,497.5Joules

6 0

3 years ago

A hoop and a solid disc are relased from rest

Murljashka [212]

Answer:

1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

<u>For the hoop:</u>

$K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0$

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

$I_h = m_hR^2$

Let's continue with the energy equations:

$m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}$

Similarly <u>for the solid disk</u> with a moment of inertia of (1/2)mR^2:

$K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}$

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

$v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2$

where a is the linear acceleration.

The relation with the angular and linear acceleration is

$a = \alpha R$

where R is the radius of the tire. Since it is not given in the question, we will leave it as R.

The angular acceleration of the small pebble is

$\alpha = 2.235/R ~m/s^2$

4 0

4 years ago

Read 2 more answers

a/ Compute the quantity of heat released by 25.0 g of steam initially at 100.0oC, when it is cooled to 34.0°C and by 25.0 g of w

Svetach [21]

Answer:

When put into steam

Explanation:

When a certain amount of steam at boiling temperature condenses (turning into water), the amount of heat released is

$Q_1 = m\lambda_v$

where in this case

m = 25.0 g = 0.025 kg is the mass of steam at 100.0°C

$\lambda_v=2.256\cdot 10^6 J/kg$ is the latent heat of vaporization of water

So,

$Q_1=(0.025)(2.256\cdot 10^6)=56400 J$

Instead, the amount of heat released when the water at 100.0°C is cooled down to 34.0°C is given by

$Q_2=mC\Delta T$

where

m = 25.0 g = 0.025 kg is the mass of water

$C=4.19\cdot 10^4 J/kg K$ is the specific heat of water

$\Delta T=100-34=66^{\circ}C$ is the change in temperature

Therefore,

$Q_2=(0.025)(4.19\cdot 10^3)(66)=6913 J$

Since $Q_1>Q_2$ , we can say that your hand will burn more in the first case.

5 0

3 years ago