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mylen [45]
3 years ago
10

A force of 300 N is used to stretch a horizontal spring with a 0.5 kg block attached to it by 0.5 m. The block is released from

rest and undergoes simple harmonic motion. Calculate spring constant, mechanical force applied and velocity when x=15 cm.
Physics
1 answer:
morpeh [17]3 years ago
6 0

Answer:

0.7000cm

Explanation:

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A (blank) is a point on a standing wave that appears to be stationary
Vikentia [17]

Hi! Your answer would be "node"

7 0
3 years ago
Find the electron and hole mobilities, and the resistivity of intrinsic silicon at 300K. Is intrinsic silicon a semiconductor
tino4ka555 [31]

Answer:

Resistivity = 231.481 K Ohm

Yes, Intrinsic Silicon is the semiconductor.

Explanation:

Solution:

At 300K:

Let suppose mobility of electron in intrinsic semiconductor = M_{e}

Mobility of electron in intrinsic semiconductor is:

M_{e}  = 1300 cm^{2}/volt.sec

Let suppose mobility of hole in intrinsic semiconductor = M_{h}

M_{h} = 500 cm^{2}/volt.sec

We know that, intrinsic silicon semiconductor has equal number of holes and electrons. So,

At 300 K

Intrinsic Carrier Concentration = 1.5 x 10^{10}/cm^{3} = C

And,

Conductivity of intrinsic Silicon is:

σ = C x (M_{h} + M_{e}) e

e = 1.6 x 10^{-19} C

So, plugging in the values, we get:

σ = C x (M_{h} + M_{e}) e

σ = 1.5 x 10^{10} x (500 + 1300) x 1.6 x 10^{-19}

σ = 4.32 x 10^{-6}

So, now we can find the resistivity.

Resistivity = 1/σ

Resistivity = 1/ 4.32 x 10^{-6}

Resistivity = 231.481 K Ohm

Yes, Intrinsic Silicon is the semiconductor.

7 0
2 years ago
Brainliest for the first person to answer Which is a characteristic of an electromagnetic wave?
KiRa [710]
The first one. The E and B chatacteristic are perpendicular to eachother. The direction of the wave can be found by the right hand rule.
5 0
3 years ago
A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
Delvig [45]

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

<h3>Work done by the applied force</h3>

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

<h3>Work done by frictional force</h3>

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

<h3>Net work done by all the forces on the tractor</h3>

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

4 0
1 year ago
A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals
slamgirl [31]

Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

Where P = Power, V = Voltage, I = Current.

make I the subject of the equation

I = P/V................ Equation 2

From the question,

Given: P = 0.414 W, V = 1.50 V

Substitute into equation 2

I = 0.414/1.50

I = 0.276 A

Also,

Q = It............... Equation 3

Where Q = amount of charge, t = time

make t the subject of the equation

t = Q/I.................. Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

Q = 7.899×10⁻¹⁹ C

Substitute these value into equation 4

t =  7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

5 0
3 years ago
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