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OverLord2011 [107]
3 years ago
6

Identify Label these cell parts

Physics
2 answers:
Kruka [31]3 years ago
8 0

Answer:

I don't know this answer because I am not American

navik [9.2K]3 years ago
6 0
I would say nucleus I’m not so sorry
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(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup
KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx

Where

x_{o}, x_{f} - Initial and final position, respectively, measured in meters.

F(x) - Force as a function of position, measured in newtons.

Given that F = k\cdot x and the fact that F = 25\,N when x = 0.3\,m - 0.2\,m, the spring constant (k), measured in newtons per meter, is:

k = \frac{F}{x}

k = \frac{25\,N}{0.3\,m-0.2\,m}

k = 250\,\frac{N}{m}

Now, the work function is obtained:

W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx

W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]

W = 0.313\,J

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be r(\theta) = 2\cdot \sin 5\theta. The area of the region enclosed by one loop of the curve is given by the following integral:

A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta

A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta

By using trigonometrical identities, the integral is further simplified:

A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta

A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta

A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta

A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)

A = 4\pi

The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

5 0
3 years ago
Amount of work done by a rotating object
Oduvanchick [21]
The work done by a rotating object can be calculated by the formula Work = Torque * angle.

This is analog to the work done by the linear motion where torque is analog to force and angle is analog to distance. This is Work = Force * distance.

An example will help you. Say that you want to calculate the work made by an engine that rotates a propeller with a torque of 1000 Newton*meter over 50 revolution.

The formula is Work = torque * angle.

Torque = 1000 N*m

Angle = [50 revolutions] *  [2π radians/revolution] = 100π radians

=> Work = [1000 N*m] * [100π radians] = 100000π Joules ≈ 314159 Joules of work.

 
5 0
3 years ago
Read 2 more answers
If a cart of 10 kg mass has a force of 5 newtons exerted on it, what is its acceleration? m/s2
Elza [17]

Answer:

= 0.5 m/s²

Explanation:

  • According to Newton's second law of motion, the resultant force is directly proportion to the rate of change of linear momentum.

Therefore;<em> F = ma , where F is the Force, m is the mass and a is the acceleration.</em>

<em>Thus; a = F/m</em>

<em>but; F = 5 N, and m = 10 kg</em>

<em>  a = 5 /10</em>

    <u>= 0.5 m/s²</u>

4 0
3 years ago
Read 2 more answers
g a stone with mass m=1.60 kg IS thrown vertically upward into the air with an initial kinetic energy of 470 J. the drag force a
otez555 [7]

Answer:

Height reached will be 28.35 m

Explanation:

Here we can use the work energy theorem to find the maximum height

As we know by work energy theorem

Work done by gravity + work done by friction = change in kinetic energy

-mgh - F_f h = 0 - \frac{1}{2}mv_i^2

now we will have

-1.60(9.8)(h) - 0.900(h) = - 470

-16.58 h = -470

h = 28.35 m

so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction

5 0
3 years ago
Passengers on an airplane move from rest to 86 meters per second before the airplane takes off. If the airplane takes 100 second
VikaD [51]

Utilize the formula:  V _{f} = V _{i} + a\Delta t

V _{f} = Final Velocity (86 m/s)

V _{f} = Initial Velocity (0 m/s)

a = acceleration (m/s²)

\Deltat = Time (100 seconds)

As a result,

86 m/s = 0 + (a)(100 seconds)

Using algebra, divide 86 m/s by 100 seconds:

86 m/s = 100a

a = 0.86 m/s²

Rounded to one decimal place: 0.9 m/s²

Let me know if you have any questions!

4 0
3 years ago
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