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2 years ago

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A 6.50 × 10–3 m2 piston compresses gas in a cylinder with a surface area of 9.75 × 10–2 m2. What is the force on the cylinder wa

lls if 50.0 N are applied to the piston?

1 answer:

vodomira [7]2 years ago

8 0

Yes it is 2cm and 1 quarter

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A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed

Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

$\int EdS=\frac{Q_{in}}{\epsilon_o}$ (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

$\int EdS=ES=E(4\pi r^2)$ (2)

Qin is calculate by using the charge density:

$Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho$ (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

$\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}$

Next, you use the results of (3), (2) and (1):

$E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})$

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

$E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}$

hence, the electric field is 1580594.95 N/C

7 0

2 years ago

a 2000kg car initially traveling at a speed of 15 m/s is accelerated by a constant force of 10000 n for 3 seconds. the new speed

Juliette [100K]

Mass of the car (m) = 2000 Kg
Initial velocity (u) = 15 m/s
Force (F) = 10000 N
Time (t) = 3 s
Let the acceleration be a.
By using the formula, F = ma, we get,
10000 N = 2000 Kg × a
or, a = 10000 N ÷ 2000 Kg
or, a = 5 m/s^2
Let the final velocity be v.
By using the formula, v = u + at, we get,
v = 15 m/s + 5 m/s^2 × 3 s
or, v = 15 m/s + 15 m/s
or, v = 30 m/s

<u>Answer</u><u>:</u>

<em><u>The </u></em><em><u>new </u></em><em><u>sp</u></em><em><u>e</u></em><em><u>ed </u></em><em><u>of </u></em><em><u>the </u></em><em><u>car </u></em><em><u>is </u></em><em><u>3</u></em><em><u>0</u></em><em><u> </u></em><em><u>m/</u></em><em><u>s.</u></em>

Hope you could get an idea from here.

Doubt clarification - use comment section.

8 0

2 years ago

When during new product development is Design For Manufacture and Assembly (DFMA) most effective? a) At all times b) During prod

Lina20 [59]

Answer:most likely E

Explanation:

Why would anybody do something after design there done with there work after that

3 0

3 years ago

The force it would take to accelerate a 900-kg car at a rate of 3 m/s2 is

Vlad1618 [11]

F = ma

F = (3)(900)

F = 2700 N

4 0

2 years ago

Read 2 more answers

An object is moving along the x axis. At t = 5.6 s, the object is at x = +3.0 m and has a velocity of +5.7 m/s. At t = 8.5 s, it

Debora [2.8K]

The average acceleration between t = 5.6 s and t = 8.5 s is 2.31 m/s²

<h3>What is acceleration?</h3>

Acceleration is defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

An object is moving with initial velocity u =5.7 m/s and its final velocity v= -1.0 m/s.

Time taken for the change in speed, t= 8.5 - 5.6 = 2.9 seconds

The acceleration is given by

a = (-1 - 5.7)/ 2.9

a = - 2.31 m/s²

|a | = 2.31 m/s²

Thus, the object's acceleration is 2.31 m/s²

Learn more about acceleration.

brainly.com/question/12550364

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3 0

2 years ago