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4 years ago

Determine the amount of work done when a crane lifts a 100-N block form 2m above the ground to 6m above the ground

Physics

1 answer:

Dima020 [189]4 years ago

4 0

Data given:

F=100N

Δx=6m-2m=4m

A=?

Formula needed:

A=F×s

Solution:

A=100N×4m

A=400J

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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

3 years ago

A compressed spring does not have elastic potential energy.

Alexus [3.1K]

Answer:

False

Explanation:

The second you let go its gonna release kinetic energy that's why it's potential

6 0

3 years ago

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force

galina1969 [7]

Answer:

μ =tanθ

Explanation:=

The ratio of the force of static friction and the normal reaction is equal to tanθ. F=mgsinθ. R = mgcosθ.

μ=tanθ

6 0

3 years ago

Read 2 more answers

Which of the following statements correctly describes the position of the intake and exhaust valves during most of the power sta

I am Lyosha [343]

Both valves are closed during the power stroke.

While the fuel is burning in the cylinder, you want
all the force of the expanding gases to push the
piston down ... you don't want any of the gases
or their pressure escaping.

If either of the valves was open, even just a crack,
then part of the gases would go blooey out the valve,
and some pressure would be lost that's supposed to be
pushing the piston.

5 0

3 years ago

Read 2 more answers

In the model of the hydrogen atom due to Niels Bohr, the electron moves around the proton at a speed of 3.3 × 106 m/s in a circl

irga5000 [103]

Answer:

$1.5048\times 10^{-23}\ Am^2$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

r = Radius of circle = $5.7\times 10^{-11}\ m$

v = Velocity of proton = $3.3\times 10^6\ m/s$

Magnetic moment is given by

$M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2$

The magnetic moment associated with this motion is $1.5048\times 10^{-23}\ Am^2$

5 0

3 years ago