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3 years ago

5

I can raise a bucket of cement mix of mass 12kg through a vertical height of 8m in 10 seconds.Calculate the average power used i

n raising the bucket against the gravitational force?

1 answer:

Andrei [34K]3 years ago

5 0

Explanation:

Power = work / time

Power = force × distance / time

P = (12 kg × 10 m/s²) (8 m) / (10 s)

P = 96 Watts

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What is a short circuit?

Marizza181 [45]

Answer:

A short circuit is an electrical circuit that allows a current to travel along an unintended path. It has no or very low electrical impedances. The opposite of a short circuit

Explanation:

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on the surface of theearth ,the weight of a boy is 400N but on a mountainpeak his weight is 360N,Calculatethe value of ''g' on t

andreyandreev [35.5K]

The value of g at sea level is 9.81 ms^-2.

The boy's mass is constant wherever he is in the universe but his weight will depend on the strength gravity where he is.

By proportion its value on the mountain peak is (360 /400) * 9.81

= 0.9 * 9.81 = 8.83 ms^-2 to nearest hundredth, (answer).

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3 years ago

What is the distance from the crest to the equilibrium of a wave called? A. amplitude B. period C. frequency D. phase

Goryan [66]

its the first one. A.

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3 years ago

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1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

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3 years ago

A 5.00 g bullet traveling 355 m/s is stopped by lodging in the side of a building. The heat produced is shared between the build

steposvetlana [31]

Answer:

Explanation:

Given

mass of bullet $m=5\ gm$

speed of bullet $v=355\ m/s$

bullet is stopped by building and heat produced is shared between building and bullet

Kinetic Energy of bullet is converted into Thermal energy

Kinetic Energy of bullet $K=\frac{1}{2}mv^2$

$K=0.5\times 5\times 10^{-3}\times (355)^2$

$K=315.06\ J$

So 315.06 J of Energy is converted in to thermal energy

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3 years ago