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3 years ago

6

At the circus, a 100-kilogram clown is fired 15 meters per second from a 500-kilogram cannon. What is the recoil speed of the ca

nnon?

1 answer:

Ahat [919]3 years ago

4 0

Recoil speed= 3m/s, method shown in photo

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A positively charged metal sphere, A, is held close to but not touching and identical uncharged sphere, sphere B. Sphere A is no

Yuri [45]

Answer:

The sphere C carries no net charge.

Explanation:

When brougth close to the charged sphere A, as charges can move freely in a conductor, a charge equal and opposite to the one on the sphere A, appears on the sphere B surface facing to the sphere A.
As sphere B must remain neutral (due to the principle of conservation of charge) an equal charge, but of opposite sign, goes to the surface also, on the opposite part of the sphere.
If sphere A is removed, a charge movement happens in the sphere B, in such a way, that no net charge remains on the surface.
If in such state, if the sphere B (assumed again uncharged completely, without any local charges on the surface), is touched by an initially uncharged sphere C, due to the conservation of charge principle, no net charge can be built on sphere C.

3 0

3 years ago

The diagram below shows a periodic wave. Which two points on the wave are 180 degrees out of phase?

Ksju [112]

Answer:

A and C are 180 deg out of phase (opposite points on a 360 deg wave)

3 0

3 years ago

Suppose you need your silicon circuit element to run continuously for 3 minutes before it shuts off long enough to cool back dow

Zigmanuir [339]

The maximum rate at which energy can be added to the circuit element mathematically given as

$MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}$

<h3>What is the maximum rate at which energy can be added to the circuit element?</h3>

Generally, the equation for P is mathematically given as

$P=\ln s \frac{\Delta T}{\Delta t}$

Therefore

$Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}$

$\frac{p}{lnS}=\frac{7.4 \times 10^{-3}}{23 \times 10^{-6} \times 705}$

$\frac{p}{lnS}=0.456^{\circ \mathrm{c}} / \mathrm{sec}$

Max temp Change

$MaxT=5.6^{\circ} \mathrm{C}$

$\text { time }=3 \times 60$

t=180s

In conclusion, Max Energy Rate

$MER =23 \times 10^{-6} \times \frac{301 \times 5.6}{180}$

$MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}$

Read more about Energy

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3 0

2 years ago

During an investigation, a scientist heated 123.6 g of copper carbonate till it decomposed to form a black residue. The total ma

iris [78.8K]

When copper carbonate is heated, it decomposes to form a black residue. This black residue is actually copper(II) oxide. Along with this black residue, carbon-dioxide is released. The complete reaction is the following:

$CuCO_3\to CuO(s)+CO_2(g)$

The law of conservation of mass holds true. 123.6 g of copper carbonate decomposes to 79.6 g of copper oxide and (123.6 g-79.6 g=44 g) 44 g of carbon-dioxide.

3 0

3 years ago

A car is stationary. It accelerates at 0.8 ms^2

lara31 [8.8K]

Answer:

the car’s final displacement is 60 m

Explanation:

Given;

initail velocity of the car, u = 0

acceleration of the car, a = 0.8 m/s²

time of motion, t = 10 s

The first displacement of the car:

$x_1 = ut + \frac{1}{2} at^2\\\\x_1 = 0 + \frac{1}{2} (0.8)(10)^2\\\\x_1 = 40 \ m$

The second displacement of the car;

acceleration, a = 0.4 m/s², time of motion, t = 10 s

$x_2 = ut + \frac{1}{2} at^2\\\\x_2 = 0 + \frac{1}{2} (0.4)(10)^2\\\\x_2 = 20 \ m$

The final displacement of the car;

x = x₁ + x₂

x = 40 m + 20 m

x = 60 m

Therefore, the car’s final displacement is 60 m

8 0

3 years ago