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3 years ago

10

Plants in a tropical rain forest usually have

1 answer:

Karolina [17]3 years ago

6 0

The correct response is B .

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When a pendulum is at the position all the way to the left when it is swinging (at the top of the arc), what is true of the kine

JulsSmile [24]

Choice - B is the correct one.

At the top of the arc, at one end of the swing:
-- it's not going to get any higher, so the potential energy is maximum
-- it stops moving for an instant, so the kinetic energy is zero

At the bottom of the arc, in the center of the swing:
-- it's not going to get any lower, so the potential energy is minimum
-- it's not going to move any faster, so the kinetic energy is maximum

7 0

3 years ago

Cathy, a 460-N actress playing Peter Pan, is hoisted above the stage in order to "fly" by a stagehand pulling with a force of 60

SCORPION-xisa [38]

Answer:7.67

Explanation:

Given

Weight of Cathy $W_1=460\ N$

Force exerted by rope $F=60\ N$

Mechanical advantage is the ratio of load by Pulling effort

$M.A.=\frac{Load}{Pulling\ effort}$

$M.A.=\frac{460}{60}$

$M.A.=7.67\ N$

8 0

3 years ago

A gas is confined in a cylinder with a piston. What happens when work is done on the gas.

goblinko [34]

It combusts in to a vapor

5 0

3 years ago

Read 2 more answers

If the atomic number of an element is 26, then what else is also 26?

Pachacha [2.7K]

Answer:

B

Explanation:

7 0

3 years ago

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Suppose we could shrink the Earth without changing its mass. At what fraction of its current radius would the free-fall accelera

spin [16.1K]

Answer:

$R' = \frac{1}{\sqrt{3}}R$

Explanation:

The acceleration due to gravity on the surface of the Earth is given by:

$g=\frac{GM}{R^2}$

where

G is the gravitational constant

M is the mass of the Earth

R is the radius of the Earth

Here we want to find the new Earth radius R' for which the gravitational acceleration at the surface, g', would be 3 times the current value of g:

$g' = 3g$

So we would have

$\frac{GM}{R'^2}=3(\frac{GM}{R^2})$

Solving the equation for R', we find

$R'^2 = \frac{1}{3}R^2\\R' = \frac{1}{\sqrt{3}}R$

7 0

3 years ago

Read 2 more answers