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3 years ago

Create a multimedia presentation about your favorite element.

Physics

2 answers:

Volgvan3 years ago

7 0

Answer:

Do the element oxygen

Explanation:

...

Nostrana [21]3 years ago

6 0

Answer:

,,,

Explanation:

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Gina applied 1,780 J of work to move a sofa 3.8 meters. What amount of force did she exert?

Gnoma [55]

Answer:

468.42N

Explanation:

Here,

Work done(W)=1780J

Displacement(s)=3.8m

As we have,

Work=Force×displacement

Now,

$\hookrightarrow 1780J=F\times3.8m\\\\\hookrightarrow \frac{1780J}{3.8m} =F\\\\\hookrightarrow F=468.42N$

7 0

2 years ago

If a ball is dropped from rest and falls 8 m to the ground, what is the speed just before it

Natalija [7]

Answer:

12.5m/s

Explanation:

(Assuming the question was asking for the speed just before it hit the ground)

We can use the first key equation of accelerated motion

Vf^2 = Vi^2+2aΔd

Vf^2 = 0 + 2(9.8)(8) (plugged in values, initial velocity is 0 since the ball was at rest)

Vf^2 = 156.8

Vf = 12.5 (squared both sides)

4 0

3 years ago

What is the repulsive force between two pith balls that are 10.0 cm apart and have equal charges of -40.0 NC?

galina1969 [7]

Answer:

So the repulsive force between the pith ball will be $1.44\times 10^{-3}N$

Explanation:

We have given that the pith ball have the equal charge q = -40 nC = $-40\times 10^{-9}C$

Distance between the charges = 10 cm =0.1 m

According to coulombs law $F=\frac{KQ_1Q_2}{R*2}$

$F=\frac{9\times 10^9\times -40\times 10^{-9}\times -40\times 10^{-9}}{0.1^2}=1440000\times 10^{-9}=1.44\times 10^{-3}N$

So the repulsive force between the pith ball will be $1.44\times 10^{-3}N$

5 0

3 years ago

The pressure at the bottom of a full barrel of water is Poriginal. Determine what happens to the pressure when the radius or hei

erastovalidia [21]

Answer:

C) Pnew=Poriginal/9

Explanation:

From the given conditions:

Original area, $A_o=\pi.r_o\,^2$

New area of the barrel bottom panel after increasing by a factor of 3:

$r_{new}=3r_o$

On increasing the radius the area now becomes:

$A_{new}=\pi.(3r_o)^2$

$A_{new}=9\pi.(r_o)^2=9A_o$

As we know that there will be no increase in the down-force since the height remains constant.

Pressure is given as:

$P_{original}=\frac{Force}{A_o}$

∴New pressure:

$P_{new}=\frac{Force}{9A}$

$\Rightarrow P_{new}=\frac{P_{original}}{9}$

4 0

3 years ago

For a Soap Box racer, on a 302 m long course, start from rest, and accelerating at a constant 1.27 m/s?,

Naddik [55]

B Ya. cause A is no

7 0

3 years ago