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bearhunter [10]

3 years ago

8

A 600-g block is dropped onto a relaxed vertical spring that has a spring constant k =190.0 N/m. The block becomes attached to t

he spring and compresses the spring 37.5 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by the gravitational force on it?

1 answer:

charle [14.2K]3 years ago

5 0

Answer:

Work done will be 2.205 j

Explanation:

We have given that the spring is compressed b 37.5 cm

So d = 0.375 m

Mass of the block m = 600 gram = 0.6 kg

Acceleration due to gravity $g=9.8m/sec^2$

Gravitational force on the block $F=mg=0.6\times 9.8=5.88N$

Now we know that work done is give by $W=Fd=5.88\times 0.375=2.205J$

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A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th

pav-90 [236]

Answer: $F_{net}=3.383\times 10^{-7}\ N$

Explanation:

Given

Mass of first object $m_1=225\ kg$

Mass of second object $m_2=525\ kg$

Distance between them $d=3.8\ m$

$m_3=61\ kg$ object is placed between them

So force exerted by $m_1$ on $m_3$

$F_{13}=\frac{Gm_1m_3}{1.9^2}$

$F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}$

$F_{13}=2.5374141274×10^{−7}\ N$

Force exerted by $m_2\ on\ m_3$

$F_{23}=\frac{Gm_2m_3}{1.9^2}$

$F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}$

$F_{23}=5.920632964\times 10^{-7}\ N$

So net force on $m_3$ is

$F_{net}=F_{23}-F_{13}$

$F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}$

$F_{net}=3.383\times 10^{-7}\ N$

i.e. net force is towards $m_2$

(b)For net force to be zero on $m_3$ , suppose

So force exerted by $m_1$ and $m_2$ must be equal

$F_{13}=F_{23}$

$\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}$

$\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}$

$\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}$

$\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}$

$\Rightarrow 3.8-x=1.52752x$

$\Rightarrow 3.8=2.52x$

$\Rightarrow x=1.507\ m$

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2 years ago

What is the effect of pressure on the solubility of gases in liquids?

34kurt

The solubility of gases in liquids increases with the increase in pressure.

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2 years ago

A forklift raises a 1,020 N crate 3.50 m up to a shelf. How much work is done by the forklift on the crate?

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Does the under water pressure depend on the molecules of the air or what does it depend on

miskamm [114]

Answer:the pressure depends on gas and it will be half as much underwater

Explanation:

Water pressure increases with the depth of the water. This is because the weight of the column of water above the object increases. But a large, shallow pond may have more water in it than a small, deep pond.

This is due to an increase in hydrostatic pressure, the force per unit area exerted by a liquid on an object. The deeper you go under the sea, the greater the pressure of the water pushing down on you. For every 33 feet (10.06 meters) you go down, the pressure increases by one atmosphere .

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3 years ago

A particular 12 V car battery can send a total charge of 110 A·h (ampere-hours) through a circuit, from one terminal to the othe

DiKsa [7]

<h2>Answer:</h2>

(a) 3.96 x 10⁵C

(b) 4.752 x 10⁶ J

<h2>Explanation:</h2>

(a) The given charge (Q) is 110 A·h (ampere hour)

Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;

=> Q = 110A·h

=> Q = 110 x 1A x 1h [1 hour = 3600 seconds]

=> Q = 110 x A x 3600s

=> Q = 396000A·s

=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C

Therefore, the number of coulombs of charge is 3.96 x 10⁵C

(b) The energy (E) involved in the process is given by;

E = Q x V -----------------(i)

Where;

Q = magnitude of the charge = 3.96 x 10⁵C

V = electric potential = 12V

Substitute these values into equation (i) as follows;

E = 3.96 x 10⁵ x 12

E = 47.52 x 10⁵ J

E = 4.752 x 10⁶ J

Therefore, the amount of energy involved is 4.752 x 10⁶ J

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3 years ago