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3 years ago

The use of a simple machine will reduce the amount of work needed to move an object.true false

Physics

1 answer:

Eddi Din [679]3 years ago

3 0

False they simple machines do not reduce the amount of work you do it would just reduce the amount of force you use.

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A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in

Kruka [31]

Answer:

1020g

Explanation:

Volume of can= $1100cm^3=1100\times 10^{-6}m^3$

$1cm^3=10^{-6}m^3$

Mass of can=80g= $\frac{80}{1000}=0.08kg$

1Kg=1000g

Density of lead= $11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3$

By using $1g/cm^3=10^3kg/m^3$

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can and lead inside it they are floating in the water.

Buoyancy force = $F_b=Weight of can+weight of lead$

$\rho_wV_cg=m_cg+m_lg$

Where $\rho_w=10^3kg/m^3=$ Density of water

$m_c=$ Mass of can

$m_l=$ Mass of lead

$V_c=$ Volume of can

Substitute the values then we get

$1000\times 1100\times 10^{-6}=0.08+m_l$

$1.1-0.08=m_l$

$m_l=1.02 kg=1.02\times 1000=1020g$

$1 kg=1000g$

Hence, 1020 grams of lead shot can it carry without sinking water.

4 0

4 years ago

Force is measured by: a) ammeter b) spring balance c) two pan balance d) ruler

laila [671]

Answer:

force can also be considered as weight so

Spring balance is used for measuring Force.

Explanation:

8 0

3 years ago

Read 2 more answers

What is the average speed of particles of atoms at room temperature?

Paha777 [63]

Answer:

300 meters per second. That's equal to about 670 miles per hour.

Explanation:

Not only are air particles incredibly small, they are always moving. And they move fast. At room temperature, they are going about 300 meters per second. That's equal to about 670 miles per hour.

8 0

4 years ago

Energy flows in a circuit in which direction?

victus00 [196]

Answer:

It flows in one direction. It goes out the positive side and back into the negative side.

Explanation:

There are images that can help you understand better. I hope this helped a little :)

7 0

3 years ago

The intensity of sunlight reaching the earth is 1360 w/m2. Assuming all the sunligh is absorbed, what is the radiation pressure

krok68 [10]

(a) $1.15\cdot 10^9 N$

For an electromagnetic wave incident on a surface, the radiation pressure is given by (assuming all the radiation is absorbed)

$p=\frac{I}{c}$

where

I is the intensity

c is the speed of light

In this problem, $I=1360 W/m^2$ ; substituting this value, we find the radiation pressure:

$p=\frac{1360 W/m^2}{3\cdot 10^8 m/s}=4.5\cdot 10^{-6} Pa$

the force exerted on the Earth depends on the surface considered. Assuming that the sunlight hits half of the Earth's surface (the half illuminated by the Sun), we have to consider the area of a hemisphere, which is

$A=2pi R^2$