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4 years ago

How many planets are there?

Physics

1 answer:

Leno4ka [110]4 years ago

6 0

There are 7 planets in all

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What is the amount of heat, in Sl units, necessary to melt 1 lb of ice?

likoan [24]

Answer:

Heat required to melt 1 lb of ice is 151.469 KJ

Explanation:

We have given mass of ice = 1 lb

We know that 1 lb = 0.4535 kg

Latent heat of fusion for ice =334 KJ/kg

Amount if heat for fusion of ice is given by

$Q=mL$ , here m is mass of ice and L is latent heat of fusion

So heat $Q=mL=0.4535\times 334=151.469kj$

So heat required to melt 1 lb of ice is equal to 151.469 KJ

3 0

3 years ago

which of the following is the name of the process scientists use to gain knowledge abut the physical world

raketka [301]

The scientific method is the process that scientists use to answer question about the world.

7 0

3 years ago

An astronaut is on the moon. He drops a hammer from a height of 3.2metres and it takes 2.0 seconds to reach the lunar landscape.

Anvisha [2.4K]

Answer:

1/6 m/s^2 ( about 1/6th gravity of Earth ( 9.81 m/s^2)

Explanation:

Displacement = yo + vo t - 1/2 a t^2

- 3.2 = 0 + 0 - 1/2 a(2.0)^2

- 3.2 = -2a

a = 3.2 / 2 = 1.6 m/s^2

6 0

2 years ago

Another name for semi-arid climate is

pantera1 [17]

Answer: an steppe climate

Explanation:

7 0

3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

6 0

3 years ago