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3 years ago

Which of the four fundamental forces holds the planets in their orbits around the sun?

1 answer:

8 0

Gravity is the force that pulls us to the surface of the Earth, keeps the planets in orbit around the Sun and causes the formation of planets, stars and galaxies.

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What does charge does a balloon have when its rubbed on your hair?

emmainna [20.7K]

Answer:

im pretty sure thats called friction, or did you mean something else?

3 0

3 years ago

2. Can you rearrange the equation F = kx to get k on one side of the equation

Bumek [7]

k = f/x

x = f/k

( / means divide )

5 0

3 years ago

A block of mass 2 kg slides down ramp 1 with a height of 3 m. It then slides for 1 m on a surface with a kinetic coefficient of

Lady bird [3.3K]

Answer:

PE = 58.8,J

Explanation:

The potential energy at the top of ramp1 can be found by the application of the formula PE = mgh

Where PE =potential energy

m = mass of block = 2kg

g = acceleration due to gravity = 9.80m/s²

h = height of the top of the ramp from its bottom where the block is initially positioned = 3m

PE = 2×9.8×3 = 58.8J

3 0

3 years ago

A machine produces 4,000 J of work in 5 seconds. how much power does the machine produce.

ad-work [718]

Explanation:

Power is defined as the work done per unit time or

$P = \dfrac{\Delta W}{\Delta t}$

$\;\;\;\;= \dfrac{4000\:\text{J}}{5\:\text{s}} = 800\:\text{Watts}$

5 0

2 years ago

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

5 0

2 years ago

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