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Anna [14]
4 years ago
15

A brick in the shape of a cube with sides 10 cm has a density of 2500 kg/m^3. What is its weight? a.) 250 N

Physics
1 answer:
Arisa [49]4 years ago
8 0

Answer:

c.) 25 N

Explanation:

 We find the volume of the brick, knowing that the volume of a cube is given by the formula:

l=0,1 metros \\V=l^{3}\\V=(0,1\: \: m)^{3}=0,001\: \: m^{3}

being l the side of the cube, which in this case is 10 cm or 0,1 m. Now we find the mass of the object, knowing the density and the Volume of the cube:

m=V*d\\m=(0,001 \:\:m^{3})(2500 \: \: \frac{Kg}{m^{3}})=2,5 \:\: Kg

We find the weight by multiplying the mass of the object with the gravity constant.

W=m*g=(2.5 \:Kg)*(9,81 \:m/s^{2} )=24,5 N\approx25\: N

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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km. The First Cosmic Speed i.e. the speed of a satellite on a low lying
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10,378.82 m/s is the second cosmic speed.

69,801 km is the radius of the synchronous orbit of a satellite.

Given

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First Cosmic speed = 7.34 km/sec

1) Second cosmic speed i.e. the minimum speed required for a satellite to break free permanently from the planet is also known as the escape velocity of a satellite.

It can be calculated by

v = √2GM/r where,

v= Escape velocity of the satellite

G = Gravitational constant

M = Mass of planet

r = Radius of planet

v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 10,378.82 m/s

2) Speed of the satellite at the given period

v = 2πr/T where,

T= Time period of rotation = 16.6 × 3600 seconds

r = v×T/2π

r = (7,338.93 x 16.6 x 3600 s) / (2π)

r = 69,801 km

Hence

The <u>Second Cosmic Speed</u> i.e. the minimum speed required for a satellite to break free permanently from the planet is 10,378.82 m/s.

And the radius of the synchronous orbit of a satellite is 69,801 km.

Learn more about cosmic speed here brainly.com/question/15351190

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2 years ago
An electric current is made up of _____ charged electron.
pav-90 [236]

Answer:

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3 years ago
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A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi
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Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

m_{m} = mass of metal = 400 g

c_{m} = specific heat of metal = ?

T_{mi} = initial temperature of metal = 100 °C

m_{a} = mass of aluminum cup = 100 g

c_{a} = specific heat of aluminum cup = 900.0 J/kg ∙ K

T_{ai} = initial temperature of aluminum cup = 15 °C

m_{w} = mass of water = 500 g

c_{w} = specific heat of water = 4186 J/kg ∙ K

T_{wi} = initial temperature of water = 15 °C

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Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}

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4 years ago
Which phenomenon generates electric current?
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Answer:

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W_s = \frac{1}{2}kc^2

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Energy:

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\frac{dv}{dx}=0 if:

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