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Ilya [14]
3 years ago
7

WHICH IS THE HARDEST SUBSTANCE IN EARTH?

Physics
2 answers:
PIT_PIT [208]3 years ago
8 0
Diamond is the hardest naturally occurring substance found on Earth. 
wlad13 [49]3 years ago
3 0
Diamond is the hardest substance in Earth.
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An object has a mass of 0.250 kg. What is the gravitational force of on the object by the earth?
sesenic [268]

Answer:

2.4525 N

Explanation:

The earths gravity is 9.81 N/Kg

And so to work this out you would multiply 9.81 by 0.250 which equals to 2.4525N

7 0
3 years ago
A 150-kg object takes 1.5 minutes to travel a 2,500- meter straight path. It begins the trip traveling 120 m/s and decelerates t
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The answer is it is -1.11 m/s^2.
4 0
3 years ago
The value of the surface area of the cylinder is equal to the value of the volume of the cylinder. Find the value of x.
max2010maxim [7]

The volume of a cylinder is given by the formula v=pi r^2h, where r is the radius of the cylinder and h is the height.

<h3>What Does a Cylinder's Surface Area Look Like?</h3>

The overall area or region that the surface of a cylinder covers is referred to as its surface area. A cylinder's total surface area includes both the area of the curved surface and the area of the two flat surfaces because there are two flat surfaces and one curved surface. A cylinder's surface area is measured in square units like m2, in2, cm2, yd2, etc.

<h3>What is the cylinder's total surface area?</h3>

The sum of the curved surface areas makes up the cylinder's overall surface area.

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5 0
1 year ago
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0
3 years ago
Convert the speed of light, 3.0 x 108 m/s, to km/s.
zhuklara [117]

Answer:

Answer:

I don't know

Explanation:

I don't know

Explanation:

Answer:

I don't know

Explanation:

I don't know

3 0
3 years ago
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