All categories

3 years ago

Numbered 8 nowwwwwww

Physics

1 answer:

amm18123 years ago

7 0

Torque = r x F

|F| = mg = 60 * 10 N = 600 N ( assuming g ~ 10m/s^2)

distance of fulcrum = torque / Force = 90/600 m = .15 m.

You might be interested in

Physicists and engineers from around the world came together to build the largest accelerator in the world, the Large Hadron Col

Vsevolod [243]

Answer:

Explanation:

From the question,

The kinetic energy of protons

K = 7eV.

Mass of proton and the speed of light are the given parameters.

Please kindly go through the files I attached for a step by step explanation of the answer required.

5 0

3 years ago

A gas has a pressure of 48atm in a 15.5L container. It was found that at 25∘C the gas occupied a volume of 25L and had a pressur

EleoNora [17]

Answer:

130.165636364°C

Explanation:

P = Pressure

V = Volume

n = Number of moles

R = Gas constant = 0.082 L atm/mol K

From ideal gas law we have

$PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{22\times 25}{0.082\times (25+273.15)}\\\Rightarrow n=22.496451696\ moles$

$PV=nRT\\\Rightarrow T=\dfrac{PV}{nR}\\\Rightarrow T=\dfrac{48\times 15.5}{22.496451696\times 0.082}\\\Rightarrow T=403.315636364\ K$

The initial temperature is $403.315636364-273.15=130.165636364\ ^{\circ}C$

3 0

4 years ago

Two electrons in a vacuum exert force of F = 3.8E-09 N on each other. They are then moved such that they are separated by x = 8.

iren [92.7K]

Answer:

F_n = 5.65E-11 N

d = 1.20682E-31 m

Explanation:

F = 3.8E-09 N

where

m = Mass of electron = 9.109E−31 kilograms

G = Gravitational constant = 6.67E-11 m³/kgs²

x = Distance between them

$F=G\frac{m^2}{x^2}\\\Rightarrow 3.8E-09=G\frac{m^2}{x^2}$

For $F_n$

$F_n=G\frac{m^2}{x^2}\\\Rightarrow F_n=G\frac{m^2}{(8.2x)^2}\\\Rightarrow F_n=G\frac{m^2}{67.24x^2}$

Dividing the above equations we get

$\frac{F}{F_n}=\frac{G\frac{m^2}{x^2}}{G\frac{m^2}{67.24x^2}}\\\Rightarrow \frac{F}{F_n}=67.24\\\Rightarrow F_n=\frac{F}{67.24}\\\Rightarrow F_n=\frac{3.8E-09}{67.24}\\\Rightarrow F_n=5.65E-11\ N$

F_n = 5.65E-11 N

$F=G\frac{m^2}{x^2}\\\Rightarrow x=\sqrt{\frac{Gm^2}{F}}\\\Rightarrow x=\sqrt{\frac{G}{F}}m\\\Rightarrow x=\sqrt{\frac{6.67E-11}{3.8E-09}}9.109E-31\\\Rightarrow x=1.20682E-31\ m$

d = 1.20682E-31 m

8 0

3 years ago

What is the amount of heat required to raise the temperature of 200.0 g of aluminum by 10 c?

mash [69]

Q = 175.8J = 1.8• 10 2 J

5 0

3 years ago

In which of the two situations described is more energy transferred?

Furkat [3]

Answer:

More energy is transferred in situation A

Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m· $H_f$

Where;

$H_f$ = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0

3 years ago