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3 years ago

The formulas for these two ?

Physics

1 answer:

Usimov [2.4K]3 years ago

3 0

Answer:

These are called kinematics equations.

Explanation:

These equations are used when there is either a constant velocity motion or a constant acceleration motion.

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a 12.0L container is filled with a gas to a pressure of 2660 torr at 0°C. At what temp. will the pressure inside the container b

Alexxx [7]

$\frac{p1}{t1} = \frac{p2}{t2}$

Pressure Law: constant volume

Convert all Temperatures to Kelvin

0°C= 273K

$\frac{2660}{273} = \frac{3040}{t2} \\ \\ 3040 \times 273 = t2 \times 2660 \\ \\ 829920 \div 2660 = t2 \\ \\ t2 = 312$

answer= 312Kelvin

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3 years ago

The temperature decreases 5°C in one hour and then decreases another 3°C in the next hour. Does this fit the definition of weath

Rus_ich [418]

Answer:

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8 0

3 years ago

f the CD rotates clockwise at 500 rpmrpm (revolutions per minute) while the last song is playing, and then spins down to zero an

vazorg [7]

Answer:

The magnitude of the angular acceleration is $a = 20.14 rad/s^2$

Explanation:

From the question we are told that

The angular speed of CD is $w_{CD} = 500 rpm = \frac{500 rpm}{\frac{60 \ s }{1 \ min} } * \frac{2 \pi }{ 1 \ rev} = 52.37 rad/s$

time taken to decelerate is $t_{CD} = 2.60\ s$

The final angular speed is $w_f= 0 \ rad/s$

The angular acceleration is mathematically represented as

$a = \frac{w_f - w_{CD}}{t}$

substituting values

$a = \frac{0 - 52.37}{2.60}$

$a = - 20.14 rad/s^2$

The negative sign show that the CD is decelerating but the magnitude is

$a = 20.14 rad/s^2$

3 0

3 years ago

A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the

kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂ : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F : Force ( N)

d : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the bar

mm = 1.5 kg : mass of the monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances from forces the point 2

d₁₂ = (3-0.6) m = 2.4m : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium of moments at the point 2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0

3 years ago

A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of

miskamm [114]

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= $\frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}$

= 1.72 x 10³ N.

8 0

3 years ago