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adoni [48]
3 years ago
13

1. When does raising the temperature of a gas increase its pressure? when volume is increased and the number of particles is con

stant when volume is increased and the number of particles is increased when volume and the number of particles are constant
Physics
1 answer:
Neporo4naja [7]3 years ago
3 0

Answer:

when volume and the number of particles are constant

Explanation:

Gay Lussac law states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Gay Lussac's law is given by;

PT = K

\frac{P1}{T1} = \frac{P_{2}}{T_{2}}

The ideal gas law is the equation PV = nRT

Where;

P is the pressure.

V is the volume.

n is the number of moles of substance.

R is the ideal gas constant.

T is the temperature.

Generally, raising the temperature of an ideal gas would increase its pressure when volume and the number of particles are constant.

This ultimately implies that, when volume and the number of particles are held constant, there would be a linear relationship between the temperature and pressure of a gas i.e temperature would be directly proportional to the pressure of the gas. Thus, an increase in the temperature of the gas would cause an increase in the pressure of the gas at constant volume and number of particles.

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A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
2 years ago
An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard that is at a distance o
Minchanka [31]

Answer:

B. 0.16 m

Explanation:

The vertical distance by which the player will miss the target is equal to the vertical distance covered by the dart during its motion.

Since the dart is thrown horizontally, the initial vertical velocity is zero:

v_y = 0

While the horizontal velocity is

v_x = 15 m/s

The horizontal distance covered is

d_x = 2.7 m

Since the dart moves by uniform motion along the horizontal direction, the time it takes for covering this distance is

t=\frac{d_x}{v_x}=\frac{2.7 m}{15 m/s}=0.18 s

along the vertical direction, the motion is a uniformly accelerated motion with constant downward acceleration g=9.8 m/s^2, so the vertical distance covered is given by

d_y = \frac{1}{2}gt^2=\frac{1}{2}(9.8 m/s^)(0.18 s)^2=0.16 m

8 0
3 years ago
a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in
Kryger [21]

The average power is 3.0\cdot 10^6 W

Explanation:

First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:

W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2

where :

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the car, with

m = 2000 kg is the mass of the car

v = 60 m/s is the final speed of the car

K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the car, with

u = 30 m/s is initial speed of the car

Soolving:

W=\frac{1}{2}(2000)(60)^2 - \frac{1}{2}(2000)(30)^2=2.7\cdot 10^6 J

Now we can find the power required for the acceleration, which is given by

P=\frac{W}{t}

where

t = 9 s is the time elapsed

Solving:

P=\frac{2.7\cdot 10^6}{9}=3.0\cdot 10^6 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0
3 years ago
If we increase the force applied to an object, and all other factors remain the same, the amount of work will
soldier1979 [14.2K]
The question doesn't give us enough information to answer.
The answer depends on the mass of the object, how long the force
acts on the object, the OTHER forces on the object, and whether the
object is free to move.

-- If you increase the force with which you push on a brick wall,
the amount of work done remains unchanged, namely Zero.

-- If you push on a pingpong ball with a force of 1 ounce for 1 second,
the ball accelerates substantially, it moves a substantial distance, and
so the work done is substantial.

-- But if you push on a battleship, even with a much bigger force ...
let's say 1 pound ... and keep pushing for a month ... the ship accelerates
microscopically, moves a microscopic distance, and the work done by
your force is microscopic. 
3 0
3 years ago
Read 2 more answers
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