When electrons are moving we call it ______

A gas expands from an initial volume of 30.0 L to a final volume of 65.0 L at a constant pressure of 110kPa. How much work is do

Brilliant_brown [7]

Answer:

3850 J

Explanation:

$V_{i}$ = initial volume of the gas = 30 L = 0.03 m³

$V_{f}$ = final volume of the gas = 65 L = 0.065 m³

$P$ = Constant pressure of the gas = 110 kPa = 110000 Pa

$W$ = Work done by the gas

Since the pressure is constant, Work done by the gas is given as

$W = P (V_{f} - V_{i})\\W = (110000) (0.065 - 0.03)\\W = 3850 J$

8 0

4 years ago

While running a 100m race, a runner runs from the 20m to 30m mark in 77 frames of a video record. If the video camera recorded d

Marrrta [24]

Answer:

Speed of the runner during video interval is 6.49 m/s

Explanation:

According to the problem,

Number of frames recorded by camera in 1 second = 50

Time takes by camera to record 1 frame = (1/50) s

Time taken by camera to record 77 frames, t = $\frac{1}{50}\times 77$ s

Distance covered by the runner during the video recording, d = 10 m

Speed, v = $\frac{Distance}{time}=\frac{d}{t}$

Substitute the values of d and t in the above equation.

v = $\frac{10}{\frac{77}{50} }$

v = 6.49 m/s

3 0

4 years ago

A 1500 kg weather rocket accelerates upward at 10m/s2. It explodes 2.0 s after liftoff and breaks into two fragments, one twice

ladessa [460]

Answer:

Approximately $\rm 19.8\; m\cdot s^{-1}$ (downwards.)

Assumptions:

the rocket started from rest;
the gravitational acceleration is constantly $\rm -9.8\; m \cdot s^{-2}$ ;
there's no air resistance on the rocket and the two fragments.
Both fragments traveled without horizontal velocity.

Explanation:

The upward speed of the rocket increases by $\rm 10\; m \cdot s^{-1}$ . If the rocket started from rest, the vertical speed of the rocket should be equal to $\rm 20\; m \cdot s^{-1}$ .

The mass of the rocket (before it exploded) is 1500 kilograms. At 20 m/s, its momentum will be equal to $\rm 20 \times 1500 = 30,000\; kg \cdot m\cdot s^{-1}$ .

What's the initial upward velocity, $u$ , of the lighter fragment?

The upward velocity of the lighter fragment is equal to $v = 0$ once it reached its maximum height of $x = \rm 530\; m$ .

$v^2 - u^2 = 2g \cdot x$ .

$\begin{aligned}u &= \sqrt{v^2 - 2g\cdot x} \\ &= \sqrt{-2 (-9.8) \times 530}\\ &\approx \rm 101.922\; m \cdot s^{-1}\end{aligned}$ .

Mass of the two fragments:

Lighter fragments: $\displaystyle \frac{1}{1 + 2} \times 1500 =\rm 500\; kg$ .
Heavier fragment: $\displaystyle \frac{2}{1 + 2} \times 1500 =\rm 1000\; kg$ .

Initial momentum of the lighter fragment:

$m \cdot v = \rm 10192.2\; kg \cdot m \cdot s^{-1}$ .

If there's no air resistance, momentum shall conserve. The momentum of the lighter fragment, plus that of the heavier fragment, should be equal to that of the rocket before it exploded.

The initial momentum of the heavier fragment should thus be equal to the momentum of the two pieces, combined, minus the initial momentum of the lighter fragment.

$\rm 30000 - 10192.2 = 19807.8\;kg \cdot m \cdot s^{-1}$ .

Velocity of the heavier fragment:

$\displaystyle \rm \frac{19807.8\;kg \cdot m \cdot s^{-1}}{1000\; kg} \approx 19.8\; m \cdot s^{-1}$ .

5 0

4 years ago

Read 2 more answers

A proton and an electron enter perpendicular to the direction of the magnetic field. The speed of the proton is twice the speed

Nezavi [6.7K]

Answer:

The force on the proton will be twice in comparison with the force experienced by the electron.

Explanation:

The magnetic force acting on the moving charge particle is perpendicular to the velocity as well as the magnetic field. The factors upon which the magnitude of the force is depending and proportional are as follows:

1 Magnitude of the charge particle.

2 Magnitude of the velocity of moving charge.

3 Magnitude of magnetic field.

4 Sine of angle between the velocity and magnetic field.

As far as direction is concerned, it can be find out by right hand rule.

Lets consider,

Speed of electron = Ve

Speed of proton vp= 2ve

Magnitude of the force F = qvBsin∅

Force acting on electron Fe = qe.ve.Bsin∅

Force acting on proton Fp = qp.vp.Bsin∅

Fp = -2 qe.ve.Bsin∅

Fp = -2Fe

Negative sign shows the direction of Force on proton is opposite to the direction of force on electron.

3 0

3 years ago

A box sits at the edge of a spinning disc. The radius of the disc is 0.5 m, and it is initially spinning at 5 revolutions per se

Furkat [3]

Answer:

a) α = 0.375 rad/s²

b) at = 0.1875 m/s²

c) ac =79 m/s²

d) θ = 52 rad

Explanation:

The uniformly accelerated circular movemeis a circular path movement in which the angular acceleration is constant.

Tangential acceleration is calculated as follows:

at = α*R Formula (1)

Centripetal acceleration is calculated as follows:

ac =ω² *R Formula (2)

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (3)

θ= ω₀*t + (1/2)*α*t² Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

R : radius of the circular path (m)

at: tangential acceleration, (m/s²)

ac: centripetal acceleration, (m/s²)

Data:

R= 0.5 m : radius of the disk

t₀=0 , ω₀ = 5 rev/s

1 revolution = 2π rad

ω₀ = 5*(2π)rad/s =10π rad/s = 31.42 rad/s

ωf = 2*(2π)rad/s =4π rad/s = 12.57 rad/s

t = 8 s

(a) angular acceleration of the box

We replace data in the formula (3)

ωf= ω₀ + α*t

2 = 5 + α*(8)

2 -5 = α*(8)

-3 = (8)α

α=3 /8

α = 0.375 rad/s²

(b) Tangential acceleration of the box

We replace data in the formula (1)z

at =(α)*R

at = (0.375)*(0.5)

at = 0.1875 m/s²

c) Centripetal acceleration of the box at t = 8 s

We replace data in the formula (2)

ac =ω² *R

ac =(12.57)² *(0.5)

ac = 79 m/s²

d) Radians that the box has rotated over after t = 8 s

We replace data in the formula (4)

θ = ω₀*t + (1/2)*α*t²

θ = (5)*(8)+ (1/2)*( 0.375)*(8)²

θ = 52 rad

9 0

3 years ago

When electrons are moving we call it _______