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3 years ago

At the atomic level, what is meant by saying something is electrically charged?

Physics

1 answer:

Scorpion4ik [409]3 years ago

7 0

Answer:

When something is electrically charged, at the atomic level it either has more total electrons than protons (in which case it is negatively charged), or it has fewer total electrons than protons (in which case it is positively charged).

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A truck weighs 500 N on Earth. What's the mass of the truck?

Jobisdone [24]

The answer is 50 kg plz mark as brainliest

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3 years ago

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In a new lab experiment, two parallel vertical metal rods are separated by L = 1.4 m . A R = 2.0-Ω resistor is connected from th

artcher [175]

Answer:

Explanation:

Let v be the terminal velocity of the bar .

emf induced in the bar of length L

= B L v where B is the value of magnetic field.

current i in the circuit containing resistance R

i = induced emf / R

BLv / R

Magnetic force in upward direction in the bar

F = BiL

= BL x BLv / R

B²L²v / R

For attainment of uniform velocity

magnetic force = weight

B²L²v / R = mg

so current

i = BLv / R

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3 years ago

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A flywheel in a motor is spinning at 530 rpm when a power failure suddenly occurs. The flywheel has mass 40.0kg and diameter 75.

aleksley [76]

Answer:

w = 25.05 rad / s , α = 0.7807 rad / s² , θ = 1972.75

Explanation:

This is a kinematic rotation exercise, let's start by looking for the acceleration when the engine is off

θ = w₀ t - ½ α t²

α = (w₀t - θ) 2/t²

let's reduce the magnitudes to the SI system

w₀ = 530 rev / min (2pi rad / 1 rev) (1 min / 60 s) = 55.5 rad / s

θ = 250 rev (2pi rad / 1 rev) = 1570.8 rad

let's calculate the angular acceleration

α = (55.5 39 - 1570.8) 2/39²

α = 0.7807 rad / s²

having the acceleration we can calculate the final speed

w = w₀ - ∝ t

w = 55.5 - 0.7807 39

w = 25.05 rad / s

the time to stop w = 0

0 = wo - alpha t

t = wo / alpha

t = 55.5 / 0.7807

t = 71.09 s

the angle traveled

w² = w₀⁹ - 2 α θ

w = 0

θ = w₀² / 2α

let's calculate

θ = 55.5 2 / (2 0.7807)

θ = 1972.75

5 0

3 years ago

The average human walks at a speed of 5km per hour if your PE teacher asks you to walk for 30 minutes in gym class how far would

Leokris [45]

Answer:

2.5km

Explanation:

Assuming you also walk with the average human walking speed of 5 km/h. We know that speed=distance/time and now making distance the subject, distance is a product of speed and time. The time is 30 minutes, expressed in hours will be 30/60=0.5 hours. Substituting 5km/h for speed and 0.5 hours for time then the diatance covered will be

Distance=5*0.5=2.5 km

6 0

3 years ago

A worker pushes a 50 kg crate a distance of 7.5 m across a level floor. He

Taya2010 [7]

a) 73.5 N

b) 551.3 J

c) -551.3 J

d) 0 J

e) 0 J

f) 0 J

g) 0 J

Explanation:

There are two forces acting on the crate:

- The push of the worker, F, in the forward direction

- The frictional force, $F_f=\mu mg$ , in the backward direction, where

$\mu=0.15$ is the coefficient of friction

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

According to Newton's second law of motion, the net force on the crate must be equal to the product of mass and acceleration, so:

$F-F_f=ma$

However, the crate here is moving with constant velocity, so its acceleration is zero:

$a=0$

So the previous equation becomes:

$F-F_f=0$

And we can find the magnitude of the applied force:

$F=F_f=\mu mg=(0.15)(50)(9.8)=73.5 N$

The work done by the applied force on the crate is

$W_F=Fd cos \theta$

where:

F is the magnitude of the force

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here we have:

F = 73.5 N

d = 7.5 m

$\theta=0^{\circ}$ (the force is applied in the same direction as the displacement)

Therefore,

$W_F=(73.5)(7.5)(cos 0^{\circ})=551.3 J$

The work done by friction on the crate is:

$W_{F_f}=F_f d cos \theta$

where in this case:

$F_f=73.5 N$ is the magnitude of the force of friction

d = 7.5 m is the displacement of the crate

$\theta=180^{\circ}$ , because the displacement is forward and the force of friction is backward, so they are in opposite direction

Therefore, the work done by the force of friction is:

$W_{F_f}=(73.5)(7.5)(cos 180^{\circ})=-551.3 J$

To find the normal force, we analyze the situation of the force along the vertical direction.

We have two forces on the vertical direction:

- The normal force, N, upward

- The force of gravity, $mg$ , downward, where

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

Since the crate is in equilibrium in this direction, the vertical acceleration is zero, so the two forces balance each other:

$N-mg=0\\N=mg=(50)(9.8)=490 N$

The work done by the normal force is:

$W_N=Nd cos \theta$

In this case, $\theta=90^{\circ}$ , since the normal force is perpendicular to the displacement of the crate; therefore, the work done is

$W_N=(490)(7.5)(cos 90^{\circ})=0$

The work done by the gravitational force is:

$W_g=F_g d cos \theta$

where:

$F_g=mg=(50)(9.8)=490 N$ is the gravitational force

d = 7.5 m is the displacement of the crate

$\theta=90^{\circ}$ is the angle between the direction of the gravitational force (downward) and the displacement (forward)

Therefore, the work done by gravity is

$W_g=(490)(7.5)(cos 90^{\circ})=0 J$

The total work done on the crate can be calculated by adding the work done by each force:

$W=W_F+W_{F_f}+W_N+W_g$

where we have:

$W_F=+551.3 J$ is the work done by the applied force

$W_{F_f}=-551.3 J$ is the work done by the frictional force

$W_N=0$ is the work done by the normal force

$W_g=0$ is the work done by the force of gravity

Substituting,

$W=+551.3+(-551.3)+0+0=0 J$

So, the total work is 0 J.

According to the work-energy theorem, the change in kinetic energy of the crate is equal to the work done on it, therefore:

$W=\Delta E_K$

where

W is the work done on the crate

$\Delta E_K$ is the change in kinetic energy of the crate

In this problem, we have:

$W=0$ (total work done on the crate is zero)

Therefore, the change in kinetic energy of the crate is:

$\Delta E_K = W = 0$

5 0

4 years ago