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denpristay [2]
3 years ago
8

Suppose that you have a rod that is positively charged, a rod that is negatively charged and a stand that allows rotation. If yo

u are handed a polystyrene rod of unknown charge-state, how could you determine if it has a positive charge, a negative charge or if it is uncharged? Explain what happens to the unknown rod if it has a positive charge, a negative charge or if it is uncharged.
Physics
1 answer:
amm18123 years ago
3 0

Answer:

Explanation:

A charge is produced when an atom losses or gains an electron. The law of static electricity states that like charges repels, while unlike charges attracts.

1. To determine the charge on the polystyrene rod.

Place the polystyrene rod on the non-conducting rotating stand, and bring the positively charged rod close to it. If attraction occurs, it shows that it is oppositely charged. If repulsion occurs, it shows that it is positively charged.

Bringing a negatively charged rod close to the rotating polystyrene rod would attract it if the charge is opposite. But if the charge on the two rods are the same, repulsion occurs.

2a. When the polystyrene rod is positively charged, it would attract the negatively charged rod but repel the positively charged rod.

b. When the polystyrene rod is negatively charged, it would repel the negatively charged rod but attract the positively charged rod.

c. When the polystyrene rod is uncharged, no reaction would be observed when either the positively charged or negatively charged rod is brought close to it.

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A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi
Mamont248 [21]

Answer:

a=368.97\ m/s^2

Explanation:

Given that,

Initial angular velocity, \omega=0

Acceleration of the wheel, \alpha =7\ rad/s^2

Rotation, \theta=14\ rotation=14\times 2\pi =87.96\ rad

Let t is the time. Using second equation of kinematics can be calculated using time.

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s

Let \omega_f is the final angular velocity and a is the radial component of acceleration.

\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s

Radial component of acceleration,

a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2

So, the required acceleration on the edge of the wheel is 368.97\ m/s^2.

6 0
3 years ago
True or false? A protostellar cloud spins faster as it contracts
Aleks04 [339]

Answer:

No. The protostellar cloud spins faster in the collapsing stage (stage 1) and becomes much slower in the contraction stage (stage 2)

Explanation:

Once the cloud is so dense that the heat which is being produced in its center cannot easily escape, pressure rapidly rises, and catches up with the weight, or whatever external force is causing the cloud to collapse, and the cloud becomes stable, as a protostellar cloud.

The protostellar cloud will become more dense over thousands of years. This stage of decreasing size is known as a contraction, rather than a collapse. In the contraction stage the cloud has become much slower, and because weight and pressure are more or less in balance. In the first stage of formation, the decrease of size is very rapid, and compressive forces completely overwhelm the pressure of the gas, and we say that the cloud is collapsing.

3 0
3 years ago
(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us
konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

5 0
3 years ago
HELP BRANLIEST
Vsevolod [243]

Explanation:

- Newton's first law of motion:

"An object at rest (or in uniform motion) remains at rest (or in uniform motion) unless acted upon an unbalanced force

In this situation, we can apply Newton's first law to the keys of the keyboard that are not hit by the fingers of the man. In fact, as no force act on the keys, they remain at rest.

- Newton's second law of motion:

"The acceleration experienced by an object is proportional to the net force exerted on the object; mathematically:

F=ma

where F is the net force, m is the mass of the object, and a its acceleration"

In this case, we can apply Newton's second law to the keys of the keyboard that are hit by the man: in fact, as they are hit, they experience a downward force, and therefore they experience a downward acceleration.

"Newton's third law of motion:

"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

Here We can apply Newton's third law to the pair of objects finger-key: in fact, as the finger apply a force on the key (action force), then the key exerts a force back on the finger (reaction force), equal and opposite.

3 0
3 years ago
A coin is placed on a large disk which rotates uniformly at a rate of 1 rot/s. The coefficient of friction between the coin and
s2008m [1.1K]

Answer:

r = 0.02 m

Explanation:

from the question we have :

speed = 1 rps = 1x 60 = 60 rpm

coefficient of friction (μ) = 0.1

acceleration due to gravity (g) = 9.8 m/s^{2}

maximum distance without falling off (r) = ?

to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

mv^2 / r =  m x g x μ

v^2 / r =  g x μ  .......equation 1

where

velocity (v) = angular speed (rads/seconds) x radius

angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm

angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds

now

velocity = 6.28 x r = 6.28 r

now substituting the value of velocity into equation 1

v^2 / r =  g x μ

(6.28r)^2 / r = 9.8 x 0.1

39.5 x r = 0.98

r = 0.02 m

6 0
3 years ago
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