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2 years ago

An initially neutral glass rod is rubbed with silk. It becomes positively charged by_____.a) electrons are created on the rod.

Physics

1 answer:

Dimas [21]2 years ago

3 0

Answer:

e) electrons are transferred from the rod to the silk.

Explanation:

An initially neutral glass rod contains equal number of electron and proton.

If the rod becomes positively charged after being rubbed with silk, then the rod must have lost some its electron to the silk since electrons are more mobile than protons, leaving the rod with excess positive charge (protons), and the silk will be negatively charged (excess electron).

Thus, the rod becomes positively charged by transfer of electrons from rod to the silk.

e) electrons are transferred from the rod to the silk.

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In this lab you will use a cart and a track to explore Newton's second law of motion. You will vary two different variables and

Savatey [412]

Answer:

Newtons II law:

It is defined as "the net force acting on the object is a product of mass and acceleration of the body" . Also it defines that the "acceleration of an object is dependent on net force and mass of the body".

Let us assume that,a string is attached to the cart, which passes over a pulley along the track. At another end of the string a weight is attached which hangs over the pulley. The hanging weight provides tension in the spring, and it helps in accelerating the cart. We assume that the string is massless and no friction between pulley and the string.

Whenever the hanging weight moves downwards, the cart will accelerate to right side.

For the hanging weight/mass

When hanging weight of mass is m₁ and accelerate due to gravitational force g.

Therefore we can write F = m₁ .g

and the tension acts in upward direction T (negetive)

Now, Fnet = m₁ .g - T

= m₁.a

So From Newtons II law F = m.a

3 0

2 years ago

Read 2 more answers

Two objects (45.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h

Oksanka [162]

a) The acceleration of the objects is $3.56 m/s^2$

b) The tension in the string is 280.8 N

Explanation:

We start by writing the equations of motion for the two masses attached to the pulley.

For the heavier mass, we have:

$m_1 g - T = m_1 a$ (1)

where

$m_1 = 45.0 kg$ is the mass

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

T is the tension in the string

a is the acceleration of the system (here we assumed that the heavier mass accelerates downward)

For the lighter mass, we have

$T-m_2 g = m_2 a$ (2)

where

T is the tension in the string

$m_2 = 21.0 kg$ is the mass

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

a is the acceleration of the system (here we assumed that the lighter mass accelerates upward)

From (1) we get

$T=m_1g - m_1 a$

And substituting into (2),

$(m_1 g - m_1 a)-m_2 g = m_2 a\\(m_1 -m_2)g = (m_1+m_2)a\\a=\frac{m_1 - m_2}{m_1+m_2}g=\frac{45-21}{45+21}(9.8)=3.56 m/s^2$

From the previous part of the problem we got an expression for the tension in the string:

$T=m_1g - m_1 a$

Where we have

$m_1 = 45.0 kg$

$g=9.8 m/s^2$

$a=3.56 m/s^2$ is the acceleration, found in part a)

Susbtituting, we find

$T=(45.0)(9.8)-(45.0)(3.56)=280.8 N$

Learn more about forces and acceleration:

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4 0

2 years ago

In an experiment, students release a block from rest at the top of an inclined plane. The block slides down the plane through a

mote1985 [20]

Answer:

B) Friction

Explanation:

The main source of error is the omission of the effect from friction between block and incline, which is directly proportional to the mass of the block. The force of gravity is constant. The friction force dissipates part of the gravitational potential energy, generating a final speed less than calculated under the consideration of a conservative system. Air resistance is neglected at low speeds like this case.

8 0

2 years ago

An ant travels 2.78 cm [W] and then turns and travels 6.25 cm [S 40 degrees E]. What is the ant's total displacement?

Sunny_sXe [5.5K]

Answer is 6.84 approx
reason:-
(2.78^2+6.25^2)^1/2=6.84 approx

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2 years ago

A machine produces photo detectors in pairs. Tests show that the first photo detector is acceptable with probability 3/5. When t

klasskru [66]

Answer:

a. $a. \ \frac{7}{25}$

$b.\ \ \ P(D_1D_2)=\frac{6}{25}$

Explanation:

a. Find the probability that exactly one photo detector of a pair is acceptable:

Let $A_i=i^{th}$ photo is accepted and the probability $D_i=i^{th}$ is defected.

Therefore:

$P(A_i)=3/5,\ P(A_2|A_1)=4/5,\ \ P(A_2|D_1)=2/5\\\\\\=P(A_1D_2)+P(D_1A_2)\\\\=\frac{3}{5}\times\frac{1}{5}+\frac{2}{5}\times\frac{2}{5}\\\\=\frac{7}{25}$

#The probability of exactly one photo detector of a pair is accepted is 7/25

b.Find the probability that both photo detectors in a pair are defective,P(D1D2):

$P(D_1D_2)=\frac{2}{5}\times \frac{3}{5}\\\\=\frac{6}{25}$

Hence, from out tree diagram,the probability that both photo detectors in a pair are defective is 6/25

4 0

2 years ago