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Naya [18.7K]
2 years ago
13

Hanging by a thread. Two metal spheres hang from nylon threads and attract each other when brought close together. (i) What can

you say about the charges on the two spheres? (ii) The spheres are now brought into contact. Will they continue to attract each oth
Physics
1 answer:
elena-14-01-66 [18.8K]2 years ago
5 0

Answer:

Explained

Explanation:

i)Two spheres hanging from nylon threads attract each other because either the two spheres are charged with opposite sign or only one of the spheres is charge so the other would be charge by induction of the charged sphere and hence attract each other.

ii)However, when they are touched the charges will be rearranged among the two sphere such that the two sphere have exact same magnitude and sign of charge and now they will repel each other or the  magnitude of charges on the two spheres become zero and they neither attract or repel each other.

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A spaceship of frontal area 10 m2 moves through a large dust cloud with a speed of 1 x 106 m/s. The mass density of the dust is
Step2247 [10]

Answer:

The decelerating force is 3\times 10^{- 11}\ N

Solution:

As per the question:

Frontal Area, A = 10\ m^{2}

Speed of the spaceship, v = 1\times 10^{6}\ m/s

Mass density of dust, \rho_{d} = 3\times 10^{- 18}\ kg/m^{3}

Now, to calculate the average decelerating force exerted by the particle:

Mass,\ m = \rho_{d}V                                (1)

Volume, V = A\times v\times t

Thus substituting the value of volume, V in eqn (1):

m = \rho_{d}(Avt)

where

A = Area

v = velocity

t = time

m = \rho_{d}(A\times v\times t)                  (2)

Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t

From Newton's second law of motion:

F = \frac{dp}{dt}

Thus differentiating w.r.t time 't':

F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}

where

F_{avg} = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N

4 0
1 year ago
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfo
dolphi86 [110]

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³  / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

= 2.33 x 10³ kg / m³ .

3 0
2 years ago
The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,
KATRIN_1 [288]

Answer:

The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.

C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'

Explanation:

See attached picture.

3 0
2 years ago
The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

7 0
1 year ago
A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On the w
slamgirl [31]

Answer:

The answer is 3.

Explanation:

The answer to this question can be found by applying the right hand rule for which the pointer finger is in the direction of the electron movement, the thumb is pointing in the direction of the magnetic field, so the effect that this will have on the electrons is the direction that the middle finger points in which is right in this example.

So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.

I hope this answer helps.

4 0
2 years ago
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