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horrorfan [7]
2 years ago
14

three neutral metal cans mounted on isulating stands are touching a negatively charge ballon is brought near can a can b is then

removed. what is the charge of a? what is the charge of can c?
Physics
1 answer:
Nimfa-mama [501]2 years ago
7 0
Charge on can A is positive. 
Charge on can C is negative.  
Punctuation and capitalization are very useful things to pay attention to and this question would be a lot easier to understand if you had actually used both capitalization and punctuation. If I'm understanding the question, you have 3 metal can that are insulated from the environment and initially touching each other in a straight line. Then a negatively charged balloon is brought near, but not touching one of the cans in that line of cans. While the balloon is near, the middle can is removed. Then you want to know the charge on the can that was nearest the balloon and the charge on the can that was furthermost from the balloon. 
 As the balloon is brought near to can a, the negative charge on the balloon repels some of the electrons from can a (like charges repel). Some of those electrons will flow to can b and in turn flow to can c. Basically you'll have a charge gradient that's most positive on that part of the can that's closest to the balloon, and most negative on the part of the cans that's furthest from the balloon. You then remove can B which causes cans A and C to be electrically isolated from each other and prevents the flow of elections to equalize the charges on cans A and C when the balloon is removed. So you're left with a deficiency of electrons on can A, so can A will have a positive overall charge, and an excess of electrons on can C, so can C will have a negative overall charge.
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A sock with a mass of 0.03 kg is stuck to the inside of a clothes dryer spins
ValentinkaMS [17]

Answer:

15.71 m/s

Explanation:

We are given;

Time; t = 0.2 s

Radius; r = 0.5 m

The circumference will give us the distance covered.

Formula for circumference is 2πr

Thus; Distance = 2πr = 2 × π × 0.5 = π

Linear speed = distance/time = π/0.2 = 15.71 m/s

5 0
2 years ago
A child's toy consists of a m = 36 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monke
kolezko [41]

Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

Part D - E = 1/2kd² + 1/2mv² + mgh

Explanation:

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).

The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².

Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

8 0
2 years ago
A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and
scoray [572]

Answer:

F_{allow} = 208.15kN

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

K_{allow} =\frac{K_c}{N}

We know that K_c is 33Mpa*m^{0.5} and our Safety factor is 2,

K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

<em>For the crack;</em>

\alpha = 0.5*L_c = 0.5*4mm = 2mm

<em>For the panel</em>

\gamma = 0.5*W = 0.5*250mm = 125mm

To find now the goemetry factor we need to use this equation

\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1

That allow us to determine the allowable nominal stress,

\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}

\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}

\sigma_{allow} = 208.15Mpa

So to get the force we need only to apply the equation of Force, where

F_{allow}=\sigma_{allow}*L_c*W

F_{allow} = 208.15*250*4

F_{allow} = 208.15kN

That is the maximum tensile load before a catastrophic failure.

4 0
2 years ago
The muzzle velocity of a 50.0g shell leaving a 3.00 kg rifle is 400m/s what is the recoil velocity of the rifle
serg [7]

Here if we consider bullet and gun as a system then we can say that momentum of the system will remain conserved

so we will have

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now we know that

m_1 = 50 g = 0.050 kg

m_2 = 3 kg

v_{1i} = v_{2i} = 0

v_{1f} = 400 m/s

now we will have

0.050(0) + 3(0) = 0.050(400) + 3(v)

20 + 3v = 0

v = - \frac{20}{3} = - 6.67 m/s

so gun will recoil with speed 6.67 m/s

6 0
2 years ago
A 13-cm-diameter cd has a mass of 25 g . part a what is the cd's moment of inertia for rotation about a perpendicular axis throu
faltersainse [42]
<span>Answer: For a disc, the moment of inertia about the perpendicular axis through the center is given by 0.5MR^2. where M is the mass of the disc and R is the radius of the disc. For the axis through the edge, use parallel axis theorem. I = I(axis through center of mass) + M(distance between the axes)^2 = 0.5MR^2 + MR^2 (since the axis through center of mass is the axis through the center) = 1.5 MR^2</span>
8 0
2 years ago
Read 2 more answers
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