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Finger [1]
2 years ago
10

Two similar metal spheres, A and B, have charges of +2.0 x10 –6 coulomb and +1.0 x10 –6 coulomb, respectively, as shown in the d

iagram below. What is the charge on each sphere after the spheres make contact and are then separated?
Physics
1 answer:
Law Incorporation [45]2 years ago
4 0

Answer:

+1.5×10^-6C

................

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1) A person having a mass of 59.1 kg stands before a flight of 30 stairs each of which is 25.0 cm high. He runs up 20 stairs, tu
poizon [28]

Answer:

P = 147,75 W

Explanation:

A man whose mass is 59.1kg climbs up 30 steps of a stair each step is 25 cm high

Height at 30 steps , h=30×2.5=  7.5 m

Change in potential energy , =mgh=59.1×10×7.5 = 4432.5 J

So, Work done by the man , W= 4432.5J

Power used , P= \frac{W}{T}

P = 4432.5 /30

P = 147,75 W

Solve any question of Work, Energy and Power with:-

brainly.com/question/3854047

#SPJ1

3 0
1 year ago
A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un
KonstantinChe [14]

Answer:

32s

Explanation:

We must establish that by the time the police car catches up to the speeder, both have travelled a certain distance during the same amount of time. However, the police car experiences accelerated motion whereas the speeder travels at a constant velocity. Therefore we will establish two formulas for distance starting with the speeder's distance:

x=vt=23.3\frac{m}{s}t

and the police car distance:

x=vt+\frac{at^{2}}{2}=0+\frac{2.75\frac{m}{s^{2}} t^{2}}{2}=0.73\frac{m}{s^{2}}

Since they both travel the same distance x, we can equal both formulas and solve for t:

0 = 0.73\frac{m}{s^{2}}t^{2}-23.3\frac{m}{s} t\\\\0=t(0.73\frac{m}{s^{2}}t-23.3\frac{m}{s} )\\\\

Two solutions exist to the equation; the first one being t=0

The second solution will be:

0.73\frac{m}{s^{2}}t=23.3\frac{m}{s}\\\\t=\frac{23.3\frac{m}{s}}{0.73\frac{m}{s^{2}}}=32s

This result allows us to confirm that the police car will take 32s to catch up to the speeder

7 0
3 years ago
Which example show harassment?
koban [17]

C. making fun of a peer because she is Asian

hope this helps

4 0
3 years ago
Read 2 more answers
A 103 kg horizontal platform is a uniform disk of radius 1.71 m and can rotate about the vertical axis through its center. A 68.
Andreyy89

Answer:

I_{total}=220.64 kg*m^{2}

Explanation:

The moment of inertia of the system is equal to the each population and the platform inertia so

Inertia disk

I_{disk}=\frac{1}{2}*m_{disk}*(r_{p})^{2}

Inertia person

I_{p}=\frac{1}{2}*m_{p}*(r_{p})^{2}

Inertia dog

I_{d}=\frac{1}{2}*m_{d}*(r_{d})^{2}

The Inertia of the system is the sum of each mass taking into account that all exert the force of inertia:

I_{total}=I_{disk}+I_{p}+I_{d}

I_{total}=\frac{1}{2}*103kg*(1.71)^{2}+\frac{1}{2}*68.9kg*(1.09)^{2}+\frac{1}{2}*27.7kg*(1.45)^{2}

I_{total}=220.64 kg*m^{2}

5 0
3 years ago
The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the
Naily [24]

To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

V_d = \frac{I}{nAq}

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

I = 25 A

A= 1.2*20 *10^{-6} m^2

q= 1.6*10^{-19}C

N = 8.47*10^{19} mm^{-3}

V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}

V_d = 7.68*10^{-5}m/s

The hall voltage is given by

V=\frac{IB}{ned}

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}

V = 3.84*10^{-6}V

5 0
3 years ago
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