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Radda [10]
3 years ago
11

What is the value of x in the equation Ix - 3y = 30, when y = 15?

Physics
1 answer:
ankoles [38]3 years ago
8 0

Answer: x= 75

Explanation:1x -3y = 30

And y = 15

Substitute y accordingly into the equation, so that

x - 3(15) = 30

x-45=30

Add 45 to both sides if the equation

x-45+45=30+45

x=75

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Jim thinks that he can throw a baseball six meters per second. What does the six meters refer to?
SVETLANKA909090 [29]

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Six meters refers to the distance the baseball travels in one second.

Explanation:

6 m/s means that every second, the baseball travels 6m. So, 6 meters is the distance traveled in one second.

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Starting from rest, a car undergoes a constant acceleration of 6 m/s^2. How far will the car travel in the two seconds?
oksano4ka [1.4K]

If time is specified, the distance may be estimated in constant acceleration using the formula: X=(at2)/2 if the beginning velocity is 0. (A automobile begins from a stop...) As a result, X=(6*10*10)/2=600/2 = 300 m.

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An electrical current

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A pingpong ball has 2 kg m/s of momentum when thrown 8 m/s find the mass of the ball
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0.25 kg

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4 0
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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

8 0
3 years ago
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