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Finger [1]
3 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]3 years ago
4 0

Answer:

+1.5×10^-6C

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

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A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

3 0
3 years ago
All moving objects have momentum. __ momentum refers to the momentum of objects moving in a straight line and __ momentum refers
Inessa [10]

Answer: linear,angular

Explanation:

3 0
3 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
HELP!!!!!!
NemiM [27]
B. Gabriella is slowing down at the same rate that Kendall is speeding up, and Franklin is not accelerating.
5 0
3 years ago
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Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on
Zolol [24]

Answer:71 dB

Explanation:

Given

sound Level \beta _1=124 dB

distance r_1=5.01 m

From sound Intensity

\beta =10dB\log (\frac{I_1}{I_0})

124=10dB\log (\frac{I_1}{I_0})

12.4=\log (\frac{I_1}{I_0})

I_1=(1\times 10^{-12})\times 10^{12.4}

I_1=2.51 W/m^2

we know Intensity I\propto ^\frac{1}{r^2}

I_1r_1^2=I_2r_2^2

I_2=I_1(\frac{r_1}{r_2})^2

I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2

I_2=1.24\times 10^{-5} W/m^2

Sound level corresponding to I_2

\beta _2=10\log (\frac{I_2}{I_0})

\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})

\beta _2=70.93\approx 71 dB

6 0
3 years ago
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