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Natali5045456 [20]
2 years ago
7

As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -

1.1 V during one revolution. Assuming the shuttle is a conducting sphere of radius 10 m, estimate the amount of charge it collects.
Physics
1 answer:
Svetllana [295]2 years ago
8 0

Answer:

The amount of charge the space shuttle collects is -1.224nC

Explanation:

The magnitude of Electric potential is given as;

V = kq/r

where;

V is the electric potential in volts

k is coulomb's constant

r is the radius of the sphere or distance moved by the charge

given;  V = -1.1 V, k = 8.99 x 10⁹ Nm²/C², r = 10m

Substituting this values in the above equation, we estimate the amount of charge space shuttle collects.

q = (V*r)/k

q = (-1.1 *10)/(8.99 x 10⁹ )

q = -1.224 X 10⁻⁹ C

q = -1.224nC

Therefore, the amount of charge the space shuttle collects is -1.224nC

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Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass
zvonat [6]

Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

Mass of one otter = 8.50 kg

Speed = 6.00 m/s

Mass of other = 5.75 kg

Speed = 5.50 m/s

(a). We need to calculate the magnitude and direction of the velocity of these free-spirited otters right after they collide

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v

Put the value into the formula

8.50\times(-6.00)+5.75\times5.50=(8.50+5.75)\times v

v=\dfrac{-19.375}{14.25}

v=-1.35\ m/s

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2

Put the value into the formula

K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2

K.E_{i}=239.96\ J

We need to calculate the final kinetic energy

Using formula of kinetic energy

K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2

Put the value into the formula

K.E_{f}=\dfrac{1}{2}\times(8.50+5.75)\times(-1.35)^2

K.E_{f}=12.98\ J

We need to calculate the mechanical energy dissipates during this play

Using formula of loss of mechanical energy

\Delta K.E=K.E_{f}-K.E_{i}

Put the value into the formula

\Delta K.E=12.98-239.96

\Delta K.E=-226.98\ J

Negative sign shows the loss of mechanical energy

Hence, (a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

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The correct answer would be B. It is the compound Na2S that forms an ionic bond from the given ion above. From the choices, it is the only compound that exist and is available. Metals cannot react with another metal as well as nonmetal to nonmetal. 
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2 years ago
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2 years ago
a film of transparent material 120 nm thick and having refractive index 1.25 is placed on a glass sheet having refractive index
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Answer:

a) 600nm

b) 300nm

Explanation:

the path difference = 2t  

t = thickness of the film

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L = wavength of light in air

n = refractive index of glass

(a)

for destructive interference 2t = L'/2 = L/2n

L = 4*t*n

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(b)

for constructive interference 2t = L' = L/1.25

L = 2tn

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Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

Explanation:

Magnetic field around a long current carrying wire is given by

B=\frac{\mu _o I}{2\pi r}

where B= magnetic field

           \mu _o= permeability of free space

           I= current in the long wire and

           r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  

Now if I'=3I and r'=2r then magnetic field B' is given by

B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B

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