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Serga [27]
3 years ago
15

The electric potential ( relative to infinity ) due to a single point charge Q is 400 V at a point that is 0.6 m to the right of

Q. The electric potential (relative to infinity) at a point that is 0.90 m to the left of 0 is:_____.
A. + 400 V.
B. -400 V.
C. + 200 V.
Physics
1 answer:
UkoKoshka [18]3 years ago
5 0

Answer:

The potential at a distance of 0.9 m is 266.67 V.

Explanation:

Charge = Q

Potential is 400 V at a distance 0.6 m .

Let the potential is V at a distance 0.9 m.

Use the formula of potential.

V = \frac{Kq}{r}\\\\\frac{V}{400}=\frac{0.6}{0.9}\\\\V = 266.67 V

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You wish to place a spacecraft in a circular orbit around the earth so that its orbital speed will be 4.00×103m/s. What is this
Levart [38]

Answer:

The radius of orbit=2.49\times 10^7 m

Explanation:

We are given that

Orbital speed=4.00\times 10^3m/s

We have to find the radius of orbit of spacecraft.

We know that

Gravitational constant=6.67\times 10^{-11}m^3/kgs^2

Mass of earth=5.972\times 10^{24} kg

Orbital speed=\sqrt{\frac{GM}{r}}

Where G= Gravitational constant

M=Mass of earth

r=Radius of orbit

Substitute the values in the formula

4\times 10^3=\sqrt{\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{r}

Squaring on both sides

16\times 10^6=\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{r}

r=\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{16\times 10^6}

r=2.49\times 10^7 m

Hence, the radius of orbit=2.49\times 10^7 m

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4 years ago
True or false. both the incident and reflected ray lie in the same plane. ​
olga2289 [7]

Answer:

I think its false...........

4 0
4 years ago
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A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot
FromTheMoon [43]

Answer:

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

Explanation:

To solve the problem/ we first write the differential equation governing the motion. So,

m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0

with m = 1 slug and k = 9 lb/ft, the equation becomes

\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0

The characteristic equation is

D² + 9 = 0

D = ±√-9 = ±3i

The general solution of the above equation is thus

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity  v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ    where k is an integer

3t  = ±π/6 + 2kπ - 4π/3

t  = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3  or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6. So,

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

6 0
3 years ago
The surface area of the earth's crust is 5.10*10^8km^2. The average thickness of the earth's crust is 35km. The mean density of
Mars2501 [29]

Answer:

The answer is V = 1.785 \times 10^{10}\ Km^3.

Explanation:

You are asking only for the total volume, so the important data here is the surface area of the earth's crust,

A = 5.10 \times 10^8\ Km^2

and the average thickness of the earth's crust,

h = 35\ Km.

Imagine that you can stretch the surface area as if it were a blanket, now if you want to calculate its volume, you just need to multiply its area by its thickness,

V = A*h = 1.785 \times 10^{10}\ Km^3.

4 0
4 years ago
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