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aev [14]
3 years ago
10

Use the following equation to help you answer the question. The peak intensity of radiation from a star named Sigma is 2 x 10 6

nm. What is the average surface temperature of Sigma rounded to the nearest whole number
Physics
2 answers:
puteri [66]3 years ago
8 0

Answer:

1449 K

Explanation:

The surface temperature of a star is related to its peak wavelength by Wien's displacement law:

T=\frac{b}{\lambda}

where

T is the surface temperature

b is Wien's displacement constant

\lambda=2\cdot 10^{-6} m

So the surface temperature of the star is

T=\frac{2.898 \cdot 10^{-3} Km}{2\cdot 10^{-6} m}=1449 K

Kruka [31]3 years ago
7 0

Answer:

1,450 K

Explanation:

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Nuetrik [128]

Answer:

It takes approximately 52 min to drive from San Jose to San Francisco.

Explanation:

5 0
3 years ago
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An electron is in a vacuum near Earth's surface and located at y = 0 m on a vertical y axis. At what value of y should a group o
Bingel [31]

Answer:

the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.

Explanation:

As we know that the gravitational force on electron at y = 0 is counter-balanced by the weight of the electron

So we have

\frac{kq_1q_2}{r^2} = mg

here we have

q_1 = e

q_2 = 23 e

m = 9.11 \times 10^{-31} kg

also we know that

e = 1.6 \times 10^{-19} C

so we will have

\frac{(9\times 10^9)(1.6 \times 10^{-19})(23\times 1.6 \times 10^{-19})}{r^2} = (9.11 \times 10^{-31})(9.81)

\frac{5.3 \times 10^{-27}}{r^2} = 8.94 \times 10^{-30}

r = 24.35 m

so the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.

7 0
3 years ago
1. *A car is going over the top of a hill whose curvature approximates a circle of radius 200 m. At
Greeley [361]

Answer:

The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight is approximately 19.81 m/s

Explanation:

The given parameters are;

The curvature of the hill, r = 200 m

Due to the velocity, v, the occupants weight = 20% less than the normal weight

The outward force of an object due to centripetal (motion) force is given by the following equation;

F_c = \dfrac{m \times v^2}{r}

Where;

r = The radius of curvature of the hill = 200 m

Given that the weight of the occupants, W = m × g, we have;

F_c = 0.2 \times W = 0.2 \times m \times g

\therefore 0.2 \times m \times g = \dfrac{m \times v^2}{r}

v = √(0.2 × g × r)

By substitution, we have;

v = √(0.2 × 9.81 × 200) ≈ 19.81

The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight ≈ 19.81 m/s.

3 0
3 years ago
The position x of a particle moving along x axis varies with time t as x = Asin(wt) , where A and w are constants . The accelera
Elodia [21]

Let's see

\\ \rm\Rrightarrow x=Asin(\omega t)\dots(1)

Now we know the formula of acceleration

\\ \rm\Rrightarrow \alpha=-A\omega^2sin(\omega t)

\\ \rm\Rrightarrow \alpha=-Asin(\omega t)\times \omega^2

  • From eq(1)

\\ \rm\Rrightarrow \alpha=-x\omega^2

Or

\\ \rm\Rrightarrow \alpha=-\omega^2x

6 0
2 years ago
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A car accelerates at a rate of 13m/s^2[S]. If the car's initial velocity is 120km/h[N]. What will its final velocity be in m/s,
Delvig [45]

Answer:

the final velocity of the car is 59.33 m/s [N]

Explanation:

Given;

acceleration of the car, a = 13 m/s²

initial velocity of the car, u = 120 km/h = 33.33 m/s

duration of the car motion, t = 2 s

The final velocity of the car in the same direction is calculated as follows;

v = u + at

where;

v is the final velocity of the car

v = 33.33 + (13 x 2)

v = 59.33 m/s [N]

Therefore, the final velocity of the car is 59.33 m/s [N]

6 0
3 years ago
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