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2 years ago

PLEASE HELP: Why are shooting stars bright?

Physics

1 answer:

mrs_skeptik [129]2 years ago

4 0

the answer is c I hope this helps

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An electric light is plugged into a 120-V outlet. If the current in the bulb is 0.50 A, how much electrical energy does the bulb

ioda

Answer:

= 54,000 Joules or 54 kJ

Explanation:

Electrical energy is given by the formula;

E = VIt; where V is the potential difference in volts, I is the current and t is the time in seconds.

Therefore;

Electrical energy = 120 V × 0.50 A × 15 ×60 seconds

= 54,000 Joules

Thus; the electrical energy is 54,000 joules or 54 kJ

7 0

2 years ago

The wavelength of a light wave will affect the light’s

Tom [10]

Frequency and color.

6 0

2 years ago

I need help!! i’ll give 20 points

larisa86 [58]

Answer:

780 m to travel north

Explanation:

6 m over = 750

53 degree so it will take about 2 min to reach the destination

8 0

2 years ago

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

2 years ago

Look at the equation. What detail is missing? 3 m/s2= (33 m/s - X)/30 S <br>

Tems11 [23]

Answer:

The starting velocity.

Explanation:

We must understand that this equation comes from the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity = 33 [m/s]

Vo = starting velocity [m/s]

a = acceleration = 3 [m/s²]

t = time = 30 [s]

So, these values can be assembly in the following way:

$v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}$

6 0

2 years ago