Help with this voltage drop? I'd also like to know how you did it.

Physics

1 answer:

grandymaker [24]3 years ago

4 0

No one answeres physic questions, iveneeded help all morning

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When an object radiates heat, the strength of this radiation far from the object decreases when distance from the source increas

rodikova [14]

Answer:

B I am not sure about the answer but it may be B

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2 years ago

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In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. The sharp upward

11111nata11111 [884]

Answer:

u = 11.6 m/s

Explanation:

The end of a launch ramp is directed 63° above the horizontal. A skier attains a height of 10.9 m above the end of the ramp.

Maximum height, H = 10.9

Let v is the launch speed of the skier. The maximum height attained by the projectile is given by :

$H=\dfrac{u^2\ sin^2\theta}{g}$

$10.9=\dfrac{u^2\ sin^2(63)}{9.8}$

u = 11.6 m/s

So, the launch speed of the skier is 11.6 m/s. Hence, this is the required solution.

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3 years ago

If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star

torisob [31]

There's no such thing as "stationary in space". But if the distance
between the Earth and some stars is not changing, then (A) w<span>avelengths
measured here would match the actual wavelengths emitted from these
stars. </span><span>

</span><span>If a star is moving toward us in space, then (A) Wavelengths measured
would be shorter than the actual wavelengths emitted from that star.

</span>In order to decide what's actually happening, and how that star is moving,
the trick is: How do we know the actual wavelengths the star emitted ?

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3 years ago

A boy throws a ball with an initial velocity of 19.6 m/s. What maximum height does the ball reach?

Wewaii [24]

<h2>Hello!</h2>

The answer is: 19.59 m

Since there is no information about the launch type, we can assume that the ball is thrown vertically upward.

When the ball reaches the maximum height, just at that moment, the velocity turns to 0, and after that moment, the ball starts falling, so:

We will use the following formula:

$Vf^2=Vi^2+2*g*s$