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KiRa [710]
3 years ago
15

Un Esquiador Inicia un Salto Horizontal con una velocidad Inicial de 25 m/s. La Altura al Final de la Rampa es de 80 m del punto

de contacto con el suelo. ¿Cuánto tiempo permanece el esquiador en el aire? ¿Cuán lejos viaja horizontalmente?
Physics
1 answer:
Evgesh-ka [11]3 years ago
5 0

Responder:

Explicación:

Para que podamos calcular el tiempo que le tomará al esquiador permanecer en el aire, encontraremos el tiempo de vuelo como se muestra;

T = 2Usin theta / 2

theta = 90 grados

U = 25 m / s

T = 25sin90 / 2 (9,8)

T = 25 / 19,62

T = 1,27 segundos

Por lo tanto, los cielos usarán 1.27 segundos en el aire.

La distancia horizontal es el rango;

Rango R = U√2H / g

R = 25√2 (80) /9,8

R = 25√160 / 9,8

R = 25 * √16,326

R = 25 * 4.04

R = 101,02 m

Por tanto, la distancia horizontal recorrida por el esquiador es 101,02 m

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Answer:

The Flemings left hand rule is used to find the magnitude of a magnetic force

Explanation:

Fleming's left hand rule states that if the first three fingers are held mutually at right angles to one another, then the fore finger points into the direction of magnetic field the middle finger in the direction of current while the thumb points in the direction of force.

Mathematically

Magnetic Force F= BILsinθ

Where

B= magnetic field density Tesla

I= current

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3 0
3 years ago
A 20-kg child running at 1.4 m/s jumps onto a playground merry-go-round that has mass 180 kg and radius 1.6m. She is moving tang
Dominik [7]

Answer:

ωf = 0.16 rad/s

Explanation:

Moment of inertia of the child = mr² = 20(1.6²) = 51.2 kg•m²

Moment of Inertia of the MGR = ½mr² = ½(180)1.6² = 230.4 kg•m²

(ASSUMING it is a uniform disk)

Initial angular momentum of the child = Iω = I(v/r) = 51.2(1.4/1.6) = 44.8 kg•m²/s

Conservation of angular momentum

44.8 = (51.2 + 230.4)ωf

ωf = 0.15909090...

4 0
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

<h2>Why?</h2>

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

Where:

Vf= Final velocity = 0

Vi= Initial velocity = \frac{19.6m}{s}

g = Gravity Acceleration = \frac{9.81m}{s^{2} }

s = Traveled distance

0=19.6^2+2*-9.81*s\\s=\frac{19.6^2}{2*9.81}=\frac{384.16}{19.62}=19.59m

Have a nice day!

8 0
3 years ago
You are moving at a speed 2/3 c toward randy when randy shines a light toward you. at what speed do you see the light approachin
yarga [219]
I see the light moving exactly at speed equal to c.

In fact, the second postulate of special relativity states that:
"The speed of light in free space has the same value c<span> in all inertial frames of reference."
</span>
The problem says that I am moving at speed 2/3 c, so my motion is a uniform motion (constant speed). This means I am in an inertial frame of reference, so the speed of light in this frame must be equal to c.
3 0
3 years ago
The desert sand is very hot during the day and very cold at night. What does this tell you about its specific heat capacity?
erma4kov [3.2K]

Answer: The specific heat capacity is very low.

Explanation:

The specific heat capacity of a body is defined as the heat energy required by a body to cause a unit change in its temperature. The value is over low that is why it is easier for the desert sand to easily get very hot during the day. Conversely, it is very easy for the desert sand to lose it's heat a cool breeze pass over it in the night making it very cold in the night. This value also defines how long the desert sand can retain heat. Therefore, the desert sand has a low specific heat capacity.

8 0
3 years ago
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