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Vitek1552 [10]
4 years ago
13

What is the speed of a 55.0 kg skydiver who has 7.81 x 104 J of kinetic energy?

Physics
1 answer:
KATRIN_1 [288]4 years ago
5 0

Answer:

v=53.3m/s

Explanation:

Ek=1/2mv²

7.81×10⁴=1/2×55.0v²

v= the square root of 7.81×10⁴/0.5×55.0

v=53.3m/s

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The colour of star depend on its temperature, why?​
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<em>Another key factor that determines a star's colour is its temperature. As stars become hotter, the overall radiated energy increases, and the peak of the curve changes to shorter wavelengths. To put it another way, when a star heats up, the light it produces moves toward the blue end of the spectrum.</em>

4 0
3 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate &lt;br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
3 years ago
When do we land on Mars ?
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Answer:

I think like 2024 or 25

Explanation:

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5 0
3 years ago
Read 2 more answers
At the surface, atmospheric pressure is 1.013 × 10^5 Pa. People can normally snorkel down to a depth of roughly one meter. What
natulia [17]

Answer:

1.01 × 10⁵ Pa  

Explanation:

At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.

We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.

Pressure at a depth is given by :

P=\rho gh

Where

\rho is the density of air, \rho=1.225\ kg/m^3

So,

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Total pressure, P = Atmospheric pressure + 12 Pa

= 1.013 × 10⁵ Pa + 12 Pa

= 1.01 × 10⁵ Pa

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3 years ago
A distance of 1.0 meter separates the centers of
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Neither set of choices is correct.

If the distance is tripled, then the forces decrease to

1/9 Fg. and. 1/9 Fe.

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3 years ago
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