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viva [34]
3 years ago
5

gAn Olympic diver is on a diving platform 5.40 m above the water. To start her dive, she runs off of the platform with a speed o

f 1.17 m/s in the horizontal direction. What is the diver's speed, in m/s, just before she enters the water
Physics
1 answer:
lina2011 [118]3 years ago
4 0

Answer:

The diver's speed, in m/s, just before she enters the water = 10.19 m/s

Explanation:

Assuming the horizontal velocity of the diver remains constant by neglecting the air resistance.

V^2 = U^2 + 2as

Where, V^2 = the diver's final velocity before impact with the water

            U^2 = Initial diver's velocity as she leaves the diving platform

                a = acceleration due to gravity

                s = the displacement

V =  √U^2 + √2as

= √(5.40 m/s)^2 + √2(9.81 m/s/s) (1.17 m)  

= √ 29.16 + √ 22.9554

=  5.40 + 4.79

= 10.19 m/s

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Answer:

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It is given that,

Mass of the object, m = 350 g = 0.35 kg

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Explanation:

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8 0
2 years ago
A baseball of radius r = 5.2 cm is at room temperature T = 20.8 C. The baseball has emissivity of ε = 0.86 and the Stefan-Boltzm
Marrrta [24]

Answer:

P = 12.37 \frac{J}{s} = 12.37 Watts

Explanation:

Previous concepts

The Thermal radiation is one of "3 mechanisms who allows to bodies exchange energy".

The thermal radiation formula is given by:

\frac{P}{A} = \epsilon \sigma T^4

Where \sigma = 5.67 x10^{-8} \frac{J}{sm^2 K^4}

If we solve for P we got:

P = A \epsilon \sigma T^4

Since we have a baseball ball considered as a sphere the superficial area is given by:

A = 4\pi r^2

Solution to the problem

And if we replace this into our equation of P we got:

P = (4\pi r^2) \epsilon \sigma T^4

And we can reorder this like that:

P = 4 \epsilon \pi \sigma r^2 T^4

We can convert the radius to meters and we got:

r= 5.2 cm*\frac{1m}{100 cm}=0.052 m

Now we can convert the temperature to Kelvin and we got:

T = 20.8 +273.15 = 293.95 K

\epsilon = 0.86 the emissivity given

And now we can replace into the formula for P and we got:

P = 4*0.86*\pi *(5.67x10^{-8} \frac{J}{s m^2 K^4}) (0.052m)^2 (293.95 K)^4

P = 12.37 \frac{J}{s} = 12.37 Watts

6 0
3 years ago
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