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3 years ago

Velocity of the ball after 3.63 s

Physics

1 answer:

MrRa [10]3 years ago

5 0

Answer:

Explanation:

This question is regarding gravity under motion. Then we can use motion equation to find the final velocity .

Inertial velocity is zero when it's release from rest.

V = U+at

V = 0+ 9.8×3.63

V = 35.574 m/s

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shtirl [24]

5.6 g/ml. That is the density.

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3 years ago

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Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black

oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

$Q = FA\sigma\Delta T^4$

Where,

F =View Factor

A = Cross sectional Area

$\sigma =$ Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

$L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K$

The view factor between two coaxial parallel disks would be

$\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33$

$\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75$

Then the view factor between base to top surface of the cylinder becomes $F_{12} = 0.26$ . From the summation rule

$F_{13} = 1-0.26$

$F_{13} = 0.74$

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

$\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}$

$\dot{Q_3} = 2\dot{Q_{13}}$

$\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)$

$\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)$

$\dot{Q_3} = 780.76W$

Therefore the rate heat radiation is 780.76W

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2 years ago

When 1 kg of water and 1 kg of wood absorb the same amount of heat, the change in temperature of the wood is greater than the ch

Strike441 [17]

Intermolecular forces in water are greater than those in wood. APEX

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3 years ago

Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirig

MissTica

Answer:

v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

Em_f = K = ½ m v_f²

Emo = Em_g

½ m v₀² + m g y₀ = ½ m v_f²

v_f² = v₀² + 2 g y₀

let's calculate

v_f = √(10² + 2 9.8 25)

v_f = 24.3 m / s

6 0

2 years ago

someone help pls. Two students, Mia and Peter, leave school to meet at the local coffee shop. Peter decides to jog to the coffee

cluponka [151]

Answer:

1) The distance further it takes Peter to arrive at the Coffee shop than Mia is 1.24 km

2) Mia's average speed is 6.00 km/hour

Peter's average speed is 8.48 km/hour

4) Mia's average velocity = Peter's average velocity = 6.00 km/hour

Explanation:

The given information from the diagram are;

The distance Peter jogs from school to the flower shop = 2.00 km

The distance Peter jogs from the Flower shop to the Coffee shop = 2.24 km.

The distance Mia walks from school directly to the Coffee shop = 3.00 km

The time it takes both Peter and Mia to arrive at the coffee shop = 30 minutes = 0.5 hour

1) The total distance Peter travels from school to the Coffee shop = 2.00 km + 2.24 km = 4.24 km

The distance Mia travels from school to the Coffee shop = 3.00 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 4.24 km - 3.00 km = 1.24 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 1.24 km

2) $Average \ speed = \dfrac{Total \ distance \ traveled}{Total \ time \ taken \ in \ the \ journey}$

$Therefore, \ Mia's \ average \ speed = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Mia's average speed = 6.00 km/hour

$Peter's \ average \ speed = \dfrac{4.24 \ km}{0.5 \ hour}= 8.48 \ km/hour$

Peter's average speed = 8.48 km/hour

4) $Average \ velocicty = \dfrac{Displacement }{Time \ taken}$

The displacement from the School to the Coffee shop is 3.00 km for both Mia and Peter

The time it takes both Peter and Mia to arrive at the Coffee shop from the school is 30 minutes = 0.5 hour

$Therefore, \ Mia's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Mia's average velocity = 6.00 km/hour

$Peter's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Therefore, Peter's average velocity is also = 6.00 km/hour

6 0

2 years ago