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zalisa [80]
2 years ago
8

assuming birdman flies at a speed of 22 m/s how high should birdman fly to hit the bucket if the bucket is placed 92 m from the

start of the field
Physics
1 answer:
Ludmilka [50]2 years ago
5 0

Answer:

I will suppose that: The initial velocity of the birdman is horizontal. Now, the only force acting on birdman will be the gravitational force, so we can write the acceleration of birdman as

Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/1406862-assuming-birdman-flies-at-a-height-of-78m-how-fast-should.html#answer1893970

Explanation:

hope \: it \: helps

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A steel rod has a radius of 10mm and a
ss7ja [257]
Given a steel rod:
Radius = r = 10 mm = 0.010 m.
Length L = 1 m.
Tensile Force = F = 100 kN.
Young's Modulus = Y = 2.0 &* 10¹¹ Nm⁻²

a) Stress = Force/cross sectional area
σ = F / A
= 100,000 /(π 0.010²) Pa
= 318.309 MPa

b) Y = (F/A) / (ΔL/L)
Elongation ΔL = L F / (A Y)
= L σ / Y
= 1 * 318.309 * 10⁶ / 2.0 * 10¹¹ m
= 1.591 mm

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4 0
1 year ago
A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperat
Nostrana [21]

Answer:

Approximately 13\; \rm g of steam at 100\; \rm ^\circ C (assuming that the boiling point of water in this experiment is 100\; \rm ^\circ C\!.)

Explanation:

Latent heat of condensation/evaporation of water: 2260\; \rm J \cdot g^{-1}.

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to \rm J \cdot g^{-1}.

Specific heat of water: 4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}.

Specific heat of copper: 0.39\; \rm J \cdot g^{-1}\cdot K^{-1}.

The temperature of this calorimeter and the 250\; \rm g of water that it initially contains increased from 20\; \rm ^\circ C to 50\; \rm ^\circ C. Calculate the amount of energy that would be absorbed:

\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J  \end{aligned}.

\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J  \end{aligned}.

Hence, it would take an extra 585\; \rm J + 31500\; \rm J = 32085\; \rm J of energy to increase the temperature of the calorimeter and the 250\; \rm g of water that it initially contains from 20\; \rm ^\circ C to 50\; \rm ^\circ C.

Assume that it would take x grams of steam at 100\; \rm ^\circ C ensure that the equilibrium temperature of the system is 50\; \rm ^\circ C.

In other words, x\; \rm g of steam at 100\; \rm ^\circ C would need to release 32085\; \rm J as it condenses (releases latent heat) and cools down to 50\; \rm ^\circ C.

Latent heat of condensation from x\; \rm g of steam: 2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J.

Energy released when that x\; {\rm g} of water from the steam cools down from 100\; \rm ^\circ C to 50\; \rm ^\circ C:

\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J  \end{aligned}.

These two parts of energy should add up to 32085\; \rm J. That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from 20\; \rm ^\circ C to 50\; \rm ^\circ C.

(2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J.

Solve for x:

x \approx 13.

Hence, it would take approximately 13\; \rm g of steam at 100\; \rm ^\circ C for the equilibrium temperature of the system to be 50\; \rm ^\circ C.

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2 years ago
The picture to the right shows which wave behavior?
kifflom [539]

Answer:

It is refraction

Explanation:

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