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Fantom [35]
3 years ago
11

Which of the following describes the least amount of work being done? A. wind turning a windmill B. a batter hitting a baseball

C. steam turning a turbine D. a ball hanging from a tall pole
Physics
2 answers:
HACTEHA [7]3 years ago
8 0
Answer:

Step bye step:
Mandarinka [93]3 years ago
6 0

Answer:

a ball hanging from a tall pole

Explanation:

work is force times a distance in the direction of the force. The ball just hanging there has no motion, so is associated with no work. Work was probably done in the hanging process, but none after.

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3 years ago
How are surface temperature and length of year on a planet related to the planet’s distance from the Sun?
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The farther from the sun an object is the larger the distance a planet has to orbit the sun. So if a planet is far from the sun the years would be longer. And if a planet is close to the sun it would be much hotter than a planet that is far from the sun.
8 0
3 years ago
A wave with 2.0 m amplitude has a frequency of 500 Hz is travelling at a speed of 200 m/s. What is the wavelength?
Serggg [28]

Answer: 0.4m

Explanation:

Given that:

Amplitude of wave = 2.0 m

Wavelength (λ)= ?

Frequency F = 500Hz

Speed V = 200 m/s

The wavelength is measured in metres, and represented by the symbol λ.

So, apply the formula:

Wavespeed V= Frequency F xwavelength λ

200m/s = 500Hz x λ

λ = 200m/s / 500Hz

λ = 0.4m

Thus, the wavelength is 0.4 metres

3 0
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Some coastal regions of the world have cooler summers and warmer winters than inland regions at the same latitude. What accounts
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8 0
3 years ago
A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s a
anastassius [24]

a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

5 0
2 years ago
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