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Radda [10]
3 years ago
13

A 25-foot ladder is leaning against a house. The base of the ladder is pulled away from the house at a rate of 2 feet per second

. How fast is the top of the ladder moving down the wall when the base is given below?
Physics
1 answer:
grandymaker [24]3 years ago
4 0

A ladder 25 feet long is leaning against a house.  The base of the ladder is pulled away at a rate of 2 ft/sec.  

a.)  How fast is the top of the ladder moving down the wall when the base of the ladder is 12 feet from the wall?

Answer:

dy/dt = -1.094ft/sec

Explanation:

Given that:

dz/dt = 0,

dx/dt = 2,

dy/dt = ?

Hence, we have the following

Using Pythagoras theorem

We have 25ft as the hypotenuse, y as the opposite or height of wall, and x as the base of the triangle

X² + y² = z²,

12² + y² = 25²,

y² = 25² - 12²

y = √481

Therefore, we have the following:

2x dx/dt + 2y dy/dt,

= 2z dz/dt,

= 12 (2) √481 dy/dt,

= √481 dy/dt = -24,

= dy/dt = -1.094ft/sec

Therefore, final answer is -1.094ft/sec

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kotegsom [21]

Answer:

v > ii > iii > i > iv

Explanation:

The speed of light does not depend on the wavelength of light rather the medium through which it passes through.

v) The speed of light in a vacuum is maximum = 300000 km/s

ii) The speed of light in air = 2999910.026 km/s

iii) The speed of light in water = 2255639.097 km/s

i) The speed of light in glass = 1973684.210 km/s

iv) The speed of light in diamonds = 1239669.421 km/s

Hence the ranking will be

v > ii > iii > i > iv

5 0
3 years ago
Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

λ = \frac{8.47*2}{3}

λ = 5.65m

3 0
4 years ago
I can raise a bucket of cement mix of mass 12kg through a vertical height of 8m in 10 seconds.Calculate the average power used i
Andrei [34K]

Explanation:

Power = work / time

Power = force × distance / time

P = (12 kg × 10 m/s²) (8 m) / (10 s)

P = 96 Watts

5 0
3 years ago
45 joules were expended to move a box weighing 30 newtons. How many meters was it moved?
slava [35]

Answer:

1.5 meter is the answer

7 0
3 years ago
A cycle travels along a circular track of diameter 42 m. Calculate the distance travelled and the displacement of the cycle in (
DENIUS [597]

Answer:

(a) i) The distance travelled by the cycle in half round is approximately 65.97 m

ii) The displacement is 42 m

(b) (i) The distance travelled in one round is approximately 131.95 m

(ii) The displacement of the cycle in one round is 0

Explanation:

The diameter of the track through which the cycle travels, D = 42 m

(a) i) Half round is the motion of half the length of the circular path

The distance travelled by the cycle in half round = The length of half the circular track = (1/2) × π × D

∴ The distance travelled by the cycle in half round = (1/2) × π × 42 m = 21·π m ≈ 65.97 m

ii) The displacement half round = The change in the location of the cycle = The difference between the start and stop locations of the cycle on a straight line after half round

The angle at the center of the circular path the cycle turns in half round  = 180°

Therefore, the path between the start and stop location of the cycle in half round = The diameter of the circular track

The displacement of the cycle in half round = The diameter of the circular track = 42 m

(b) (i) The distance travelled in one round = The perimeter of the circular track = π·D

∴ The distance travelled in one round = π × 42 m ≈ 131.95 m

(ii) The displacement of the cycle in one round = The change in the location of the cycle

The start and stop location of the cycle after moving one round is the same, therefore, there is no change in the location of the cycle.

Therefore we have;

The displacement of the cycle in one round = 0 (no change in location of the cycle)

7 0
3 years ago
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