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borishaifa [10]
3 years ago
11

A rowboat crosses a river with a velocity of 3.30 mi/h at an angle 62.5° north of west relative to the water. the river is 0.50

5 mi wide and carries an eastward current of 1.25 mi/h. how far upstream is the boat when it reaches the opposite shore?
Physics
1 answer:
shepuryov [24]3 years ago
4 0

<u>Answer:</u>

 Rowboat arrives 0.558 miles to the upstream when it reaches the opposite shore.

<u>Explanation:</u>

   Velocity of rowboat = 3.30 mi/h

   Angle with north of west = 62.5⁰

   Width of river = 0.505 mile.

   Velocity of river = 1.25 mi/hr eastward.

Let us take east as positive X -axis and North as positive Y-axis.

   So angle of boat with horizontal axis = (90+62.5) = 152.5⁰

  Horizontal speed of boat = 3.30*cos 152.5 = -2.93 mi/h

  Vertical speed of boat = 3.30*sin 152.5 = 1.52 mi/h

  Horizontal speed of river = 1.25 mi/h

  Time taken to cross river = Width of river/Vertical speed = 0.505/1.52 = 0.33 hour.

  Velocity along upstream of river = -2.93+1.25 = -1.68 mi/hr

  So distance moved along upstream = -1.68*0.33 = -0.558 miles ( negative sign means along west direction)

  It reaches 0.558 miles to the upstream when it reaches the opposite shore.

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Electromagnetic waves are used extensively in modern technology. Many devices are built to emit and/or receive EM waves at a ver
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Answer: Option D and E

Explanation: Across the electromagnet spectrum, the frequency increases but the wavelength reduces.

This implies that radio waves have the longest wavelength but the smallest frequency, while gamma rays have the shortest wavelength but the highest frequency.

This implies that there is an inverse relationship between wavelength and frequency.

Of all the technologies listed above, wireless internet has the highest frequency hence making it have the shortest wavelength.

Also by comparing values from the data given to us above, it is possible for some wireless internet device to operate within the same frequency range of the microwave.

3 0
3 years ago
A helicopter is lifting two crates simultaneously. One crate with a mass of 160 kg is attached to the helicopter by cable A. The
motikmotik

Answer

given,

mass of crate attached by cable A  = 160 Kg

mass of crate attached by cable B  = 73 Kg

acceleration of helicopter = 1.4 m/s²

tension in the cable when the move up

F = m (a + g)

tension in cable B

F = 73 x (1.4 + 9.8)

F = 73 x 11.2

F = 817.6 N

tension in cable A

F = (160 + 73 ) x (1.4 + 9.8)

F = 233 x 11.2

F = 2609.6 N

3 0
3 years ago
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
3 years ago
Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile? Answ
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Answer: 272.82 drop/tile

Explanation:

Given that the Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile?

Tiles/ft^2 × drop/tiles = drop/ft^2

Tiles will cancel out. Leaving the answer to be drop/ ft^2

Substitutes all the magnitude of the above units.

17 × drop/tiles = 4638

Make drop/tiles the subject of formula

Drop/tiles = 4638/17

Drop/tiles = 272.82

Therefore, 272.82 drop/tile drops fall on each tile? 

8 0
3 years ago
A bolt is to be tightened with a torque of 7.0 N · m. If you have a wrench that is 0.55 m long, what is the least amount of forc
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To solve this problem we will apply the given concept for torque which explains the relationship between the force applied and the distance to a given point. Mathematically this relationship is given as

\tau = F*d \rightarrow F = \frac{\tau}{d}

Where,

\tau = Torque

F = Force

d = Distance

Our values are given as,

\tau = 7.0Nm , d = 0.55m

Therefore replacing we have that the force is

F = \frac{7}{0.55}

F = 12.72N

Therefore the least amount of force that you must exert is 12.72N

4 0
3 years ago
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