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Natali [406]
3 years ago
9

A boat is heading across a river at a velocity of 25 mph. The river is flowing downstream at 10 mph. What is the actual velocity

of the boat?
Physics
1 answer:
AnnyKZ [126]3 years ago
8 0
Principle: Pythagorean theorem

Given:
Let a = boat velocity = 25 miles per hour
      b = river velocity = 10 miles per hour

Required: c = actual velocity

Solution:
     
     c^2 = a^2 + b^2
     c^2 = (25)^2 + (10)^2
     c^2 = 725
     (c^2)^0.5 = (725)^0.5
      c = 26.93 miles per hour

<em>ANSWER: Actual boat velocity is 26.93 miles per hour</em>
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Which property of sound waves decreases as the square of the distance from the source increases?
Vitek1552 [10]

Answer:

Intensity

Explanation:

The intensity of a sound wave is equal to the ratio between to the power emitted by the source divided by the area of the spherical surface through which the wave propagates:

I=\frac{P}{4\pi r^2}

where

P is the power

4\pi r^2 is the area of the spherical surface

r is the distance from the source

As we see from the formula, the intensity is inversely proportional to the square of the distance from the source:

I\propto \frac{1}{r^2}

so, intensity is the correct answer.

3 0
3 years ago
The movement of water between the atmosphere, land, oceans, and even living things makes up the water cycle. Precipitation is on
Anna [14]

I have no idea

Explanation:

4 0
3 years ago
Water flows with constant speed through a garden hose that goes up to 27.5 cm high. if the water pressure is 132kpa at the botto
sergejj [24]

Answer:

The pressure at the top of the step is 129.303 kilopascals.

Explanation:

From Hydrostatics we find that the pressure difference between extremes of the water column is defined by the following formula, which is a particular case of the Bernoulli's Principle (v_{bottom}\approx v_{top}):

p_{bottom}-p_{top} = \rho\cdot g\cdot \Delta h (1)

p_{bottom}, p_{top} - Total pressures at the bottom and at the top, measured in pascals.

\rho - Density of the water, measured in kilograms per cubic meter.

\Delta h - Height difference of the step, measured in meters.

If we know that p_{bottom} = 132000\,Pa, \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} and \Delta h = 0.275\,m, then the pressure at the top of the step is:

p_{top} = p_{bottom}-\rho\cdot g\cdot \Delta h

p_{top} = 132000\,Pa-\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.275\,m)

p_{top} = 129303.075\,Pa

p_{top} = 129.303\,kPa

The pressure at the top of the step is 129.303 kilopascals.

6 0
3 years ago
Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an
joja [24]

Answer:

Approximately 9.7\; \rm m \cdot s^{-2}.

Explanation:

Assuming that there is no other force on this vehicle, the 16000\; \rm N force from the road would be the only force on this vehicle. The net force would then be equal to this 16000\; \rm N\! force. The size of the net force would be 16000\; \rm N\!\!.

Let m denote the mass of this vehicle and let \Sigma F denote the net force on this vehicle.  

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, a, would be:

\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}.

For this vehicle, \Sigma F = 16000\; \rm N whereas m = 1650\; \rm kg. The acceleration of this vehicle would be:

\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}.

8 0
3 years ago
A book is to be produced with pages of thickness 0.125mm.If the book is to be exactly 1cm thick what is the maximum number of pa
miskamm [114]

Answer:

1cm=10mm

10cm/0.125mm

80pages

5 0
3 years ago
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