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3 years ago

Waves begin to "feel bottom" when the depth of water is

2 answers:

8 0

I think the correct answer would be one half the wavelength. Waves would "feel bottom" when the water is at the depth of 0.5 of the wavelength. "Feel bottom" is a term used to describe that the depth of water affects the wave properties. Hope this answers the question.

Likurg_2 [28]3 years ago

6 0

Equal to one-half the wavelength is the right answer.

Water is regulated by many approaches such as depth gauges (based on force) a sounding line and by sonar range. Prior to the modern era, sailors would cast metal weighted materials into the ocean. Submarines made a new demand for depth estimation, to need to know the depth of boats in the water. However, waves start to "feel bottom" when the depth of water is equivalent to one-half the wavelength. As the wave approaches a lightweight bank, its track is adjusted. This involution made the tides to become inactive. As the flow becomes steeper, the width of the wave diminishes.

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The eyepiece of a light microscope has a magnification level of 10x. If you were looking at a paramecium under the lowest-power

tester [92]

Answer:

The total magnification will be 40x.

Explanation:

As we know if the magnification of the eyepiece of a microscope be $m_{e}$ and that of the objective be $m_{o}$ , then the total magnification is given by

$m = m_{e} \times m_{o}$

Given, $m_{e} = 10x$ and $m_{o} = 4x$

Therefore total magnification is $m = 10 \times 4 = 40x$

4 0

3 years ago

How much heat does a freezer need to remove from 1kg of water at 40°C to make ice at 0°C? (You will need to use the specific hea

Montano1993 [528]

Answer:

Explanation:

Q = m c Δt

heat withdrawn to lower temperature by 40° C

Q = 1 X 1000 X 40 = 40,000 calories.

Q = mL

heat withdrawn to freeze water at 0°C to ice

= 1 x 80000 = 80,000 calories

total heat to be withdrawn = 120,000 calories. = 120,000 x 4.2 = 504 ,000 Joules.

3 0

3 years ago

A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat

rusak2 [61]

Power is needed for (1) acceleration and (2) lifting the loaded chairs. These two parts can be calculated separately and then added together.

(1) Power for acceleration:

The final speed of the lift is

V=(10 km/h)(1 h×1000 m60 sec×1 km)=2.887 m/s.

Then the power needed is

Pa=12m(V2−V20)/Δt=12(50×250 kg)(2.778 m/s)2=9.6 kW.

(2) Power for lift

Assume that the acceleration is constant (i.e. power supply is constant), its value will be

a=ΔVΔt=2.778 m/s5 s=0.556 m/s2.

Then the vertical lift during acceleration will be

(12at2)×(2001000)=1.36 m.

Hence, the power needed to increase the potential energy of the lift is

Pg=mgΔhΔt=(50×250 kg) (9.89 m/s2)(1.36 m)/(5 s)=3.41 kW.

Then the total Power required is

Ptotal=Pa+Pg=9.6+34.1=43.7 kW.

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4 0

2 years ago

what is the magnitude of the gravitational force acting on the earth due to the sun? express your answer in newtons.

Alborosie

The gravitational force the sun experiences from the earth is 3.48×10²²N, which is exactly the same as the force the sun experiences from the earth.

Gravity is a force that develops as a result of the attraction between mass-containing objects. The mass of the object has a direct relationship to the strength of this attraction. r equals the separation of two objects.

F = G (M₁M₂)/r²

Where, F the gravitational force

G=6.67×10⁻¹¹Nm²kg⁻² gravitational constant

M₁=5.98×10²⁴kg mass of earth

M₂= 1.99×10³⁰ kg the mass of the sun

r =15×10¹⁰ m is the distance between sun and earth

Putting all the values in above equation,

F = 6.67×10⁻¹¹Nm²kg⁻²(5.98×10²⁴kg 1.99×10³⁰ kg)/15×10¹⁰ m

On solving the above equation we get,

F = 3.48×10²²N

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5 0

1 year ago

A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back

Kobotan [32]

Answer:

a) 5 m/s

b) 17.8542 m/s

c) 24.7212 m/s

0.229

Explanation:

t = Time taken

u = Initial velocity = 5 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=5-9.81\times t\\\Rightarrow \frac{-5}{-9.81}=t\\\Rightarrow t=0.51 \s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=5\times 0.51+\frac{1}{2}\times -9.81\times 0.51^2\\\Rightarrow s=1.27\ m$

So, the stone would travel 1.27 m up

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.27+0^2}\\\Rightarrow v=5\ m/s$

Velocity as the rock passes through the original point is 5 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.27\times 2}{9.81}}\\\Rightarrow t=0.51\ s$

Time taken to reach the original point is 0.51+0.51 = 1.02 seconds

So, total height of the rock would fall is 30+1.27 = 31.27 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 16.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{16.27\times 2}{9.81}}\\\Rightarrow t=1.82\ s$

Time taken by the stone to reach 15 m above the ground is 1.82+0.51 = 2.33 seconds

$v=u+at\\\Rightarrow v=0+9.81\times 1.82\\\Rightarrow v=17.8542\ m/s$

Speed of the ball at 15 m above the ground is 17.8542 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 31.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{31.27\times 2}{9.81}}\\\Rightarrow t=2.52\ s$

$v=u+at\\\Rightarrow v=0+9.81\times 2.52\\\Rightarrow v=24.7212\ m/s$

Speed of the stone just before it hits the street is 24.7212 m/s

F = Force

m = Mass = 100 kg

g = Acceleration due to gravity = 9.81 m/s²

s = Displacement = 4 km = 4000 m

P = Power = 1 hp = 745.7 Watt

t = Time taken = 20 minutes = 1200 seconds

μ = Coefficient of sliding friction

F = μ×m×g

⇒F = μ×100×9.81

W = Work done = F×s

P = Work done / Time

⇒P = F×s / t

$745.7=\frac{\mu \times 981\times 4000}{1200}\\\Rightarrow \mu=\frac{747.5\times 1200}{981\times 4000}\\\Rightarrow \mu=0.229$

Coefficient of sliding friction is 0.229

8 0

3 years ago