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2 years ago

In a harbor, you can see sea waves traveling around the edges of small stationary boats. Why does this happen?

1 answer:

8 0

Below are the choices that can be found in the other sources:

A. diffraction
B. refraction 
C. reflection 
D. transmission

The answer is diffraction. It means that the process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the wave forms produced.

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Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed a speed bump (called the "Holly h

Bezzdna [24]

: 30.50 km/h = 30.50^3 m / 3600s = 8.47 m/s

At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg)

So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road ..

Fn = mg - mv²/R
Fn = m(g - v²/R) .. .. 1800kg (9.80 - 8.47²/20.20) .. .. .. ►Fn = 11 247 N (upwards)
(b)
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero.

ie .. mg = mv²/R .. .. v² = Rg .. .. 20.20m x 9.80 = 198.0(m/s)²

►v = √198 = 14.0 m/s

3 0

2 years ago

550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c

Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work

7 0

2 years ago

A 63.0 kg astronaut is on a spacewalk when the tether line to the shuttle breaks. the astronaut is able to throw a spare 10.0 kg

Llana [10]

There are other forces at work here nevertheless we will imagine it is just a conservation of momentum exercise. Also the given mass of the astronaut is light astronaut.

The solution for this problem is using the formula: m1V1=m2V2 but we need to get V1:

V1= (m2/m1) V2

V1= (10/63) 12 = 1.9 m/s will be the final speed of the astronaut after throwing the tank.

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2 years ago

Read 2 more answers

The graph represents the heating of water in a pot. At 150 seconds, the water has just reached a boil. If the heat is left on, w

a_sh-v [17]

The temperature will remain constant, at around 100 C, and the volume of water in the pot will decrease, as it turns into steam and floats away from the pot.

4 0

2 years ago

You have been abducted by aliens and find yourself on a strange planet. Fortunately, you have a meter stick with you. You observ

defon

Answer:

$36.4276\ m/s^2$

Explanation:

m = Mass of stick

L = Length of stick = 1 m

h = Center of mass of stick = $\dfrac{1}{2}=0.5\ m$

g = Acceleration due to gravity

T = Time period = 0.85 s

Time period is given by

$T=2\pi\sqrt{\dfrac{I}{mgh}}$

Moment of inertia is given by

$I=m\dfrac{L^2}{3}$

$T=2\pi\sqrt{\dfrac{m\dfrac{L^2}{3}}{mgh}}\\\Rightarrow T=2\pi\sqrt{\dfrac{L^2}{3gh}}\\\Rightarrow g=\dfrac{4\pi^2L^2}{3hT^2}\\\Rightarrow g=\dfrac{4\pi^2\times 1^2}{3\times 0.5\times 0.85^2}\\\Rightarrow g=36.4276\ m/s^2$

The acceleration of gravity on this planet is $36.4276\ m/s^2$

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2 years ago

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