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BARSIC [14]
4 years ago
15

77. There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the to

p of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long a time will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335.0 m/s on this day.
Physics
1 answer:
lisabon 2012 [21]4 years ago
8 0

Answer:70 m/s or 6.84 s

Explanation:

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A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while
zlopas [31]

Answer:

new energy of the capacitor is half that of initial energy

Explanation:

As we know that the energy stored in the capacitor is given as

U = \frac{1}{2}CV^2

here we know that capacitance of parallel plate capacitor is given as

C = \frac{\epsilon_0 A}{d}

V = voltage of battery

so now we have capacitor remains connected to the same battery and the separation between the plates is doubled

so we have

C' = \frac{\epsilon_0 A}{2d}

so now the energy stored between the plates is

U = \frac{1}{2}(\frac{C}{2})V^2

so new energy of the capacitor is half

8 0
4 years ago
Place the single weight with a known mass on the spring and release it. Eventually, the weight will come to rest at an equilibri
aliya0001 [1]

Answer:

X=m*g/K

Explanation:

Since the elastic force of the spring is balancing the force of gravity:

Fe = m*g

Now, on the spring by Hook's law, the magnitude of the elastic force will be:

Fe = K*X     where K is the elastic constant of the spring and X is the distance the spring is strectches measured from its original lenght to its current length.

Replacing this value:

K*X = m*g   Solving for X:

X = m*g/K    This value is directly proportional to the object's weight and inversely proportional to the spring's constant.

3 0
4 years ago
Can someone help me figure out how to find tension in a rope? The question is:
Simora [160]
You are welcome.......

8 0
3 years ago
Identify two everyday phenomena that exhibit diffraction of sound and explain how diffraction of sound applies. Identify two eve
Naya [18.7K]

Answer:

Explanation:

Diffraction is the term used to describe the bending of a wave around an obstacle. It is one of the general properties of waves.

1. Diffraction of sound is the bending of sound waves around an obstacle which propagates from source to a listener. Two of the daily phenomena that exhibit diffraction of sound are:

i. The voices of people talking outside a building can be heard by those inside.

ii. The sound from the horn of a car can be heard by people at certain distances away.

When sound waves are produced, the surrounding air molecules are required for its transmission. This is because sound wave is a mechanical wave which requires material medium for its propagation. When a source produces a sound, the sound waves bend around obstacles on its path to reach listeners.

2. Light waves are electromagnetic waves which can undergo diffraction. Diffraction of light is the bending of the rays of light around an obstacle. Two of the daily phenomena that exhibit diffraction of light are:

i. The shadow of objects which has the umbra and penumbra regions.

ii. The apparent color of the sky.

A ray of light is the path taken by light, and the combination of two or more rays is called a beam. A ray or beam of light travels in a straight line, so any obstacle on its path would subject the light to bending around it during propagation. These are major applications in pin-hole cameras, shadows, rings of light around the sun etc.

Some significant differences between diffraction of light and that of the sound are:

i. Diffraction of light is not as common as that of sound.

ii. Sound propagates through a wider region than light waves.

iii. Sounds are longitudinal waves, while lights are transverse waves.

7 0
4 years ago
A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of
tekilochka [14]

The largest mass of cargo the balloon can lift is 791.06 kg

First, we need to calculate the mass of helium.

Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.

The volume of the balloon also equals the volume of helium present.

Now, the mass of helium m = density of helium, ρ × volume of helium, V

m = ρV

Since ρ = 0.179 kg/m³

m = ρV

m = ρ4πr³/3.

m = 0.179 kg/m³ × 4π(7.15 m)³/3

m = 0.179 kg/m³ × 4π(365.525875 m³)/3

m = 0.179 kg/m³ × 1462.1035π m³/3

m = 261.7165265π/3 kg

m = 822.207/3 kg

m = 274.07 kg

Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.

The weight of this mass W = Mg where g = acceleration due to gravity.

The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.

W' = m'g

Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'

We know that the volume of air displaced, V' = volume of balloon, V

So, V' = V = 4πr³/3.

Since the density of air, ρ' = 1.29 kg/m³,

m' = ρ'V

m = 1.29 kg/m³ × 4π(7.15 m)³/3

m = 1.29 kg/m³ × 4π(365.525875 m³)/3

m = 1.29 kg/m³ × 1462.1035π m³/3

m = 1886.113515π/3 kg

m = 5925.4/3 kg

m = 1975.13 kg

So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.

So, the net mass m" = W"/g = 791.06g/g = 791.06 kg

This net mass is the largest mass of cargo that the balloon can lift.

Thus, the largest mass of cargo the balloon can lift is 791.06 kg

Learn more about balloons here:

brainly.com/question/21890581

8 0
3 years ago
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