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drek231 [11]
4 years ago
7

The graph shows how the vertical velocity of a parachutist changes from the moment the parachutist jumps from the aircraft until

landing on the ground. Using the idea of forces, explain why the parachutist reaches a terminal velocity and why opening the parachute reduces the terminal velocity.
Physics
1 answer:
soldi70 [24.7K]4 years ago
4 0

Answer:

BUT WHY

Explanation:

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three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25
ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0
3 years ago
.
Dmitrij [34]
Take 10m/s^2 for the gravitational acceleration, as we know this is a free fall, we can use the equation: d=1/2*g*t^2

Substitute g=10m/s^2, t=5s, d=125m
7 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
Help me with this problem please? c:
deff fn [24]
30*3*60=5400
300*2=600
1500*3=4500
200*10*60=120000

the computer uses more energy. 
(you can multiply all results by 60 to get the energy in J, but it's not necessary for the comparison)
5 0
3 years ago
Titania, the largest moon of the planet Uranus, has 1/8 the radius of the earth and 1/1700 the mass of the earth.What is the acc
garri49 [273]

Answer:

(a). The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). The average density of Titania is 1656.47 kg/m³

Explanation:

Given that,

Radius of Titania R_{t}= \dfrac{1}{8}R_{e}

Mass of Titania M_{t}= \dfrac{1}{1700}M_{e}

We need to calculate the acceleration due to gravity at the surface of Titania

Using formula of the acceleration due to gravity on earth

g_{e}=\dfrac{GM_{e}}{R_{e}^2}

The acceleration due to gravity on Titania

g_{t}=\dfrac{GM_{t}}{R_{t}^2}

Put the value into the formula

g_{t}=\dfrac{G\times\dfrac{1}{1700}M_{e}}{(\dfrac{1}{8}R_{e})^2}

g_{t}=\dfrac{64}{1700}\times G\dfrac{M_{e}}{R_{e}^2}

g_{t}=0.004705 g_{e}

g_{t}=0.03764\times9.8

g_{t}=0.37\ m/s^2

The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). We assume Titania is a sphere

The average density of the earth is 5500 kg/m³.

We need to calculate the average density of Titania

Using formula of density

\rho=\dfrac{m}{V}

\rho_{t}=\dfrac{M_{t}}{\dfrac{4}{3}\pi\times R_{t}^2}

\rho_{t}=\dfrac{\dfrac{M_{e}}{1700}}{\dfrac{4}{3}\pi\times(\dfrac{R_{e}}{8})^2}

\rho_{t}=\dfrac{512}{1700}\times\rho_{e}

\rho_{t}=\dfrac{512}{1700}\times5500

\rho_{t}=1656.47\ kg/m^{3}

The average density of Titania is 1656.47 kg/m³

Hence, (a). The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). The average density of Titania is 1656.47 kg/m³

8 0
3 years ago
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