1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
earnstyle [38]
3 years ago
5

Describe how the velocity and acceleration of a skydiver changes as she falls from the plane back to the ground

Physics
2 answers:
Andrews [41]3 years ago
8 0
The velocity and acceleration of the skydiver will increase as she falls to the ground. According to Newton’s Law of gravity.
lidiya [134]3 years ago
8 0

Answer:

Explanation:

The velocity and acceleration of a skydiver increases as she falls from the plane back to the ground. When she will come more closer to earth the force of gravity will increase on her and her velocity will increase so the acceleration. According to the newtons law of gravitation gravity will increase as the distance between the two body decreases.

You might be interested in
How to find new pressure of gas if bottle explodes
KIM [24]
If the container explodes there is no pressure, becuase all your gas has escaped its container, there for, you ain’t got no gas
6 0
3 years ago
Suppose that you are headed toward a plateau 55 meters high. If the angle of elevation to the top of the plateau is 40degrees​,
Iteru [2.4K]

Answer:

x=65.55m

Explanation:

Let x be the distance to the shore

From trigonometry properties:

tan(40^{o} )=\frac{55m}{x} \\x=\frac{55m}{tan(40^{o} )} \\x=65.55m

3 0
3 years ago
an object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m. determine the image height (in cm)
Juli2301 [7.4K]

Given :

An object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m.

To Find :

the image height (in cm).

Solution :

By lens formula :

\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}

Here, u = - 100 cm

f =  150 cm

\dfrac{1}{v} - \dfrac{1}{-100} = \dfrac{1}{150}\\\\v = - 300 \ cm

Now, magnification is given by :

m = \dfrac{v}{u} = \dfrac{h_i}{h_o}\\\\h_i = \dfrac{300}{100}\times 1\\\\h_i = 3 \ m

Therefore, the image height is 3 m or 300 cm.

5 0
2 years ago
The sun is going down, and most of the land is dark, but we can still see silhouettes and outlines of objects because some light
miskamm [114]
The Sun is going down, and most of the land is dark, still we can see silhouettes and outlines of objects because some light is still scattered in the atmosphere. I hope this helps you.
7 0
3 years ago
¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
ValentinkaMS [17]

Answer:

60 Ω

Explanation:

R(com) = 15 Ω

1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn

1/15 = 1/20 + 1/R2

1/R2 = 1/15 - 1/20

1/R2 = (4 - 3) / 60

1/R2 = 1/60

R2 = 60 Ω

así, la combinada de resistencia necesaria es 60 Ω

5 0
3 years ago
Other questions:
  • As energy in the form of heat is added to an ice cube ,it begins to melt.what causes melting?
    14·2 answers
  • Which occurs during transcription?
    11·2 answers
  • What is gluteus maximum
    12·2 answers
  • Light is traveling through the different media shown. In which medium does light travel fastest?
    8·2 answers
  • a street light is mounted at the top of a 15 foot pole. A man 6 ft tall walks away from the pole wit a speed of 7 ft/s along a s
    8·1 answer
  • 2. A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point
    10·2 answers
  • Which describes freefall
    5·1 answer
  • Parker wound a wire around a large iron nail. He then created an electromagnet by connecting the ends of the wire to different b
    9·2 answers
  • A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from po
    5·2 answers
  • Match each space mission with the correct milestone.
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!