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
oksano4ka [1.4K]
2 years ago
12

Can you help me on this? it is a phyisical science question.

Physics
1 answer:
LenKa [72]2 years ago
8 0

Answer:

Electron

Explanation:

You might be interested in
A active-satellite, one that sends out electromagnetic waves, can be the energy down toward the ocean surface and determine how
Anna71 [15]

Answer: This is called backscatter which refers to the ability of big waves to reflect the energy in order to give back the signal .

Explanation:

What is meant by backscatter?

Backscatter is the process where by the waves or signal is reflected back to the original direction and get scattered in all directions.

Backscatter allows us to receive signal and be able to view all the channels that are connected through the satellite.

4 0
3 years ago
What are some ways in which body fat can be measured
MissTica
I only know about the Water tank which is the most accurate. You place your body in it, and weights are added I think. Somehow some measurements are gathered to get your body fat weight. Online calculators exist, as well as electronic waves that are sent int your body, the echo is recorded and analyzed.
8 0
3 years ago
The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about
nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

6 0
3 years ago
Drag the right word to the word that means the opposite
Ainat [17]
What do you mean? Is there a picture ?
7 0
2 years ago
How do i solve this?
kenny6666 [7]

Answer:

hmmm i dont know....

Explanation:

i just wanted free point. TANKS YOU SIR!!

7 0
3 years ago
Other questions:
  • What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons fro
    14·1 answer
  • The 2004 landings of the Mars rovers Spirit and Opportunity involved many stages, resulting in each probe having zero vertical v
    15·1 answer
  • A wire is formed into a circle having a diameter of 11.1 cm and is placed in a uniform magnetic field of 2.79 mT. The wire carri
    15·1 answer
  • You and your surfing buddy are waiting to catch a wave a few hundred meters off the beach. The waves are conveniently sinusoidal
    10·2 answers
  • Why does a blue shirt look blue? Choose the BEST explanation.
    12·2 answers
  • How do scientists decide how they gather information
    7·1 answer
  • What is the total amount of force needed to keep a 6.0 kg object moving at speed
    8·2 answers
  • 2. Heather and Matthew walk with an average velocity of +0.87 m/s eastward.
    11·1 answer
  • three small balls each of mass 13.3g are suspended separately from a common point by silk threads, each 1.17m long. The balls ar
    10·1 answer
  • An echo is sound that returns to you after being reflected from a distant surface (e.g., the side of a cliff). Assuming that the
    5·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!