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
IrinaK [193]
3 years ago
15

Use the techniques to find the unit for speed​

Physics
2 answers:
stellarik [79]3 years ago
8 0

Answer:

The formula for speed is speed=<u>d</u><u>i</u><u>s</u><u>t</u><u>a</u><u>n</u><u>c</u><u>e</u>

time

Explanation:

to work out what the units are for speed,you need to know the units for distance and time.In this example,distance is in metres(m) and time is in seconds (s) , so the units for speed is metre per second (m/s).

Sedbober [7]3 years ago
8 0
Hi there sorry babe i’m just tired of seeing eieieieieieieieieijdjd
You might be interested in
1. Na jaką wysokość wzniesie się piłka kopnięta pionowo w górę z prędkością 20 m/s ? 2. Piłka o masie 20 dag spada z wysokości 1
Tresset [83]

Answer:

its gonn abe 23

Explanation:

6 0
3 years ago
Read 2 more answers
Defiance Drake, Space Adventurer, desperately needs to reach a cargo vessel 100m away, or run out of air and die in deep space.
jolli1 [7]

She misses. She should have accelerated faster in order to get to her target.

5 0
3 years ago
A nature photographer is using a camera that has a lens with a focal length of 3.06 cm. The photographer is taking pictures of a
sleet_krkn [62]

Answer:

film is at distance of 3.07 cm from lens

Explanation:

Given data

focal length = 3.06 cm

distance = 10.4 m = 1040 cm

to find out

How far must the lens

solution

we apply here lens formula that is

1/f = 1/p + 1/q

here f = 3.06 and p = 1040 so we find q

1/f = 1/p + 1/q

1/3.06 = 1/1040 + 1/q

1/ q =  0.3258

q = 3.0690 cm

so film is at distance of 3.07 cm from lens

6 0
3 years ago
Science use SI because it allows them to compare data and communication with each
Mekhanik [1.2K]
True
hope this helped!
3 0
3 years ago
Read 2 more answers
Consider a string with a length of (47.5 A) cm tied at both end (like on a stringed instrument). If the frequency of the first h
zubka84 [21]

To solve this problem it is necessary to apply the concepts related to wavelength as a function of frequency and speed, as well as to determine the wavelength as a function of length.

From the harmonic vibration generated we know that the total length of the string will be equivalent to a half of the wavelength, that is

L = \frac{\lambda}{2} \rightarrow \lambda = 2L

Where,

\lambda = Wavelength

Therefore the wavelength for us would be,

\lambda = 2*47.5cm = 95cm = 0.95m

From the relationship of speed, frequency and wavelength we know that

\lambda = \frac{v}{f} \rightarrow v = \lambda f

v = (0.95m)(245Hz)

v = 232.75 m/s

Therefore the speed of the wave is 232.75m/s

4 0
3 years ago
Other questions:
  • A car accelerates from 10 km/hr to 50 km/hr in 8 seconds. What is the acceleration?
    9·1 answer
  • A mechanic uses a jack to lift up a car. He exerts a force of 11,000 N at a distance of 3m from the axis of rotation. How much t
    9·1 answer
  • How do you calculate acceleration using a friction coefficient?
    6·1 answer
  • Suppose there is a sample of xenon in a rectangular container. The gas exerts a total force of 4.47 N perpendicular to one of th
    8·1 answer
  • Define motion also justify that rest and motion are related terms​
    11·1 answer
  • A 0.5 kg object is whirled on the end of a string that is 1.2 m long at a speed of 7.5 m/s. Calculate the angular momentum of th
    10·1 answer
  • Explain how a form of<br> energy is used in daily life
    15·1 answer
  • Put these four parts of the electromagnetic spectrum in order from shortest to longest wavelength:
    10·1 answer
  • Hi brouhjnhbhb(bhbgvgv&amp;vfc
    15·2 answers
  • 20. GP A sinusoidal wave traveling in the negative x direction (to the left) has an amplitude of 20.0cm, a wavelength of 35.0cm,
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!