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ohaa [14]
3 years ago
12

The sprinter ran 110 m in 11 seconds. What was her average speed in m/s?

Physics
2 answers:
Tanya [424]3 years ago
5 0

Answer:

10 m/s

Explanation:

110/11=10

Sunny_sXe [5.5K]3 years ago
5 0

Answer:

10 m/s

Explanation:

110/11=  10 m/s

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An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase
goldenfox [79]

Answer:

changing the magnetic field more rapidly

Explanation:

According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.

Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.

8 0
3 years ago
A charge alters the space around it. What is this alteration of space called?
iogann1982 [59]

Answer: electric field

Explanation: when a charge is placed in space, it alters the space around it by creating an electric field.

This electric field has the ability to exert a force (f) on any test charge(q) placed within this vicinity.

This is the reason why a charge can either attract or repel another charge.

6 0
3 years ago
speed of sound is 343 Ms at 20 degrees Celsius. The frequency heard from the sound is 256 Hz. what is the sounds wavelength?
Lina20 [59]
S= 343m/s
F=256Hz

WL= 343ms/256-1
WL=V/F

 = 1.339844m     
7 0
3 years ago
What are the seven types of electromagnetic waves?
Sunny_sXe [5.5K]
Radio waves. Giant satellite-dish antennas pick up long-wavelength, high-frequency radio waves. ...
Microwaves. Because cosmic microwaves can't get through the whole of Earth's atmosphere, we have to study them from space. ...
Infrared. ...
Visible light. ...
Ultraviolet light. ...
X rays. ...
Gamma rays.
5 0
3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
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