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olasank [31]
2 years ago
11

What should you do if you see lightning while you are outside exercising? Head inside Stay under a tree Lie in the grass Move qu

ickly
Physics
2 answers:
iris [78.8K]2 years ago
8 0

Answer:

The answer is C. I hope it helped you

Explanation:

Arada [10]2 years ago
5 0
If you are under a tree, they kind of act like lightning rods, so stay away from trees, so roll in dat grass. Fun Fact: Lightning comes from the ground more than it comes from the sky, its like when you rub a blanket on your head, some lil' lightnings come from your head while a little come from the blanket, its the same with grass and clouds only, 10000 volts stronger and deadly.
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Can someone help me please
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The last one: meter
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3 years ago
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Think of a famous person and gather information about him/her​
dybincka [34]

Abraham Lincoln

Explanation:

hope it helps you a little

4 0
2 years ago
The city bus traveled 30 miles in 30 minutes. What was its speed in mph?
jonny [76]
The answer will be B.60mph
4 0
2 years ago
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The drawing shows a loudspeaker A and point C, where a listeer is positioned. A second loudspeaker B is located somewhere to the
Liono4ka [1.6K]

Answer: 4.17m

Explanation:

The observer at C will hear a sound on no sound upon whether the interference is constructive or destructive.

If the listeners hears sounds it is caled constructive interference but if he hears no sound its called destructive interference.

So

d2 - d1 = (n *lamba)/ 2

Where n=1,3,5

lamda=v/f =349/62.8

lamda=5.56m

d2= d1 + nlamda/2

d2= 1 + 5.56/2

d2= 3.78m

X'= 1 cos 60= 0.5m

Y= 1 sin60= 0.866m

X"^2 + Y^2 =d2^2

X" =√(y^2 - d2^2)

X"=√(3.78^2 - 0.886^2)

X"= 3.67m

So therefore the closest that speaker A can be to speaker B so the listener does not hear any sound is X' + X"= 0.5 + 3.67

4.17m

3 0
3 years ago
ANSWER One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is
qaws [65]

Answer:

Explanation:

It is a case of oscillation by simple pendulum . Expression for simple pendulum is given as follows

T = 2\pi\sqrt{\frac{l}{g} }

where T is time period , l is length of pendulum and g is acceleration due to gravity .

\frac{1}{f} =2\pi\sqrt{\frac{l}{g} }  , f is frequency of oscillation

For the given case

\frac{1}{f_o} =2\pi\sqrt{\frac{l}{g} }

subsequently length becomes half so

\frac{1}{f} =2\pi\sqrt{\frac{l}{2g} }

dividing

\frac{f}{f_o} = \sqrt{\frac{2}{1} }

f = \sqrt{2} f_o

frequency of oscillation becomes √2 times.

4 0
3 years ago
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