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Len [333]
3 years ago
5

A cyclist maintains a constant velocity of 5 m/s headed away from point A. At some initial time, the cyclist is 247 m from point

A. What will his displacement be in meters from his starting position after 52 seconds?
​
Physics
1 answer:
worty [1.4K]3 years ago
5 0

260 meters im pretty sure because rate*time=distance

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Place a pencil or pen in front of you on the surface that you are working on. In one paragraph, using your own words, describe t
Ede4ka [16]

Answer:

READ THIS FIRST!!!

I am not going to right the parragraph (it says to use your own words, they could trace it back to me if you use my words, and you would get it trouble).

But I will tell you the forces acting on the pencil:

Gravity (pulls on the pencil)

Your hand (pushing it up canceling out gravity)

And the electromagnetic force (holding the thing together)

Hope it helped,

Have a good day

4 0
2 years ago
A world class sprinter is travelling with speed 12.0 m/s at the end of a 100 meter race. Suppose he decelerates at the rate of 2
Solnce55 [7]

Answer:

after 6 second it will stop

he travel 36 m to stop

Explanation:

given data

speed = 12 m/s

distance = 100 m

decelerates rate = 2.00 m/s²

so acceleration a = - 2.00 m/s²

to find out

how long does it take to stop and how far does he travel

solution

we will apply here first equation of motion that is

v = u + at   ......1

here u is speed 12 and v is 0 because we stop finally

put here all value in equation 1

0 = 12 + (-2) t

t = 6 s

so after 6 second it will stop

and

for distance we apply equation of motion

v²-u² = 2×a×s  ..........2

here v is 0 u is 12 and a is -2 and find distance s

put all value in equation 2

0-12² = 2×(-2)×s

s = 36 m

so  he travel 36 m to stop

3 0
3 years ago
How long did it take the flag to rotate once in a full circle
Tanzania [10]

Answer:

360 degrees is one full rotation starting at zero

4 0
2 years ago
SELECT ALL THE CORRECT ANSWERS
Aleonysh [2.5K]
I think the answer is x
3 0
3 years ago
Energy from the Sun arrives at the top of the Earth’s atmosphere with an intensity of 1.30kW/m21.30⁢kW/m2. How long does it take
vivado [14]

Answer:

T=1.384×10⁶seconds

Explanation:

Given data

p (Intensity)=1.30 kw/m²

E (Energy)=1.8×10⁹ J

A (Area)=1.00 m²

T (Time required)=?

Solution

E=PT   ................eq(i)

where E is energy

P is radiation power

T is time

Radiating Power is given as

P=pA

Where p is intensity

A is Area

Put P=pA in eq(i) we get

E=pAT

T=E/pA

T=\frac{(1.8*10^{9}}{(1.30*10^{3} )*(1.00)}  )\\T=1.38*10^{6} seconds

3 0
3 years ago
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