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77julia77 [94]
3 years ago
9

The last is 80 meter per second,please in am in exam​

Physics
2 answers:
inna [77]3 years ago
7 0

Answer:

  1. I have fond the answer

Explanation:

but my camera doesn't work

Fynjy0 [20]3 years ago
3 0

Answer:

Happy Birthday ❤️❤️!!!!

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In the 2016 Olympics in Rio, after the 50 mm freestyle competition, a problem with the pool was found. In lane 1 there was a gen
gogolik [260]

Answer:

Explanation:

still water speed is 50 m / 25.0 s = 2.00 m/s or 200 cm/s

In lane 1 the effective speed would be 201.2 cm/s

5000 cm / 201.2 cm/s = 24.85 s

The change is 25.00 - 24.85 = 0.15 s decrease in time

In lane 8, the effective speed would be 198.8 cm/s

5000 cm / 198.8 cm/s = 25.15 s

The change is 25.00 - 25.15 = 0.15 s increase in time

6 0
3 years ago
Which object has the least momentum? object a: m = 1 kg, v = 100 m/s object b: m = 10 kg, v = 12 m/s object c: m = 0.5 kg, v = 1
nirvana33 [79]
A = 1*100 = 100 Ns
b = 10 * 12 = 120 Ns
c = 0.5*1000 = 500 Ns
d = 100 * 2 = 200 Ns

a has least momentum
8 0
3 years ago
A taxi is travelling at 15 m/s. Its driver accelerates with
Hitman42 [59]
If we assume that the acceleration is constant, we can use on the kinematic equations:

Vf = Vi + a*t = 15 + 3*4 = 27 m/s
6 0
3 years ago
How do the prefixes micro,<br> nano and pico relate to each<br> other?
In-s [12.5K]

Answer:

because they are same and their properties

8 0
3 years ago
Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it
stich3 [128]

Answer:

a. 6.69m/s

b. y=4.48m

c. t=1.43secs

Explanation:

Data given, acceleration,a=35m/s^2

distance covered,d=64cm=0.64m,

a. to determine the speed, we use the equation of motion

initial velocity,u=0m/s

if we substitute values we arrive at

v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\

b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2

and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.

Hence we can write the equation above again

v^{2}=u^{2}-2a(y-2.2)\\

if we substitute values we have

v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m

c. the time it takes to arrive at 1.83m is obtain by using the equation below

1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37

if we insert the values, we solve for t , hence t=1.43secs

6 0
3 years ago
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