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Arturiano [62]
3 years ago
12

C

Physics
1 answer:
Artemon [7]3 years ago
8 0

Answer:

constant volicty of the pumper when they hit ground 7.03-/s

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A bike rider pedals with constant acceleration to reach a velocity of 7.8 m/s over a time of 4.2 s. during the period of acceler
Artyom0805 [142]

To calculate the initial velocity of the bike, we use the following equation

d=\frac{1}{2} (u+v)t.

or

u=\frac{2d}{t} -v

Here, u is initial velocity, v is final velocity, t is the time and d is the distance covered by bike.

Given, u =7.8 m/s,d= 19 m and t=4.2 s.

Substituting these values in above equation, we get

u = \frac{2 \times 19}{4.2 \ s} -7.8 m/s = 9.05 \ m/s - 7.8 \ m/s \\\\ u= 1.2 m/s.

Thus, the initial velocity of the bike is 1.2 m/s.

3 0
3 years ago
What is the frequency of the electromagnetic wave if the period is 1.54x10-15s
pogonyaev

answer✿࿐

I was not able to write it here

so I did it somewhere else and attached the picture

i hope it helps

have a nice day

#Captainpower

3 0
2 years ago
What are three (3) things you can do before exercise or sports that will keep you safe?
zaharov [31]

Answer: Drink water, practice, do some light stretches

Explanation:

3 0
3 years ago
A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liquid floats on the gasoline and has a refractive i
klio [65]

Answer:

1.08

Explanation:

This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then

path difference created will be 2μ t.

For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer ,  so for constructive interference

path diff = nλ , for minimum t , n =1

path diff = λ

2μ t. =  λ

μ = λ / 2t

= 626 / 2 x 290

= 1.08

5 0
3 years ago
A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person
vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 \frac{m}{s}

t=0.475s V=1.95 \frac{m}{s}

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

sin(\alpha) =\frac{op}{h}

op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz

c).

T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz

The period is the inverse of the time of the motion so, the T1 is faster that the T because

t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s

d).

The speed is the relation between the distance with time so:

Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}

3 0
3 years ago
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