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2 years ago

Need help ASAP Thankss + BRAINLIST only for correct answer

Physics

2 answers:

spayn [35]2 years ago

7 0

Answer:

False

true

Hope this helps!

grigory [225]2 years ago

7 0

False true (person on top is correct)

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Which of the following is an example of mechanical waves?

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3 years ago

Read 2 more answers

In doing a load of clothes, a clothes dryer uses 16 A of current at 240 V for 45 min. A personal computer, in contrast, uses 2.7

EastWind [94]

Answer:

(a) Total time will be 8.05 hour

(B) Total cost will be $417.6

Explanation:

We have given dryer uses 16 A of current so current i = 16 A

Voltage V = 240 volt

Time t = 45 minutes =45×60 = 2700 sec

So power P = VI = 240×16 = 3840 W

So energy E = power×time = 3840×45×60 = 9396 KJ

(A) Now current in the computer i = 2.7 A

Voltage V = 120 V

So power P = 120×2.7 = 324 W

Now computer is using energy of dryer

So time by which it will use energy is $t=\frac{9396000}{324}=29000sec=8.05hour$

So time will be 8.05 hour

(b) Cost per kilowatt is $0.12

So total cost =3480×0.12 = $417.6

4 0

2 years ago

In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little

horrorfan [7]

Answer:

t_up / t_down = 6.83

Explanation:

Find:

Calculate the ratio of the time he is above y_max/2 to the time it takes him to go from the floor to that height.

Solution:

- Compute the velocity v_o at y_max:

v_i^2 = v_f^2 - 2*g*y_max

0 = v_o^2 - 2*g*y_max

v_o = sqrt (2*g*y_max)

- The total time spend by athlete above height y_max / 2 is:

y - y_o = v_o*t_up - 0.5*g*t^2_up

v_o = 0.5*g*t_up

- Equate two equations:

sqrt (2*g*y_max) = 0.5*g*t_up

t_up = 2*sqrt(2*g*y_max) / g

- The total time taken by athlete to reach height y_max / 2 from ground is:

y - y_o = v_o*t_down - 0.5*g*t^2_down

g*t_^2down - 2*v_o*t_down + y_max = 0

- Solve the quadratic and evaluate t_down:

t_down = (v_o +/- sqrt (v^2_o - g*y_max)) / g

Substitute for v_o = sqrt (2*g*y_max)

t_down = (sqrt(2g*y_max) +/- sqrt(g*y_max)) / g

- We will use the minus quantity, because we need the first part of the journey from ground from the two times he passes the height of y_max/2.

Hence,

t_down = (sqrt(2g*y_max) - sqrt(g*y_max)) / g

t_down = (sqrt(g*y_max) / g) * (sqrt(2) - 1)

- Compute the ratio t_up to t_down:

t_up / t_down = 2*sqrt(2*g*y_max) / g * g / (sqrt(g*y_max)*(sqrt(2) - 1)

= 2*sqrt(2) / (sqrt(2) - 1)

= 6.83

8 0

2 years ago