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3 years ago

What is your basal metabolism

1 answer:

5 0

It's the energy your body spends to just keep you breathing and your heart beating ... just being alive, without trying to DO anything.

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Which image shows a non renewable resources?

Radda [10]

Where are the images?!?!?

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3 years ago

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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.16

ycow [4]

We assign the variables: T as tension and x the angle of the string
The centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.

(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. 
(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. 
(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. 
(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.

This minimum v is obtained when T=0 and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.

4 0

3 years ago

Read 2 more answers

3. Neha travels 4 m towards the north, 3m towards the east, and then 7 m towards south, find the displacement and total distance

Digiron [165]

Answer:

Explanation:

Explanation: total displacement =3√2m. and total distance covered=14m. I hope this is right and helps u.

7 0

3 years ago

9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the

mojhsa [17]

Answer:

Moment about SHOULDER ∑ τ = 3.17 N / m,

Moment respect to ELBOW Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

∑ τ = I α

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

∑ τ = I α

I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

I_mass = m L²

we substitute

∑ τ = (⅓ m L² + M L²) α

let's calculate

∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

Στ = I α

I = I_ forearm + I_mass

I = ⅓ m (L-0.03)² + M (L-0.03)²

let's calculate

i = ⅓ 2.3 0.47² + 0.5 0.47²

I = 0.2798 Kg m²

Στ = 0.2798 10

Στ= 2.80 N m

3 0

3 years ago

Two tuning forks of frequency 480 hz and 484 hz are struck simultaneously. what is the beat frequency resulting from the two sou

umka21 [38]

Ans: Beat frequency = $f_b$ = 4Hz

Explanation:
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. According to this interference, sound will be soft or loud.

Hence. the formula is:
Beat frequency = $f_b = |f_2 - f_1|$

Since,
 $f_1 = 480Hz$
$f_2 = 484Hz$

Therefore,
Beat frequency = $f_b = |484 - 480|$

=> Beat frequency = $f_b = 4Hz$
-i

7 0

4 years ago