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olga55 [171]
3 years ago
15

Lactic acid is produced during low-moderate exercise True False

Physics
2 answers:
patriot [66]3 years ago
7 0
True. 

       Hope this helps!!
Marizza181 [45]3 years ago
3 0
The answer to your question is true.
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Velocity (m/s)
alekssr [168]

Explanation:

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(a) <u>The</u><u> </u><u>segment</u><u> </u>A shows acceleration as velocity increases with the increase in time.

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(b) <u>The</u><u> </u><u>segment</u><u> </u>C shows the object is slowing down as the time increases in segment C, the velocity decreases and afterwards it comes to rest.

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(c) The velocity is segment B is <u>4</u><u>0</u><u>m</u><u>/</u><u>s</u><u>.</u> And in the diagram there is no change in velocity.

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(d) The acceleration of segment B is <u>zero</u><u>.</u> As there in no change in curve and it is moving with uniform velocity.

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\:

<h2>Thank you!</h2>
3 0
2 years ago
a tms (transcranial magnetic stimulation) device creates very rapidly changing magnetic fields. the field near a typical pulsed-
Digiron [165]

Part A : 4.33 Volt.

Part B : 4764 Amperes.

Given Data :  

Field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs.

Axis of his 2.1-cm-diameter wedding band is parallel to the field.

Part A :

induced emf = rate of change of magnetic flux

= d (B A )/dt

= A dB/dt

= { π (2.1 x 10^-2 / 2)^2) (2.5) } / (200 x 10^-6)

E = 4.33 Volt

Part B :

R = rho L / A

R = { (6.2 x 10^-8) ( π x 0.021) } / (4.5 x 10^-6)

R = 9.10 x 10^-4 Ohm

We know, (I = V/ R)

I = 4.33 Volts / 9.10 x 10^-4 Ohm

I = 4764 Amperes

What exactly is the function of TMS?

Transcranial magnetic stimulation (TMS) is a noninvasive procedure that uses magnetic fields to stimulate nerve cells in the brain to alleviate depression symptoms. TMS is usually used when other depression treatments have failed.

Find more on TMS at : brainly.com/question/13851741

#SPJ4

➤ Although your question isn't complete, I may have assumed that you were referring to this specific question.

The complete question is :

A TMS (transcranial magnetic stimulation) device creates very rapidly changing magnetic fields. The field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs. Suppose a technician holds his hand near the device so that the axis of his 2.1-cm-diameter wedding band is parallel to the field.

Part A: What emf is induced in the ring as the field changes? Express your answer to two significant figure E= ___ unit_

Part B: If the band is made of a gold alloy with resistivity 6.2×10−8Ω⋅m and has a cross-section area 4.5 mm2 , what is the induced current? Express your answer to two significant figures and include the appropriate units. I= ___ units__

6 0
1 year ago
The three branches of science are ____, earth, and physical
Rom4ik [11]
The three branches of science are life, earth, and physical science.
Life science deals with living organisms, so, biology is the most important of these sciences, although there are many.
Earth science deals with Planet Earth, studying its atmosphere, lithosphere, etc.
Physical science studies the inorganic world.
5 0
3 years ago
Read 2 more answers
A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.9m/s and the hatchback ca
4vir4ik [10]

Answer:

s = 589.3 m

Explanation:

Let the truck and car meet at a distance = s  m

The truck is moving at constant velocity = v

so s= v * t ---------- (1)

car:

Vi = 0 m/s

a = 3.9 m/s²

s = Vi* t + 1/2 a t²

s= 0 * t +  1/2 a t²

s =  1/2 a t²   ----------- (2)

compare equation (1)  and equation (2)

s= v * t = 1/2 a t²

⇒ v * t = 1/2 a t²

⇒ t = 2 * v/ a

⇒ t = (2 * 33.9 )/ 3.9

⇒ t = 17. 38 s

Now

from equation (1)

s= v * t

s= 33.9 * 17.38

⇒ s = 589.3 m

3 0
3 years ago
Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0
3 years ago
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