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VashaNatasha [74]
3 years ago
11

Peter was holding a Human Physiology textbook that weighs about 4lbs with his non- writing arm for one minute. His elbow remaine

d exactly at a 45° isometric position. He did not complain of any muscle fatigue or muscle weakness. a) i. Is the number of activated motor units stay the same while he was holding the book for two minutes? (0.5 pt) ii. Why? (0.5 pt) Increasing the period Peter is holding the texting will only cause the motor units to fire more frequently to keep the muscles from relaxing. b) i. Are the same motor units activated while he was holding the book for two minutes? (0.5 pt) ii. Why? (0.5 pt)
Physics
1 answer:
Slav-nsk [51]3 years ago
4 0

Answer:

a

When peter held the book for one minute the rate at which motor unites are fired increase steeply but as the duration increases the increese in the rate at which it is being fired becomes linear so the number of activated motors stay the same but are being activated at a more rapidly

b

The same motors are activated whilest he is hold the book for 2 minutes this is because for peter to hold the book in one fixed position one specific motor units  need to be activated

Note changing the motor unites would change the positon of the hand

Explanation:

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Answer:

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What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?

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3. Identify the structures of the skin by the labels A through G below
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Which of the following statements is always true?
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Answer: B

Explanation:

An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. An unbalanced force is an unopposed force that causes a change in motion.

7 0
3 years ago
Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.2 kg attached to it and has a length of 5 m. Pendulum 2 has a ba
mars1129 [50]

Given Information:

Pendulum 1 mass = m₁ = 0.2 kg

Pendulum 2 mass = m₂ = 0.6 kg

Pendulum 1 length = L₁ = 5 m

Pendulum 2 length = L₂ = 1 m

Required Information:

Affect of mass on the frequency of the pendulum = ?

Answer:

The mass of the ball will not affect the frequency of the pendulum.

Explanation:

The relation between period and frequency of pendulum is given by

f = 1/T

The period of pendulum is given by

T = 2π√(L/g)

Where g is the acceleration due to gravity and L is the length of the string

As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.

Bonus:

Pendulum 1:

T₁ = 2π√(L₁/g)

T₁ = 2π√(5/9.8)

T₁ = 4.49 s

f₁ = 1/T₁

f₁ = 1/4.49

f₁ = 0.22 Hz

Pendulum 2:

T₂ = 2π√(L₂/g)

T₂ = 2π√(1/9.8)

T₂ = 2.0 s

f₂ = 1/T₂

f₂ = 1/2.0

f₂ = 0.5 Hz

So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.

3 0
3 years ago
Read 2 more answers
. A vertical electric field is set up in space to compensate for the gravitational force on a point charge. What is the required
Delicious77 [7]

(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.

(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.

<h3>Magnitude of electric field </h3>

The magnitude of electric field is given by the following equation.

F = qE

But F = mg

mg = qE

E = mg/q

where;

  • E is the electric field
  • m is mass of the particle
  • g is acceleration due to gravity
  • q is charge of the particle
<h3>For an electron</h3>

E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)

E = 5.57 x 10⁻¹¹ N/C

<h3>For proton</h3>

E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)

E = 1.02 x 10⁻⁷ N/C

Thus, the required vertical electric field is greater when the charge is proton.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

4 0
2 years ago
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