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Aleksandr-060686 [28]

3 years ago

8

Which of the following activities would be fueled by adenosine triphosphate (ATP) and the phosphagen system?A. 400-meter sprintB

. 5-mile bike rideC. 100-meter sprintD. Marathon

1 answer:

Ksenya-84 [330]3 years ago

3 0

Answer:

C

Explanation:

The Phosphagen System or Adenosine TriPhosphate - Creatine Phosphate (ATP-CP) is available in the muscles in a limited quantity, hence the amount of energy that is released through Phosphagen system is limited.

As a result the phosphagen stored in the muscle is exhausted vigorously and can only be used for activities which require very high energy for a very short duration of time.

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A throttle position sensor waveform is going to be observed. At what setting should the volts per division be set to see the ent

Mars2501 [29]

Answer:

1 V / div

Explanation:

Solution:

- The vertical scale has eight divisions.

- If each division is set to equal 1 volt, the display will show 0 to 8 volts.

- This is okay in a 0 to 5 volt variable sensor such as a throttle position (TP) sensor.

- The volts per division (V/div) should be set so that the entire anticipated waveform can be viewed.

3 0

3 years ago

The blade of a lawn mower is rotating at an angular speed of 128 rad/s. The tangential speed of the outer edge of the blade is 3

stiv31 [10]

Answer:

0.25m

Explanation:

Using the expression that relates the angular velocity(w) and the linear velocity (v).

v= wr where;

w is the angular speed= 128rad/s

v is the linear speed = 32m/s

r is the radius

r = v/w

r = 32/128

r = 0.25m

The radius of the blade is 0.25m

4 0

3 years ago

determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund

wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is $a_G= 6.78 m/s^2$

<u>Explanation</u>:

$N_A+N_B-Mg = 0$

$-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0$

$0.8N_B +0.8N_A = 975a_G$

$N_A+N_B = 9564.75$ -------------(1)

$-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0$

$-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0$

$-2.26N_A -0.06N_B= 0$ ----------------(2)

Solving the equation (1) and(2)

$N_A + N_B = 9564.75$

$-2.26N_A-0.06N_B=0$

$N_A = -260.85N$

$N_B = 9825.60N$

$\mu_s N_B + \mu_s N_A = 975a_G$

$0.8(9825.60)+0.8(-260.85) = 975a_G$ $a_G=\frac{7651.8}{975}$ $a_G_1=7.4848m/s^2$

Next lets assume that the front wheels contact with the ground N_A = 0

$F_B = Ma_G$

$N_B = M_g$

$N_B - M_g = 0$

$N_B(b-a) –F_Bh = 0$

$F_B = 975a_G$

$N_B-975(9.8) = 0$

$N_B=9564.75N$

$9564.75(2.20 -1.82) -F_B(0.55)=0$

$\frac{3634.605}{0.55}=F_B$

$F_B = 6608.3$

$F_B = Ma_G$

$6608.3 = 975a_G$

$a_G = 6.7778 m/s^2$

$a_G_2 = 6.78m/s^2$

Choosing the critical case

$a_G = min(a_G_1 ,a_G_2)$

$a_G = min(7.848, 6.78)$

$a_G= 6.78 m/s^2$

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4 years ago

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3 years ago

In the illustration, which two simple machines are being used to enable the student to reach the door?

vodka [1.7K]

D. wheel-and-axle AND an incline plane

3 0

4 years ago

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