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Butoxors [25]
3 years ago
10

Compute the power output (watts) during one minute of treadmill exercise, given the following: Treadmill grade-10% Horizontal sp

eed 100 meters per minute Subject's body weight 60 kg (i.e., force 588.6 Newtons) 1 joule 1 newton-meter 1 watt 1 joule per second 1 kcal 426.8 kpm Multiple Choice a. 981 watts b. 9,810 wattsc. 981 wattsd. Power output cannot be calculated given the information above.
Physics
1 answer:
erma4kov [3.2K]3 years ago
6 0

Answer:

c. 981 watts

P=981\ W

Explanation:

Given:

  • horizontal speed of treadmill, v=100\ m.min^{-1}=\frac{5}{3} \ m.s^{-1}
  • weight carried, w=588.6\ N
  • grade of the treadmill, G\%=10\%

<u>Now the power can be given by:</u>

P=v.w

P=588.6\times\frac{5}{3} (where grade is the rise of the front edge per 100 m of the horizontal length)

P=981\ W

You might be interested in
The surface tension of water was determined in a laboratory by using the drop weight method. 100 drops were released from a bure
Lapatulllka [165]

Answer:

γ = 0.06563 N / m

9.78% difference

Explanation:

Solution:-

- Surface tension is the ability of any fluid to resist any external force which causes a decreases in surface area of the impact area due to inward compressive forces. These compressive forces occur due to cohesive nature of the fluid molecules.

- Mathematically, surface tension ( γ ) is defined as the force felt per unit length by the fluid.

                           γ = F / L

Where,

              F: Force imparted

              L: The length over which force is felt

- We are given the mass ( M ) of ( n = 100 ) water droplets to e 3.78 g. The mass of a single droplet ( m ) can be evaluated as follows:

                         m = M / n

                         m = 3.78 / 100

                        m = 0.0378 g        

- The force ( F ) imparted by a single drop of water from the burette can be determined from the force balance on a single droplet. Assuming the distance over which the drop falls is negligible and resistive forces are negligible. Then the only force acting on the droplet is due to gravity:

                        F = m*g

                        F = 0.0378*9.81*10^-3

                        F = 0.000370818 N      

- The length over which the force is felt can be magnified into a circular area with diameter equal to that of a single droplet ( d ). The circumferential length ( L ) of the droplet would be as follows:

                        L = π*d

                        L = π*( 0.0018 )

                        L = 0.00565 m

- Then the surface tension would be:

                        γ = F / L

                        γ = 0.000370818 / 0.00565

                        γ = 0.06563 N / m

- The tabulated value of water's surface tension is given as follows:

                       γa = 0.07275 N/m

- We will determine the percentage difference between the value evaluated  and tabulated value as follows:

                     p.diff = \frac{gamma_a - gamma}{gamma_a} * 100\\\\p.diff = \frac{0.07275- 0.06563}{0.07275} * 100 \\\\p.diff = 9.78 %

- The %difference between is within the allowable practical limits of 10%. Hence, the evaluated value ( γ = 0.06563 N / m ) can be accepted with 9.78% error.

8 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 19.0 ∘ west of north,
valina [46]

Answer:

=2.99\times10^9J

Explanation:

According to the question

net force F = 2.20×10^6 N

displacement S = 0.72\times10^3m

from figure , the horizontal forces are same in magnitude and opposite direction.

so , neglect these two forces.

we can take only vertical components of the force.

total force F' = F cos 19° + F cos 19°

= 2×F×cos 19°   ................. (1

therefore , total work is

W = F'S

= (2F cos19)×S

= (2)(2.20\times10^6 N)cos19° (0.720\times10^3 m)

=2.99\times10^9J

6 0
3 years ago
Is there any electric field inside a conductor carrying an electric current ??
denis-greek [22]
I believe so not 100% sure but i am about 50% sure 
3 0
3 years ago
Along a horizontal snow-covered track, a sled, of mass m = 105 kg, slides by the action of a horizontal force of 230 N. The coef
Andrew [12]

Answer:

Explanation:

The only thing I can figure you need here is the accleration of the sled. The equation we need to find this is Newton's Second Law that says that sum of the forces acting on an object is equal to the object's mass times its acceleration. For us, that looks like this because of the friction working against the sled:

F - f = ma but of course it's much more involved than that simple equation! We have the F value as 230 N, and we have the mass as 105, but we do not have the frictional force, f, and we need it to solve for a in the above equation. We know that

f = μF_n where μ is the coefficient of friction, and F_n is the normal force, aka weight of the object. We will use the coefficient of friction and find the weight in order to fill in for f:

F_n=mg so

F_n=(105)(9.8) so the weight of the sled is

F_n= 1.0 × 10³ with the correct number of sig dig there. Now to find f:

f = (.025)(1.0 × 10³) so

f = 25 to the correct number of sig fig. Now on to our "real" equation:

F - f = ma and

230 - 25 = 105a. We have to do the subtraction first, round, and then divide since the rules for addition and subtraction are different from the rules for dividing and multiplying.

230 - 25 will round to the tens place giving us 210. Then

210 = 105a. 210 has 2 sig figs in it while 105 has 3, so we will divide and round to 2 sig fig:

a = 2.0 m/sec²

3 0
2 years ago
What is the concentration of H^ + at apH = 2 ? Mol / L What is the concentration of H^ + ions at apH = 6 ? Mol/L How many more H
Nonamiya [84]

Answer:

The concentration of hydrogen ion at pH is equal to 2 := [H^+]=0.01 mol/L

The concentration of hydrogen ion at pH is equal to 6 : [H^+]'=0.000001 mol/L

There are 0.009999 more moles of  H^+ ions in a solution at a pH = 2 than in a solution at a pH = 6.

Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

pH=-\log [H^+]

The hydrogen ion concentration at pH is equal to 2 = [H^+]

2=-\log [H^+]\\

[H^+]=10^{-2}M= 0.01 M=0.01 mol/L

The hydrogen ion concentration at pH is equal to 6 = [H^+]

6=-\log [H^+]\\\\

[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L

Concentration of hydrogen ion at pH is equal to 2 =[H^+]=0.01 mol/L

Concentration of hydrogen ion at pH is equal to 6 = [H^+]'=0.000001 mol/L

The difference between hydrogen ion concentration at pH 2 and pH 6 :

= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L

Moles of hydrogen ion in 0.009999 mol/L solution :

=0.009999 mol/L\times 1 L=0.009999 mol

There are 0.009999 more moles of  H^+ ions in a solution at a pH = 2 than in a solution at a pH = 6.

8 0
2 years ago
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