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ohaa [14]
2 years ago
7

You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.How muc

h work do you do if you pull the wagon 110 m at a constant speed?
Physics
1 answer:
ahrayia [7]2 years ago
3 0

Work is equal to the force applied times the displacement. Since you pull the wagon at constant speed this means that there is no acceleration on the wagon as it does not change speed. F=ma. Since a=0, F=0. Therefore no work has been done in this situation

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A physics teacher pushes an environmental science teacher out of a stationary helicopter without a parachute from a height of 48
KiRa [710]

Answer:

is this phsyics teacher, is he good i-

Explanation:

4 0
2 years ago
In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis
Rus_ich [418]
M1 descending
−m1g + T = m1a 

m2 ascending
m2g − T = m2a

this gives :
(m2 − m1)g = (m1 + m2)a 

a = (m2 − m1)g/m1 + m2
   = (5.60 − 2)/(2 + 5.60) x 9.81 
   = = 4.65m/s^2
5 0
2 years ago
In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into t
Pie

Answer:

A) Q_a=74256\ J

B) Q=93562560\ J

Explanation:

Given:

  • temperature of air, T_a=-19+273=254\ K
  • temperature of lungs, T_l=37+273=310\ K
  • specific Heat exchanged from the lungs , c_l=0.47\ J.kg^{-1}.K^{-1}
  • specific heat of air, c_a=1020\ J.kg^{-1}.K^{-1}
  • mass of 1 L air, m'=1.3\ kg
  • breath rate, b=21\ breath.min^{-1}

A)

Now,

amount of heat needed to warm the air of lungs to the body temperature:

Q_a=m'.c_a.\Delta T

Q_a=1.3\times1020\times (310-254)

Q_a=74256\ J

B)

Amount of heat lost per hour:

<u>No. of breaths per hour:</u>

B=b.60

B=21\times 60

B=1260

<u>Now the total loss of energy in 1 hr.:</u>

Q=Q_a.B

Q=74256\times 1260

Q=93562560\ J

7 0
2 years ago
A 5.0-n projectile leaves the ground with a kinetic energy of 220 j. at the highest point in its trajectory, its kinetic energy
NikAS [45]
First, we get the difference between the kinetic energies such that,
             difference = (220J - 120J)
             difference = 100 J
The difference in kinetic energy is the equivalent of the potential energy which is calculated through the equation,
              PE = mgh
To calculate for the height, we derive the equation in a form,
           h = PE/mg
The product of the mass and acceleration due to gravity is the weight. 
                   h = (100 J) / (5 N)
                   h = 20 m

<em>Hence, the answer is 20 m. </em>
3 0
2 years ago
For what value m of the clockwise couple will the horizontal component ax of the pin reaction at a be zero? if a couple of that
sergejj [24]
Thank you for posting your question here at brainly. Below is the answer:

sum of Mc = 0 = -Ay(4.2 + 3cos(59)) + (275)(2.1 + 3cos(59)) + M 
<span>- Ay = (M + (275*(2.1 + 3cos(59)))/(4.2 + 3cos(59)) </span>

<span>sum of Ma = 0 = (-275)(2.1) - Cy(4.2 + 3cos(59)) + M </span>
<span>- Cy = (M - (275*2.1))/(4.2 + 3cos(59)) </span>

<span>Ay + Cy = 275 = ((M+1002.41)+(M-577.5))/(5.745) </span>
<span>= (2M + 424.91)/(5.745) </span>

<span>M = ((275*5.745) - 424.91)/2 </span>
<span>= 577.483 which rounds off to 577 </span>

<span>Is it maybe supposed to be Ay - Cy = 275</span>
8 0
2 years ago
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