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Lilit [14]
3 years ago
8

Spiderman, whose mass is 70.0 kg, is dangling on the free end of a 12.2-m-long rope, the other end of which is fixed to a tree l

imb above. By repeatedly bending at the waist, he is able to get the rope in motion, eventually getting it to swing enough that he can reach a ledge when the rope makes a θ = 58.4° angle with the vertical. How much work was done by the gravitational force on Spiderman in this maneuver?
Physics
1 answer:
Lana71 [14]3 years ago
5 0

Answer:

U = -3978.8 J

Explanation:

The work of the gravitational force U just depends of the heigth and is calculated as:

U = -mgh

Where m is the mass, g is the gravitational acceleration and h the alture.

for calculate the alture we will use the following equation:

h = L-Lcos(θ)

Where L is the large of the rope and θ is the angle.

Replacing data:

h = 12.2-12.2cos(58.4)

h = 5.8 m

Finally U is equal to:

U = -70(9.8)(5.8)

U = -3,978.8 J

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1800*30= 54000

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erik [133]

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2 years ago
A child of mass m is standing at the edge of a carousel. Both the carousel and the child are initially stationary. The carousel
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Answer:

the angular velocity of the carousel after the child has started running =

\frac{2F}{mR} \delta t

Explanation:

Given that

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change in time = \delta \ t

By using the torque around the inertia ; we have:

T = I×∝

where

R×F = I × ∝

R×F = \frac{1}{2} mR^2∝

F = \frac{1}{2} mR∝

∝ = \frac{2F}{mR}           ( expression for angular  angular acceleration)

The first equation of motion of rotating wheel can be expressed as :

\omega = \omega_0  + \alpha  \delta t

where ;

∝ = \frac{2F}{mR}    

Then;

\omega = 0+ \frac{2F}{mR} \delta t

\omega =  \frac{2F}{mR} \delta t

 

∴ the angular velocity of the carousel after the child has started running =

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7 0
3 years ago
Water flows from a large drainage pipe at a rate of 950 gal/min. What is this volume rate of flow in (a) m3/s , (b) liters/min,
Paladinen [302]

Answer:

0.05997\ m^3/s

3596.1395\ L/min

2.11647\ ft^3/s

Explanation:

1\ gal/min=\dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=0.05997\ m^3/s

Volume rate of flow is 0.05997\ m^3/s

1\ gal/min=3.78541\ L/min\\\Rightarrow 950\ gal/min=950\times 3.78541=3596.1395\ L/min

Volume rate of flow is 3596.1395\ L/min

1\ gal/min=\dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=2.11647\ ft^3/s

Volume rate of flow is 2.11647\ ft^3/s

6 0
3 years ago
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