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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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in baseball, a pitcher can accelerate 0.15kg ball from rest to 98 mi/hr in a distance of 1.7m. (a) What is the average force exe
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Answer:

F=84.68 N

Required force increases

Explanation:

By the equation

v^{2}=u^{2} +2as\\ 98^{2}*1609.344^{2}/(60*60)^{2}=0+2*a*1.7\\a = 98^{2}*1609.344^{2}/((60*60)^{2}*2*1.7)\\a = 564.50m/s^{2} \\

By newton's second law :

F=ma\\F=0.15*564.50  \\F=84.68 N

When the mass increased the force increases as the equation.

F=ma

3 0
3 years ago
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a bullet with a mass of 4.0g and a speed of 650m/s is fired at a block of wood with a mass of 0.095kg. the block rests on a fric
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Part a)

Here in this we can use momentum conservation as there is no external force on it

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} +m_2v_{2f}

here we know that

m_1 = 0.004 kg

v_{1i} = 650 m/s

m_2 = 0.095

v_{2i} = 0

v_{2f} = 23 m/s

now by above equation

0.004*650 + 0.095* 0 = 0.004*v + 0.095*23

2.6 + 0 = 0.004*v + 2.185

v = 103.75 m/s

Part b)

Final kinetic energy of the system

KE_f = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2

KE_f = \frac{1}{2}*0.004*(103.75^2) + \frac{1}{2}*(0.095)*23^2

KE_f = 46.65 J

Initial Kinetic energy of the system will be

KE_f = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2

KE_f = \frac{1}{2}*0.004*(650^2) + \frac{1}{2}*(0.095)*0^2

KE_f = 845 J

So here kinetic energy is decreased for this system

final energy is less than initial energy

6 0
3 years ago
Two small charged spheres are located on the y-axis. One is at y = 1.00 m, the other is at y = −1.00 m, and they both have a cha
yuradex [85]

Answer:

(a) 23.946 kV

(b) -0.077 J

Explanation:

(a) The electric potential is given by the following formula:

V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}   (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = q2 = 1.60*10^{-6}C

r1 and r2 are the distance from the charges to the point in which electric potential is evaluated.

Firs, you calculate the distance r1 and r2 by taking into account the position of the charges

r_1=\sqrt{(1.00)^2+(0.670)^2}m=1.20m\\\\r_2=\sqrt{(-1.00)^2+(0.670)^2}m=1.20m

Next, you replace the values of the parameters to calculate V:

V=(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}+(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}\\\\V=23946.66\ V=23.946\ kV

(b) The potential electric energy is given by:

U_T=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}\\\\r_{1,2}=2.00m\\\\r_{1,3}=1.20m\\\\r_{2,3}=1.20m\\\\U_T=(8.98*10^9)[\frac{(1.6*10^{-6})^2}{2.00m}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}]J\\\\U_T=-0.077J

5 0
3 years ago
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