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valentinak56 [21]
2 years ago
6

Two ropes in a vertical plane exert equal-magnitude forces on a hanging weight but pull with an angle of 72.0° between them. Wha

t pull does each rope exert if their resultant pull is 372 N directly upward?

Physics
1 answer:
Yanka [14]2 years ago
5 0

Answer:

T1 = 250.57[N] and T2 = 250.57[N]

Explanation:

First we have to extract all the information they give us about the problem, then make a free body diagram with all the forces that act on the body.

In the attached image we can see the free body diagram with the respective forces.

By summing up forces on the Y-axis, we can find the T1 and T2 force, these forces are equal since it is an initial condition of the problem. This equation can therefore be cleared.

The equations and analysis can be seen in the attached image

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Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0
2 years ago
A basketball rolls off a deck that is 3.2 m from the pavement. The basketball lands 0.75 m from the edge of the deck. How fast w
astra-53 [7]

Answer:

d). The value of y should be -32m

Vx=0.92 m/s

Explanation:

Using equation of motion uniform to y motion

x_{f}=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}\\x_{o}=0m\\v_{o}=0\\x_{f}=\frac{1}{2}*a*t^{2}\\x_{f}=-3.2m \\a=g=-9.8\frac{m}{s^{2}} \\

So to find t that is the same time for all the motion

t^{2}=\frac{2*x_{f}}{a} \\t=\sqrt{\frac{2*x_{f}}{a}}=\sqrt{\frac{2*-3.2m}{-9.8\frac{m}{s^{2}}}}\\t=0.808s

The value of Xf=-3.2m because the g is negative from the axis

Now in the axis 'x' to find Vx

Vx=\frac{0.75m}{0.808S}\\ Vx=0.92 \frac{m}{s}

5 0
2 years ago
For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s




Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


hence closest is in case C at 5.1 m/s




7 0
2 years ago
Read 2 more answers
Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance (in centim
Andrej [43]

Answer:Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance between two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is 340 m/s

Explanation:

5 0
2 years ago
In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P
tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites r_{sat1} = 48000 \times 10^{3} m

Distance of second satellites r _{sat2} = 64000 \times 10^{3} m

Distance of charon r_{c} = 19600 \times 10^{3} m

Time period of charon T_{c} = 6.39 days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

   T^{2} = r^{3}

   \frac{T}{r^{\frac{3}{2} } } = constant

For first satellites,

  \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} }  }

{T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat1} = 24.46 days

For second satellites,

   \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} }  }

{T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat2} = 37.67 days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0
2 years ago
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