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labwork [276]
2 years ago
10

A 500 N weight is hung at the middle of a rope attached to two buildings at the same level. If the breaks in the tension exceed

1800 N, what is the minimum angle the rope can make with the horizontal? Group of answer choices 4o 8o 11o 18o
Physics
1 answer:
dedylja [7]2 years ago
6 0

Answer:

The correct choice is 8o.

Explanation:

The given weight is hanging at the mid point of the rope and the buildings are at the same level, Obtain the given equation by equating the vertical components of force,

2Tsin(φ)=W

where φ is the angle made by rope with the horizontal.

Given T=1800 N and W=500 N, the value of φ is calculated as follows,

2*1800*sin(φ)=500

sin(φ)=0.1388

φ=7.90

φ≈8o

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IrinaVladis [17]
Convergent boundaries move towards each other. They can be classified as continental-continental, or oceanic-continental.

Oceanic-Continental: Heavier oceanic plate slides under a continental plate in a process called subduction.

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4 0
3 years ago
A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequen
MrMuchimi

The solution for this problem:

Given:

f1 = 0.89 Hz

f2 = 0.63 Hz

Δm = m2 - m1 = 0.603 kg 


The frequency of mass-spring oscillation is: 
f = (1/2π)√(k/m) 
k = m(2πf)² 

Then we know that k is constant for both trials, we have: 
k = k 


m1(2πf1)² = m2(2πf2)² 

m1 = m2(f2/f1)² 


m1 = (m1+Δm)(f2/f1)² 


m1 = Δm/((f1/f2)²-1)

 m 1 = 0.603/ (0.89/0.63)^2 – 1

= 0.609 kg or 0.61kg or 610 g

5 0
3 years ago
Nine steps for developing scientific principles are
xxTIMURxx [149]

Answer:

1)Observe a phenomenon

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3)Form a hypothesis

4)Create an experiment

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7 0
2 years ago
We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligi
antoniya [11.8K]

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

Kc=510=1/2*Ic*\omega c^2

\omega c=\sqrt{510*2/Ic}

Where the inertia is given by:

Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2

Replacing this value:

\omega c=106.46rad/s

Speed of the block will therefore be:

V_b=\omega_c*R_c=106.46*0.15=15.969m/s

By conservation of energy:

Eo = Ef

Eo = 0

Ef = 510+1/2*m_b*V_b^2-m_b*g*h

So,

0 = 510+1/2*m_b*V_b^2-m_b*g*h

Solving for h we get:

h=16.67m

3 0
3 years ago
Which of these is NOT a possible type of energy transformation?
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to be franc i really think the answer is B

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3 years ago
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