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sukhopar [10]
3 years ago
5

What is the net force experienced by the rope? Include both the magnitude and direction

Physics
1 answer:
Lapatulllka [165]3 years ago
3 0

Answer:

All the physical world objects that comers in the contact to exert the force to each other. The contact forces are different from their names and what type of force they exert.

Explanation:

The cables and the ropes are the useful objects that exert the forces that can efficiently transfer the force from a significant distance.

It is noted that tension is a type of force that the rope can not simply push it away effectively. When push happened with rope, the rope goes to slack and lose all the tension that pulls at the first place. Tension only pull objects.

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What kind of symmetry do you have
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During the daytime, I have mostly line symmetry.

During the night, I often have almost spherical symmetry.
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3 years ago
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?/1 Jorge traveled 5 miles north to school. He then traveled 3 miles west to the store. Then he left the store and traveled 5 mi
aleksklad [387]

Answer:

He's 3 miles west of school.

Explanation:

He went 5 miles up and 5 miles down which means that he really didn't go up or down.  In between that, he went 3 miles west so if the 5 milers don't count, this puts him at 3 miles west of school.

4 0
2 years ago
Ask Your Teacher The height h in feet reached by a dolphin t seconds after breaking the surface of the water is given by h = −16
kirza4 [7]

Answer:

1 second

Explanation:

h = −16t² + 32t

When, h = 16

16 = −16t² + 32t

Divide each of the numbers by 16

1 = -1t² + 2t

Rearrange the equation

1t²-2t+1 = 0

Solving by the quadratic formula, we get

t=\frac{-(-2)\pm \sqrt{(-2)^2-4\times 1\times 1}}{2\times 1}\\\Rightarrow t=\frac{2\pm 0}{2}\\\Rightarrow t=1\ s

So, time taken by the dolphin to jump out of the water and touch the trainer's hand is 1 second.

3 0
3 years ago
The water was used to power a generator, creating _______________ energy
Nikitich [7]

Answer:

electric

<em>please give brainliest</em>

3 0
2 years ago
A 295-kg object and a 595-kg object are separated by 4.10 m.
kodGreya [7K]

Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

r=d/2 = 2.05 m

The force between the mass m₁ and m₃

F_{13}=\dfrac{Gm_1m_3}{r^2}

by putting the values

F_{13}=\dfrac{Gm_1m_3}{r^2}

F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}

F₁₃=2.94 x 10⁻⁷ N

The force  between the mass m₂ and m₃

by putting the values

F_{23}=\dfrac{Gm_2m_3}{r^2}

F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}

F₂₃=5.94 x 10⁻⁷ N

The net force F

F= F₂₃- F₁₃

F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

b)

Lest take at distance x from mass m₂ net force is zero.

F_{23}=\dfrac{Gm_2m_3}{x^2}

F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}

Form above two equation

\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}

\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}

\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

4 0
2 years ago
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