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faltersainse [42]
2 years ago
13

A horizontal uniform bar of mass 2.5 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t

he end of the bar, and string 2 is attached a distance 0.6 m from the other end. A monkey of mass 1.25 kg walks from one end of the bar to the other.
Required:
Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.
Physics
1 answer:
11Alexandr11 [23.1K]2 years ago
6 0

Answer: I am not sure if you wanted me to answer this or not.

Explanation:

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an electron is released from rest in a region of space with a nonzero electric field.1. As the electron moves, does the electric
zvonat [6]

Answer:

1. a. increase

2. Because the electron has a negative charge its electric potential energy does not decrease as one might expect, but increases instead.

Explanation:

Lets first consider the relation between the electric field and electric potential.

E = -ΔV/Δs

As this equation indicates that the electric field is due to the change in potential and change in the the position of charge. Electric field is directed towards the decreasing potential and the electron moves in the opposite direction of the electric field  where potential increases. Thats why the best explanation is that the electron has a negative charge it moves towards the positive region where the electric potential energy increases.

5 0
3 years ago
What are the first three harmonics of a note produced on a 0.31 m long violin string if the waves on this string have a speed of
bezimeni [28]
Is that what you are looking for?
4 0
2 years ago
A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is
Valentin [98]

Answer:

<em>The end of the ramp is 38.416 m high</em>

Explanation:

<u>Horizontal Motion </u>

When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

\displaystyle d=v\cdot\sqrt{\frac  {2h}{g}}

If the maximum horizontal distance is known, we can solve the above equation for h:

\displaystyle h=\frac  {d^2g}{2v^2}

The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:

\displaystyle h=\frac  {70^2\cdot 9.8}{2\cdot 25^2}

\displaystyle h=\frac  {4900\cdot 9.8}{2\cdot 625}

h= 38.416 m

The end of the ramp is 38.416 m high

8 0
2 years ago
Which statement about speed and/or velocity is true?
maw [93]

Answer: The right Answer is Velocity has both speed and direction.

Explanation:

i took the test

8 0
2 years ago
Read 2 more answers
1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.
NikAS [45]

Answer:

1.

109.6 cm ,  - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{40} = \frac{1}{63} + \frac{1}{d}

d = 109.6 cm

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{-109.6}{63}

m = - 1.74

The image is real

2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}

a = 10

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{- (- (a +5))}{a}

m = \frac{(5 + 10)}{10}

m = 1.5

6 0
2 years ago
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