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nexus9112 [7]
2 years ago
15

A fisherman casts his bait toward the river at an angle of 25° above the horizontal. As the line unravels, he notices that the b

ait and hook reach a maximum height of 2.9 m. What was the initial velocity he launched the bait with? Assume that the line exerts no appreciable drag force on the bait and hook and that air resistance is negligible. A fisherman casts his bait toward the river at an angle of 25° above the horizontal. As the line unravels, he notices that the bait and hook reach a maximum height of 2.9 m. What was the initial velocity he launched the bait with? Assume that the line exerts no appreciable drag force on the bait and hook and that air resistance is negligible. 6.3 m/s 18 m/s 7.9 m/s 7.6 m/s
Physics
1 answer:
Butoxors [25]2 years ago
6 0

Answer:

v_{o}=18m/s

Explanation:

According to the exercise we know the angle which the bait was released and its maximum height

\beta =25\\y=2.9m

To find the initial y-component of velocity we need to do the following steps:

v_{y}^{2} =v_{oy}^{2}+2g(y-y_{o})

At maximum height the y-component of velocity is 0 and from the exercise we know that the initial y position is 0

0=v_{oy}^{2}-2(9.8m/s^2)(2.9m)

v_{oy}=\sqrt{2(9.8m/s^{2} )(2.9m)} =7.53m/s

Since the bait is released at 25º

v_{oy}=v_{o}sin(25)

v_{o}=\frac{7.53m/s}{sin(25)}=18m/s

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Answer:

Speed of the wave is 7.87 m/s.

Explanation:

It is given that, tapping the surface of a pan of water generates 17.5 waves per second.

We know that the number of waves per second is called the frequency of a wave.

So, f = 17.5 Hz

Wavelength of each wave, \lambda=45\ cm=0.45\ m

Speed of the wave is given by :

v=f\lambda

v=17.5\times 0.45

v = 7.87 m/s

So, the speed of the wave is 7.87 m/s. Hence, this is the required solution.

5 0
2 years ago
An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini
arsen [322]
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

  m₁:  KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

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Due to the momentum balance,

   m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

   (0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

   V = 0.2977 m/s

The kinetic energy is,
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The difference between the kinetic energies is 0.0473 J. 
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A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to
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As the potential and kinetic energies are conserved

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The speed with which the liquid sulfur left the volcano is 529.15 m/s

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2 years ago
"The predictions of Einstein’s Theory of General Relativity were tested on a double pulsar system in January of 2004. His equati
Rasek [7]

Answer:

99.95%

Explanation:

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A team of astrophysicists led by Michael Kramer, conducted a study on how these gravitational waves will impact the time in which the radio waves emitted by pulsars will reach Earth. The result of the study proved the theory of General Relativity to be accurate up to 99.95%.

8 0
2 years ago
A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
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Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
2 years ago
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