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nikklg [1K]
3 years ago
15

Suppose the Moon had twice the diameter but the same mass and same orbital distance from Earth. In that case, the high tides on

Earth would be :
Physics
1 answer:
Inessa [10]3 years ago
5 0

If the moon had twice the diameter with same mass and orbital distance from earth, then the high tides on Earth would be practically the same.

<u>Explanation:</u>

As the tides occurring in the earth is mostly due to the gravitational force of the moon acting on the earth. The distance between the earth and the moon mainly influences high tides. So, tidel force can be termed as directly proportionate to the gravitational forces acting between moon and earth.

As the gravitational forces act on Earth due to the moon will be directly proportionate to the product of masses of Earth and Moon and inversely proportionate to the distance squared (separation between Earth and Moon), due to the universal law of gravitation.

Thus, it can be stated that the diameter of the moon has no role to play in the high tides. As it is stated that the mass and distance almost remained the same, and there is only a change in the diameter of the moon, the high tides will not be exhibiting much change. Thus, on earth the high tides would be practically the same on increasing the diameter of the Moon.

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How many significant digits should the answer to the following problem have? (2.49303 g) * (2.59 g) / (7.492 g) =
Feliz [49]

The number of significant digits to the answer of the following problem is four.

<h3>What are the significant digits?</h3>

The number of digits rounded to the approximate integer values are called the significant digits.

The following problem is

(2.49303 g) * (2.59 g) / (7.492 g) =

On solving we get

= 0.86184566204

The answer is approximated to  0.86185

Thus, the significant digits must be four.

Learn more about significant digits.

brainly.com/question/1658998

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6 0
2 years ago
The Doppler effect using ultrasonic waves of frequency 2.25 × 106 Hz is used to monitor the heartbeat of a fetus. A (maximum) be
miss Akunina [59]

Answer:

v = 2.88 \times 10^7 m/s

Explanation:

As per Doppler's effect we know that the frequency of the sound that is observed by the detector is the reflected sound

This reflected sound is given as

f' = f (\frac{v - v_h}{v + v_h})

f' = 2.25\times 10^6(\frac{v - 1540}{v + 1540})

so we know that the beat frequency is

\Delta f = 240 Hz

so we will have

f - f' = \Delta f

2.25 \times 10^6 - 2.25\times 10^6(\frac{v - 1540}{v + 1540}) = 240

1 - (\frac{v - 1540}{v + 1540}) = 1.07 \times 10^{-4}

so we have

0.99989 = (\frac{v - 1540}{v + 1540})

1.99989\times 1540 = 1.067 \times 10^{-4} v

v = 2.88 \times 10^7 m/s

4 0
4 years ago
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IrinaK [193]

Answer:

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Explanation:

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5 0
3 years ago
In a historical movie, two knight on horseback start from
mixer [17]

Answer:

The knights collide 53.0 m from the starting point of sir George.

Explanation:

The equation for the position in a straight accelerated movement is as follows:

x = x0 + v0 t + 1/2 a t²

where

x = position at time t

x0 = initial position

v0 = initial speed

a = acceleration

t = time

The position of the two knights is the same when they collide. Since they start from rest, v0 = 0:

Sir George´s position:

xGeorge = 0 m + 0 m + 1/2 * 0.300 m/s² * t²

Considering the center of the reference system as Sir George´s initial position, the initial position of sir Alfred will be 88.0 m. The acceleration of sir Afred will be negative because he rides in opposite direction to sir George:

xAlfred = 88.0 m + 0 m - 1/2 * 0.200  m/s² * t²

When the knights collide:

xGeorge = x Alfred

1/2 * 0.300 m/s² * t² = 88.0 m - 1/2 * 0.200  m/s² * t²

0.150 m/s² * t² = 88.0 m - 0.100  m/s² * t²

0.150  m/s² * t² + 0.100  m/s² * t² = 88.0 m

0.250 m/s² * t² = 88.0 m

t² = 88.0 m / 0.250 m/s²

t = 18.8 s

At t = 18.8 s the position of sir George will be

x =  1/2 * 0.300 m/s² * (18.8 s)² = <u>53.0 m </u>

5 0
3 years ago
A magnifying glass has a converging lens of focal length of 13.8 cm. At what distance from a nickel should you hold this lens to
Colt1911 [192]

Answer:

19.6 cm.

Explanation:

From the question given above, the following data were obtained:

Focal length (f) = 13.8 cm

Magnification (M) = +2.37

Object distance (u) =.?

Next, we shall determine the image distance. This can be obtained as follow:

Magnification (M) = +2.37

Object distance (u) = u

Image distance (v) =?

M = v / u

2.37 = v / u

Cross multiply

v = 2.37 × u

v = 2.37u

Finally, we shall determine the object distance. This can be obtained as follow:

Focal length (f) = 13.8 cm

Image distance (v) = 2.37u

Object distance (u) =.?

1/v + 1/u = 1/f

vu / v + u = f

2.37u × u / 2.37u + u = 13.8

2.37u² / 3.37u = 13.8

Cross multiply

2.37u² = 3.37u × 13.8

2.37u² = 46.506u

Divide both side by u

2.37u² / u = 46.506u / u

2.37u = 46.506

Divide both side by 2.37

u = 46.506 / 2.37

u = 19.6 cm

Thus, the lens should be held at a distance of 19.6 cm.

7 0
4 years ago
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