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podryga [215]
3 years ago
15

Please help! I’ll give Brainliest

Physics
2 answers:
babymother [125]3 years ago
5 0

Answer:

9.8 secs

Explanation:

the ball is in the air so it takes 9.8 secs to get to the ground

UNO [17]3 years ago
5 0
9.8 is correct because that’s how long it takes the ball to hit the ground
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The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.
klemol [59]
The resistance is 27.5 ohms
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3 years ago
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A nearsighted eye has a far point of 100 cm. Objects further than 100 cm are not seen clearly. A diverging lens is used to permi
Leya [2.2K]

A diverging lens is used to permit clear vision of an object placed at infinity. The focal length of the lens is -100 cm.

<h3>What is focal length?</h3>

The focal length is half of the radius of curvature of the focal lens.

By the lens maker formula,

1/f = 1/v +1/u

where, v is the image distance and u is the object distance.

Give, the object is at infinity and the image must form at 100 cm, the the focal length will be

1/f = 1/ -100 + 1/∞

f = -100 cm

The focal length must be -100 cm for the diverging lens.

Learn more about focal length.

brainly.com/question/16188698

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6 0
2 years ago
consider a solid sphere and a solid disk wiht the same radius and the same mass. explain why the solid disk has a greater moment
andriy [413]

Answer:

Moment of inertia of the solid sphere:

I

s

=

2

5

M

R

2

.

.

.

.

.

.

.

.

.

.

.

(

1

)

Is=25MR2...........(1)

Here, the mass of the sphere is

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4 0
3 years ago
A certain AM radio wave has a frequency of 1.12 x 100 Hz. Given that radio waves travel at
riadik2000 [5.3K]

Answer: 267 m

Explanation:

2.99x10^8 m/s

———————-

1.12 x 10^6 Hz

3 0
3 years ago
Two loudspeakers emit identical sound waves along the x-axis. The sound at a point on the axis has maximum intensity when the sp
EleoNora [17]

The concept required to solve this problem is related to the wavelength.

The wavelength can be defined as the distance between two positive crests of a wave.

The waves are in phase, then the first distance is

\Delta x_1 = 20cm

And out of the phase when

\Delta x_2 = 30cm

Thus the wavelength is

\Delta x_2-\Delta x_1 = \frac{\lambda}{2}

Here,

\lambda =  Wavelength

If we rearrange the equation to find it, we will have

\lambda = 2 (\Delta x_2-\Delta x_1 )

\lambda= 2(30-20)

\lambda = 20cm

Therefore the wavelength of the sound is 20cm.

5 0
3 years ago
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