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never [62]
3 years ago
11

Help plz I’ll mark brainliest

Physics
2 answers:
slavikrds [6]3 years ago
3 0
Thinner at edges and its thick in the middle
Nat2105 [25]3 years ago
3 0
A Piece Of Glass Which Is Thinner In The Middle Than On The Edges.
Hope This Helps You.
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A woman holds a book by placing it between her hands such that she presses at right angles to the front and back covers. The boo
In-s [12.5K]

Answer:

Explanation:

In order to solve this problem we need to make a free body diagram of the book and the forces that interact on it. In the picture below you can see the free body diagram with these forces.

The person holding the book is compressing it with his hands, thus exerting a couple of forces of equal magnitude and opposite direction with value F.

Now the key to solving this problem is to analyze the equilibrium condition (Newton's third law) on the x & y axes.

To find the weight of the book we simply multiply the mass of the book by gravity.

W = m*g

W = 1.3[kg] * 9.81[m/s^2]

W = 12.75 [N]

7 0
3 years ago
The gravity of Neptune is about 1.1 times the gravity of earth. How will the mass of an object on Neptune compare with its mass
Roman55 [17]
I think the gravity doesn't affect the mass of an object. Only it's weight can be compared
7 0
3 years ago
Read 2 more answers
An open organ pipe 30 cm long and a closed organ pipe 23 cm long, both of same diameter , are each sounding its first overtone ,
cupoosta [38]

First overtone of open organ pipe is given as

f_{1o} = \frac{v}{L_1 + 2e}

first overtone of closed organ pipe is given as

f_{1c} = \frac{3v}{4(L_2 + e)}

now they are in unison so we will have

\frac{v}{L_1 + 2e} = \frac{3v}{4(L_2 + e)}

\frac{1}{30 + 2e} = \frac{3}{4(23 + e)}

90 + 6e = 92 + 4e

e = 1 cm

so end correction of both pipes is e = 1 cm

8 0
2 years ago
Choose either eclipses or lunar phases and make a model of the event. Your model can take any form, providing it can explain how
babymother [125]

Answer: See the explanation below.

Explanation: For this assignment, I chose to display how eclipses are created.

My model was made utilizing a 3D displaying device program for all intents and purposes. The items utilized are three models I made for this presentation, Earth, the moon, and the sun. These three models will be utilized for the showcase.  

The light that shines from the sun would create a shadow on the moon. The moon would then catch the light that should've arrived on Earth, making the shadow we call an eclipse. Earth gets a shadow of the moon and the remainder of Earth is lit up from the rest of the light, making an eclipse.  

The individual I demonstrated my project to was [<em>Someone you know</em>], [<em>Pronoun</em>] said it precisely took after the occasion of an eclipse. The light from the sun being shined on to the moon rather than the Earth, creating the shadow we call an eclipse.

5 0
3 years ago
Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet
Svetllana [295]
Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s
6 0
3 years ago
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