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Blababa [14]
3 years ago
10

Help pls, see picture. Will mark Brainliest

Physics
2 answers:
lilavasa [31]3 years ago
6 0

Answer: The ball is thrown straight UP so when it asked whats the ball's velocity. It'll be graph ii. But.. to me, it seems that graph ii is the only one that provides a true reaction since the ball is thrown straight up. Or it could graph iii assuming the ball is thrown by a regular person.  So when it asked for the position, I would pick iii.

Explanation: hope this made sense //Give thanks(and or Brainliest) if helpful (≧▽≦)//

Pani-rosa [81]3 years ago
5 0

Answer:

a ) option 2 is correct

b) -ve acceleration for upward motion ,0 acceleration at top point ,+ve acceleration on downward motion ...

Explanation:

mark me as brainliest ❤️

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Light travelling in one material enters another material in which it travels faster. The light wave will:
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The light will change its direction and bend away from the normal.
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Name two ways the eye adapts to dim light
Anarel [89]
There are many ways that eye adapts to dim light but i will name only 2

1) Efficiency 
2) Ambient light response 



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4 years ago
Which layer of the atmosphere has the lowest density?
stich3 [128]
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The exosphere is the lowest<span> in density of the </span>layers of the atmosphere.

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6 0
3 years ago
If the scaled-up man now stands on one leg, what fraction of the tensile strength is the stress on the femur?.
Firdavs [7]

The new cross sectional area and tensile stress is mathematically given as

A=4.8*10^{-4}m^2

X=14:100

<h3>Cross sectional area and stress on the femur</h3>

Question Parameters:

To see why this must be so, recall, that the stress on the femur for a man standing on one leg is 1.4%

we scale this man up by a factor of 10 in all dimensions

that a 70 kg person has a femur with a cross-section area

a)

Generally the New cross sectional area  is mathematically given as

100A=100*4.8*10^{-4}

A=4.8*10^{-4}m^2

b)

From the initial statement we see that  the fraction of the tensile strength is the stress on the femur  and will be

X=\frac{14}{100}

X=14$

X=14:100

For more information on fraction

brainly.com/question/1301963

Complete Question

Larger animals have sturdier bones than smaller animals. A mouse's skeleton is only a few percent of its body weight, compared to 16% for an elephant. To see why this must be so, recall, that the stress on the femur for a man standing on one leg is 1.4% of the bone's tensile strength. Suppose we scale this man up by a factor of 10 in all dimensions, keeping the same body proportions. (Assume that a 70 kg person has a femur with a cross-section area (of the cortical bone) of 4.8 x 10-4m², a typical value.)

Part A

Both the inside and outside diameter of the femur, the region of cortical bone, will increase by a factor of 10. What will be the new cross-section area? Express your answer using two significant figures.

Part B

If the scaled-up man now stands on one leg, what fraction of the tensile strength is the stress on the femur?

8 0
3 years ago
In figure 1, charge q2 experiences no net electric force. What is q1?
lukranit [14]

By using Coulomb's law, we want to find the value of q₁ given that q₂ experiences no net electric force. We will find that q₁ = 8nC

<h3>Working with Coulomb's law.</h3>

Coulomb's law says that for two charges q₁ and q₂ separated by a distance r, the force that each one experiences is:

F = k\frac{q_1*q_2}{r^2}

Where k is a constant

Here we can see that q₂ interacts with two charges, then the total force on q₂ will be:

F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2}

And we know that it must be equal to zero, so we can write it as:

F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2} = 0\\\\k*q_2*(\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2}) = 0\\

The parenthesis must be equal to zero, so we can write:

\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2} = 0

And now we can solve this for q₁ to get:

q_1  = 2nC*(\frac{(20cm)^2}{(10cm)^2} ) = 8nC

If you want to learn more about Coulomb's law, you can read:

brainly.com/question/24743340

3 0
3 years ago
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