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oksano4ka [1.4K]
2 years ago
13

When you think about and describe the characteristics of a giraffe, what type of model are you using?

Physics
1 answer:
Tpy6a [65]2 years ago
4 0

Answer:

The giraffe is the tallest of all mammals. It reaches an overall height of 18 ft (5.5 m) or more. The legs and neck are extremely long. The giraffe has a short body, a tufted tail, a short mane, and short skin-covered horns

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The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be
kap26 [50]

Answer:

v = 14.86 m/s

Explanation:

As we know that the force equation at the top is given as

\frac{mv^2}{R} = ma

now we know that

a_c = 1.5 g

so we have

\frac{v^2}{R} = 1.5 g

v = \sqrt{1.5 Rg}

so we will have

v = \sqrt{1.5(15)(9.81)}

v = 14.86 m/s

3 0
3 years ago
What is The substance that dissolves the solute.
Ludmilka [50]

Answer:

Explanation:

Do you mean the solvent? If this is off the mark, let me know in a comment.

The solvent is something that the solute is (usually) soluble in.

3 0
3 years ago
Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.
zimovet [89]
The question is asking to calculate the object's speed v1, v2, v3 at the certain time is the given of the problem, in my calculation, I would say that the speed would be  2m/s, 1.5m/s, 0.22m/s. I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarification
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Students who embrace an innate mindset might view their poor performance in math as inevitable or beyond their control
PilotLPTM [1.2K]

Answer:

True

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4 0
3 years ago
Read 2 more answers
Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp
slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

Q=Av

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

Q=1.20 m^3/s

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2

So we can re-arrange the equation to find the speed of the water:

v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

Q_1 = Q_2\\A_1 v_1 = A_2 v_2

where we have

A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s

and where A_2 is the cross-sectional area of the pipe at the second point.

Solving for A2,

A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2

And finally we can find the radius of the pipe at that point:

A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

6 0
3 years ago
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