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nikklg [1K]
3 years ago
5

In free fall the object with less air resistance falls with a greater acceleration

Physics
1 answer:
ANEK [815]3 years ago
5 0

terminal velocity ... greater speed ... acc is 10m/s/s

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What is a discussion?
kap26 [50]

Answer:

um, when you talk with other people about stuff. (I'm not trying to sound like a smarta s s I'm just giving a definition...)

Explanation:

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3 years ago
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Which of the following statements about ycarrier(x,t) is correct?
finlep [7]

Answer: Option D : is traveling rapidly but oscillating slowly.

Explanation:

ycarrier(x,t) is traveling rapidly but oscillating slowly.

8 0
3 years ago
O((>ω< ))oo((>ω< ))oo((>ω< ))oo((>ω< ))oo((>ω< ))o
Paladinen [302]

Answer:

(•_•)

Explanation:

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3 years ago
When you hit a .27 kg volleyball the contact time is 50 ms and the average force is 125 N. If you serve the volleyball (from res
yulyashka [42]

(A) P(v) = 0.135v

(B) P(h) = 0.234v

<u>Explanation:</u>

Given-

Mass of the ball, m = 0.27kg

Force, F = 125N

angle of projection, θ = 30°

Let v be the velocity of the ball.

A) vertical component of the momentum of the volleyball

We know,

P(vertical) = mvsinθ

P(V) = 0.27 X v X sin 30°

P(V) = 0.27 X v X 0.5

P(V) = 0.135v

B) horizontal component of the momentum of the volleyball

We know,

P(Horizontal) = mvcosθ

P(h) = 0.27 X v X cos 30°

P(h) = 0.27 X v X 0.866

P(h) = 0.234v

8 0
3 years ago
16. For this table of data, how should the y-axis be labeled (with units)?
vampirchik [111]

Answer:

The y-axis should be labelled as W in Newtons (kg·m/s²)

Explanation:

The given data is presented here as follows;

Mass (kg)            {}        Newtons (kg·m/s²)

3.2                      {}           31.381

4.6             {}                    45.1111

6.1              {}                    59.821

7.4              {}                    72.569

9                {}                     89.241

10.4              {}                   101.989

10.9              {}                  106.892

From the table, it can be seen that there is a nearly linear relationship between the  amount of Newtons and the  mass, as the slope of the data has a relatively constant slope

Therefore, the data can be said to be a function of Weight in Newtons to the mass in kilograms such that the weight depends on the mass as follows;

W(m) in Newtons = Mass, m in kg × g

Where;

g is the constant of proportionality

Therefore, the y-axis component which is the dependent variable is the function, W(m) = Weight of the body while the x-axis component which is the independent variable is the mass. m

The graph of the data is created with Microsoft Excel give the slope which is the constant of proportionality, g = 9.8379, which is the acceleration due to gravity g ≈ 9.8 m/s²

We therefore label the y-axis as W in Newtons (kg·m/s²)

6 0
3 years ago
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