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3 years ago

9

In a bizarre but harmless Superman fell from the top of the Eiffel How fast was Superman traveling when he hit the ground 7.80 s

econds after falling ?

1 answer:

brilliants [131]3 years ago

5 0

Vf = final velocity
Vi = initial velocity
g = acceleration due to gravity
t = time

Vf = Vi + gt

Vf = 0 + (9.8)(7.80)
Vf = 76.44 m/s

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If the object represented by the FBD below has a mass of 2.5 kg, what is the acceleration of the object?

Debora [2.8K]

Answer:

4 m/s² down

Explanation:

We'll begin by calculating the net force acting on the object.

The net force acting on the object from the left and right side is zero because the same force is applied on both sides.

Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:

Force up (Fᵤ) = 15 N

Force down (Fₔ) = 25 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fᵤ

Fₙ = 25 – 15

Fₙ = 10 N down

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass (ml= 2.5 Kg

Net force (Fₙ) = 10 N down

Acceleration (a) =?

Fₙ = ma

10 = 2.5 × a

Divide both side by 2.5

a = 10 / 2.5

a = 4 m/s² down

Therefore, the acceleration of the object is 4 m/s² down

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3 years ago

A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?

anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}$

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

$\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}$

now, replace

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}$

hence, the answer is 0.495 m/s

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1 year ago

Which of the following would most likely be an example of a mass-transit disaster?

Aleonysh [2.5K]

Answer:

i think it is c if not im sorry if im wrong

Explanation:

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3 years ago

High altitude high velocity rivers of air are called

forsale [732]

The answer to the question would be jet streams.

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3 years ago

Read 2 more answers

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

Natalija [7]

Answer:

a) F = 680 N, b) W = 215 .4 J , c) F = 1278.4 N

Explanation:

a) Hooke's law is

F = k x

To find the displacement (x) let's use the elastic energy equation

$K_{e}$ = ½ k x²

k = 2 $K_{e}$ / x²

k = 2 85.0 / 0.250²

k = 2720 N / m

We replace and look for elastic force

F = 2720 0.250

F = 680 N

b) The definition of work is

W = ΔEm

W = $K_{ef}$ - $K_{eo}$

W = ½ k ( $x_{f}$ ² - x₀²)

The final distance

$x_{f}$ = 0.250 +0.220

$x_{f}$ = 0.4750 m

We calculate the work

W = ½ 2720 (0.47² - 0.25²)

W = 215 .4 J

We calculate the strength

F = k $x_{f}$

F = 2720 0.470

F = 1278.4 N

6 0

3 years ago