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3 years ago

14

A balloon is ascending at a rate of +4.00 m/s to a height of 11.0 m above the ground when a package is dropped. In the absence o

f air resistance, the velocity of the ball when it hits the ground is

1 answer:

o-na [289]3 years ago

6 0

Answer:

Vf = 14.7 m/s

Explanation:

Vf² = Vi² + 2 * a * Δy

given:

a = 9.81 m/s²

Δy = 11m

Vi = 0 when upon release

Vf² = 0 + 2 (9.81) 11

Vf = 14.7 m/s

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ASAP<br> describe how energy is transferred in a mechanical wave

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3 years ago

If two objects of the same size move through the air at different speeds, which encounters the greater air resistance?a. The fas

blagie [28]

Answer:

Option A is correct.

(The faster object encounters more resistance)

Explanation:

Option A is correct. (The faster object encounters more resistance)

Air resistance depends on various factors:

Speed of the object
Cross-sectional area of the object
Shape of the object

Formula:

$F=\frac{1}{2}C_d\rho A v^{2}$

As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.

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3 years ago

A 12kg cheetah accelerates 24 m/s". What is the force the cheetah needed to run?

Kobotan [32]

Answer:

288N

Explanation:

Given parameters:

Mass of Cheetah = 12kg

Acceleration = 24m/s²

Unknown:

Force needed by the cheetah to run = ?

Solution:

The force needed by the Cheetah to run is the net force.

According to Newton's law;

Force = mass x acceleration

Insert the given parameters and solve;

Force = 12 x 24 = 288N

7 0

3 years ago

Find the first three harmonics of a string of linear mass density 2.00 g/m and length 0.600 m when the tension in it is 50.0 n.

blsea [12.9K]

As we know that Nth harmonic of the string is given by

$f = \frac{N}{2L}\sqrt{\frac{T}{m/L}}$

now here we will have

$m/L = mass density = 2 g/m$

$m/L = 0.002 kg/m$

$Length = L = 0.600 m$

$Tension = T = 50.0 N$

now from above formula we have

$f = \frac{N}{2(0.600)}\sqrt{\frac{50.0}{0.002}}$

$f = 131.8N$

now for first harmonic N = 1

$f_1 = 131.8 Hz$

for second harmonic N = 2

$f_2 = 263.5 Hz$

for third harmonic N = 3

$f_3 = 395.3 Hz$

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4 years ago

In 1977 off the coast of Australia, the fastest speed by a vessel on the water

fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$a_{ave}=1.80 m/s^{2}$

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0$ the initial velocity and $V_{f}$ the final velocity

$\Delta t=85.6 s$

Then:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}$

Since $V_{o}=0$ :

$a_{ave}=\frac{V_{f}}{\Delta t}}$

Finding $V_{f}$ :

$V_{f}=a_{ave} \Delta t$

$V_{f}=(1.80 m/s^{2})(85.6 s)$

Finally:

$V_{f}=154.08 m/s$

8 0

3 years ago