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3 years ago

You pull a 200 kg box straight up with a rope connected to a helicopter. If the

Physics

2 answers:

Tanzania [10]3 years ago

5 0

Answer:

Acceleration of the box = 0.2 m / s²

Explanation:

Given:

Mass of box = 200 kg

Magnitude of the tension = 2,000 N

Find:

Acceleration of the box

Computation:

Ma = T - Mg

200(a) = 2,000 - (200)(9.8)

200(a) = 2,000 - 1960

200(a) = 40

a = 40 / 200

a = 0.2

Acceleration of the box = 0.2 m / s²

s344n2d4d5 [400]3 years ago

4 0

Answer:

The acceleration is 0.2 m/s^2.

Explanation:

Tension, T = 2000 N

mass, m = 200 kg

Let the acceleration is a.

So, by the use of Newton's second law

Net force = mass x acceleration

T - mg = ma

2000 - 200 x 9.8 = 200 x a

2000 - 1960 = 200 a

a = 0.2 m/s^2

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In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he

lana [24]

Answer:8.28 km

Explanation:

Given

First it drifts $45^{\circ}$ 2.5 km

$r_1=2.5cos45 i+2.5sin45 j$

Secondly it drifts $60^{\circ}$ 4.70 km

$r_{12}=4.7cos60 i-4.7sin60 j$

After that it drifted along east direction 5.1 km

$r_{23}=5.1 i$

After that it drifts $55^{\circ}$ 7.2 km

$r_{34}=-7.2cos55 i-7.2sin55 j$

After that it drifts $5^{\circ}$ 2.8 km

$r_{54}=-2.8cos5 i+2.8sin5 j$

$r_{5O}$ = $\left [ 2.5cos45+4.7cos60+5.1-7.2cos55-2.8cos5\right ]\hat{i}$ + $\left [ 2.5sin45-4.7sin60-7.2sin55+2.8sin5\right ]\hat{j}$

$r_{5O}=2.299\hat{i}-7.95\hat{j}$

$|r_{5O}|=8.28 km$

for direction

$tan\theta =\frac{7.95}{2.299}=3.4580$

$\theta =73.87^{\circ}$ south of east

7 0

3 years ago

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Two cello strings, with the same tension and length, are played simultaneously. Their fundamental frequencies produce audible be

qwelly [4]

Explanation:

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Lower pitch frequency, $f_2=220\ Hz$

Fundamental frequency is, $f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu_1}}$ .....(1)

Lower frequency is, $f_2=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu_2}}$ ..............(2)

Dividing equation (1) and (2) as :

$\dfrac{f_1}{f_2}=\sqrt{\dfrac{\mu_2}{\mu_1}}$

$\dfrac{\mu_2}{\mu_1}=(\dfrac{f_1}{f_2})^2$

$\dfrac{\mu_2}{\mu_1}=(\dfrac{8}{220})^2$

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So, the ratio of linear mass density μ of the string with the higher pitch to that of the string with the lower pitch is 0.00132. Hence, this is the required solution.

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djyliett [7]

Answer:

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Pani-rosa [81]

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