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2 years ago

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

Physics

2 answers:

Tanzania [10]2 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]2 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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Answer:

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1 year ago

Read 2 more answers

The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p

beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

$v(t)=ate^{-6t}$ (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

$\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]$ (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

$a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}$

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

$v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a$

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Question 1 (1 point)

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