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Dvinal [7]
4 years ago
6

Two forces act on a 41-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other ha

s a magnitude 35 N at 32° clockwise from the positive y-axis. What is the magnitude of this object's acceleration?

Physics
1 answer:
Genrish500 [490]4 years ago
6 0

Answer:

a = 1.41 m/s²

Explanation:

Given that

mass ,m= 41 kg

F₁ = 65 N , θ = 59°

F₂ = 35 N ,θ = 32°

The component of Force F₁

F₁x= F₁cos59° i

F₁x= 65 x cos59° i = 33.47 i

F₁y= - F₁ sin 59° j

F₁y= - 65 x sin 59° j = - 55.71 j

The component of Force F₂

F₂x= F₂ sin 32° i

F₂x= 35 x sin 32° i = 18.54 i

F₂y=  F₂ cos 32° j

F₂y=  35 x cos 32° j =  29.68 j

The total force F

F= 33.47 i +   18.54 i - 55.71 j +   29.68 j

F= 52.01 i - 26.03 j

The magnitude of the force F

F=\sqrt{52.01^2+26.03 ^2}\ N

F=58.16 N

We know that

F= m a

a= Acceleration

m=mass

58.16 = 41 x a

a = 1.41 m/s²

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An angle of 60 degrees with the negative y-axis could mean 60 degrees clockwise or counterclockwise, which translates to two possible angles (starting from the positive x-axis and moving counterclockwise) of 210 degrees or 330 degrees.

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You put a 3 kg block in the box, so the total mass is now 9 kg, and you launch this heavier box with an initial speed of 5 m/s.
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Answer:

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Explanation:

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Answer:

(a)  Radius of orbit will be =7.32\times10^6m

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We have given

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(c)

Altitude h=radius\ of\ orbit-radius\ of\ earth=7.32\times 10^6-6.4\times 10^6=0.92\times 10^6m

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