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AlladinOne [14]
3 years ago
5

Find the y-component of this vector:

Physics
1 answer:
NARA [144]3 years ago
6 0

Answer:

Dy = 49.01 [m]

Explanation:

The vertical component of the vector can be determined with the sine of the angle.

Dy = 92.5*sin(32)

Dy = 49.01 [m]

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A trampoline has a spring constant of 3430 N/m how far will the trampoline sink when a 70 kg person steps on it
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A box is sliding down an incline tilted at a 12° angle above horizontal. The box is initially sliding down the incline at a spee
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Answer:

The box will cover a distance of 0.9199m before coming to rest

Explanation:

We are given;

Angle of tilt; θ = 12°

Speed of sliding down; u = 1.5 m/s

Coefficient of kinetic friction; μ = 0.34

We are told that the box is sliding down an incline tilted at a 12° angle above horizontal.

Thus,

The components of the weight of the block would be;

Fx = mg sinθ = mg sin 12

Fy = mg cosθ = mg cos 12

For, the normal force on the block, it will be counter balanced by the Y component of weight of block and so we have;

Normal force; Fn = mg cos 12

Now formula for the frictional force would be given by;

Ff = μmg cos 12

So, Ff = 0.34mg cos 12

So, the net force along the inclined plane is;

Fnet = Fx - Ff

Fnet = mg sin 12 - 0.34mg cos 12

Where Fnet = mass x acceleration.

Thus;

ma = mg sin 12 - 0.34mg cos 12

m will cancel out to give;

a = g sin 12 - 0.34g cos 12

a = 9.81(0.2079) - 0.34(9.81 × 0.9781)

a = -1.223 m/s²

According to Newton's equation of motion, we have;

(v² - u²) = 2as

s = (v² - u²)/2a

Final velocity is zero. Thus;

s = (0² - 1.5²)/(2 × -1.223)

s = -2.25/-2.446

s = 0.9199 m

Thus, the box will cover 0.9199m before coming to rest

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