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ki77a [65]
3 years ago
7

A ball is thrown with an angle of 25.0 to the horizon with a speed of 18.0 m/s. What are its horizontal and vertical components?

Physics
1 answer:
blondinia [14]3 years ago
6 0
      Best Answer:  <span> vi=24 m/s
θ=43°
x=16m

0) Use this equation because you have x component and not y.
You need time or part c so you solve it like this
x=(vi*cosθ)t
16=(24cos43°)t
t=.911552s=.912s

a) This is a trajectory and you need y so you use this equation.
y=(tanθ)x-(gx^2/(2(vi*cosθ)^2)
y=tan43°*16m-((-9.8*16m)^2)/(2(24*cos43°...
y=14.920m-39.901m
y=-24.9808m= -25.0m

b) Break it down into it's components.
vx=cosθ*vi
vx=cos43°*24
vx=17.5525m/s = 17.5m/s

c) vy= sin 43°*24-.5(-9.8)(.912s)
vy=22.2m/s

I'm not 100% positive on some of these but it should look something like the work above. </span>
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The tallest Ferris wheel in the world is located in Singapore. Standing 42 stories high and holding as many as 780 passengers, t
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Answer:

The speed of the riders on the Singapore Flyer is approximately 0.262 m/s

Explanation:

The dimensions of the tallest Ferris wheel in the world are;

The diameter of the Ferris wheel, D = 150 m

The tine it takes the Ferris wheel to make a full circle, T = 30 minutes = 30 min × 60 s/min = 1,800 seconds

The angular velocity of the Ferris wheel, ω = 2·π/T

The linear velocity of the Ferris wheel, v = r·ω = The speed of the riders

Where;

r = The radius of the Ferris wheel = D/2

D = 150 m

∴ r = 150 m/2 = 75 m

∴ v = r·2·π/T

∴ v = 75 m × 2 × π/(1,800 s) ≈ 0.262 m/s

The speed of the riders on the Singapore Flyer, v ≈ 0.262 m/s

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A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
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Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
3 years ago
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