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AlladinOne [14]
3 years ago
13

Your neighbor Paul has rented a truck with a loading ramp. The ramp is tilted upward at 35 ∘,and Paul is pulling a large crate u

p the ramp with a rope that angles 10 ∘ above the ramp. Paul pulls with a force of 350 N. (Force is measured innewtons, abbreviated N.)
What is the magnitude of the horizontal component of his force?

What is the magnitude of the vertical component of his force?

Physics
1 answer:
notsponge [240]3 years ago
7 0

Answer:

The magnitude of the horizontal component of his force is 345 N.

The magnitude of the vertical component of his force is 60.8 N.

Explanation:

Hi there!

Please, see the attached figure for a better understanding of the problem.

To find the horizontal and vertical component (Fx and Fy respectively) of Paul´s force (F), we have to use the trigonometry of right triangles.

Notice that Fx and Fy are respectively the adjacent and opposite sides  of the triangle formed by F, Fy and Fx (see figure).

F is the hypotenuse of the triangle.

We can apply the following trigonometric rules to obtain Fx and Fy:

cos angle = adjacent side / hypotenuse

sin angle = opposite side / hypotenuse

In our case:

cos 10° = Fx / F

F · cos 10° =Fx

350 N · cos 10° = Fx

Fx = 345 N

The magnitude of the horizontal component of his force is 345 N.

sin 10° = Fy/F

F · sin 10° = Fy

350 N · sin 10° = Fy

Fy = 60.8 N

The magnitude of the vertical component of his force is 60.8 N.

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A force of 20N changes the position of a body. If mass of the body is 2kg, find the acceleration produced in the body.2. A ball
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Explanation:

<em>Hello</em><em> </em><em>there</em><em>!</em><em>!</em><em>!</em>

<em>You</em><em> </em><em>just</em><em> </em><em>need</em><em> </em><em>to</em><em> </em><em>use</em><em> </em><em>simple</em><em> </em><em>formula</em><em> </em><em>for</em><em> </em><em>force</em><em> </em><em>and</em><em> </em><em>momentum</em><em>, </em>

<em>F</em><em>=</em><em> </em><em>m.a</em>

<em>and</em><em> </em><em>momentum</em><em> </em><em>(</em><em>p</em><em>)</em><em>=</em><em> </em><em>m.v</em>

<em>where</em><em> </em><em>m</em><em>=</em><em> </em><em>mass</em>

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5 0
3 years ago
A roller coaster cart of mass m = 223 kg starts stationary at point A, where h1 = 26.8 m and a while later is at B, were h2 = 14
Tresset [83]

Answer:

vB = 15.4 m/s

Explanation:

Principle of conservation of energy:

Because there is no friction the mechanical energy is conserve

ΔE = 0

ΔE : mechanical energy change (J)

K : Kinetic energy (J)

U: Potential energy (J)

K = (1/2)mv²

U = m*g*h

Where :

m: mass (kg)

v : speed (m/s)

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(K+U)final = (K+U)initial

((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:

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(1/2) (vB)² + (9.8)*(14.7) =  0 + (9.8)(26.8 )

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(vB)² = (2)(9.8)(26.8 - 14.7)

(vB)² = 237.16

v_{B} = \sqrt{237.16}

vB = 15.4 m/s : speed of the cart at B

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