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Anika [276]
3 years ago
5

A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward

with a force FS whose direction makes an angle of 30.0° with the ramp (Fig. E4.4). (a) How large a force FS is necessary for the component Fx parallel to the ramp to be 90.0 N?(b) How large will the component Fy perpendicular to the ramp be then?

Physics
1 answer:
AleksandrR [38]3 years ago
5 0

Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

   =104 N

(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

          = 52 N  

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A 2 kg block is pushed against a spring (k = 400 N/m), compressing it 0.3 m. When the block is released, it moves along a fricti
Kitty [74]

Answer:

2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{s}}=0.4

4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{k}}=0.2

Explanation:

Given values  

Mass (m) = 2kg

K = 400 N/M

Compressing it 0.3 m

<u>The law of conservation of energy</u>:

\frac{m v^{2}}{2}+\frac{k x^{2}}{2}=\text { constant }

\text { Where, } \frac{m v^{2}}{2} \text { is kinetic energy of the block. }

\frac{k \Delta l^{2}}{2} Energy of the spring deformation.

M mass of the block

x spring deformation

Therefore, if block left the spring (x = 0)

\frac{m v^{2}}{2}+0=0+\frac{k \Delta l^{2}}{2}

Where, Δl is initial spring deformation

\frac{m v^{2}}{2}=\frac{k \Delta l^{2}}{2}

\mathrm{v}^{2}=\frac{k \Delta l^{2}}{m}

\mathrm{v}=\sqrt{\frac{k}{m} \times \Delta l^{2}}

v=\sqrt{\frac{400}{2} \times(0.3)^{2}}

\mathrm{v}=\sqrt{200 \times 0.09}

<u>The law of conservation of energy</u>:

\frac{m v^{2}}{2}+m g h=\text { constant }

Where h is height

\frac{m v^{2}}{2}+0=0+m g h

\frac{m v^{2}}{2}=m g h

Cancel mass "m" each side

\mathrm{h}=\frac{v^{2}}{2 g}

Distance along incline equals

\begin{array}{ll}{\text { For friction us }} & {\left(L=\frac{h}{u_{s}}\right)} \\ {\text { For friction } u_{k}} & {\left(L=\frac{h}{u_{k}}\right)}\end{array}

\begin{array}{l}{\mathrm{u}_{\mathrm{s}}=0.4} \\ {\mathrm{U}_{\mathrm{k}}=0.2} \\ {\text { For friction } \mathrm{u}_{\mathrm{s}}}\end{array}

\begin{array}{l}{\mathrm{h}=\frac{v^{2}}{2 g u_{s}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.4}} \\ {\mathrm{L}=\frac{17.9776}{784}}\end{array}

\begin{array}{l}{L=2.29 \mathrm{m}} \\ {2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4} \\ {\text { For friction } \mathrm{u}_{\mathrm{k}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.2}}\end{array}

\begin{array}{l}{L=\frac{17.9776}{3.92}} \\ {L=4.58 \mathrm{m}} \\ {4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4}\end{array}

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. What is the height in meters of a 5'4" person?​
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Answer:

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At the time of quark confinement, when the universe was 10-6 seconds old, there is found to be one additional proton for every billion antiprotons.

<h3>What is quark confinement?</h3>

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Hence, At the time of quark confinement, when the universe was 10-6 seconds old, there is found to be one additional proton for every billion antiprotons.

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brainly.com/question/15103512

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