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adell [148]
3 years ago
8

We cab calculate work because we know the force and distance.what else would we need to know to calculate power

Physics
1 answer:
san4es73 [151]3 years ago
6 0

We would need to know how long it took to do the work.

Power = (work) / (time).

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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}

8 0
3 years ago
Two transverse waves travel along the same taut string inopposite directions. the waves are described by following equations use
Umnica [9.8K]

Answer: y'=2Asin(kx)cos(wt)

Explanation:

Let y1=A sin (kx + wt) be the first wave

y2=A sin (kx - wt) be the second wave in the opposite direction (which we showed by putting a negative sign between the terms kx and wt)

Please do note that both wave have the same attributes (that's Amplitude, wave number and angular frequency) because they are formed on the same medium by the same source just that their directions are opposite.

By super imposing these 2 waves, we have a resulting singular wave representing both wave (law of superimposition) with a resulting value of vertical displacement y'.

Thus y' = y1 + y2.

Let us do the math.

y'=A sin (kx + wt) + A sin (kx - wt)

By factoring A out, we have that

y' = A [ sin (kx + wt) + sin (kx - wt)]

For simplicity let us use the substitution

Let (kx + wt) = a and (kx - wt) =b

Hence we have that

y' = A [sin a + sin b].

From trigonometric ratio

sin a + sin b = 2sin[(a+b)/2] * cos [(a - b)/2]

By recalling that (kx + wt) = a and (kx - wt) =b

sin a + sin b = 2sin [(kx +wt +kx-wt) /2] * cos [(kx +wt - (kx-wt))/2]

Thus we have that

sin a + sin b = 2sin [(kx+wt+kx-wt)/2] * cos[(kx+wt-kx+wt)/2]

By collecting like terms in the bracket we have that

sin a + sin b = 2sin[2kx/2] * cos [2wt/2]

By dividing

sin a + sin b = 2sin(kx) cos(wt)

Now let us get the final resultant vertical displacement (y')

Recall that

y' = A [sin a + sin b]. and we already deduced that

sin a + sin b = 2sin(kx) cos(wt)

Finally,

y' = A [2sin(kx) cos(wt)] which is

y'=2Asin(kx)cos(wt)...... Final answer

4 0
4 years ago
1. ________________electricity is the type of electricity commonly used in homes and businesses throughout the world.
amid [387]
Number 1 is letter A
3 0
3 years ago
2 Points
Mademuasel [1]

According to Newton's Second Law of Motion :

The Force acting on an Object is equal to Product of Mass of the Object and Acceleration produced due to the Force.

:\implies  Force acting = Mass of the Object × Acceleration

Given : Force = 50 newton and Mass of the Object = 10 kg

Substituting the respective values in the Formula, we get :

:\implies  50 N = 10 kg × Acceleration

:\implies \mathsf{Acceleration = \dfrac{50\;N}{10\;kg}}

:\implies Acceleration of the Object = 5 m/s²

4 0
4 years ago
Megan was doing time trials on her bike around a 400m horizontal track. She took 32 seconds to travel 400m. What was her average
Mars2501 [29]
To find average speed, we divide the distance of travel (in this case, 400 metres) by the time she took, 32 seconds.  Therefore: 12.5 seconds is her average speed.
5 0
3 years ago
Read 2 more answers
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