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3 years ago

A wheel is observed to rotate 5 complete revolutions in 25 seconds. What is its angular speed

Physics

1 answer:

skelet666 [1.2K]3 years ago

8 0

<u>Answer</u>

1.26 rad/s

the first step is to find the frequency of the wheel.

The wheel rotates at rate of 5 revolutions in 25 seconds. The frequency is the number of revolutions per second.

frequency (f) = 5/25

= 1/5

= 0.2 Hz

Angular velocity = 2πf

= 2×π×0.2

= 0.4π rad/s or

= 1.26 rad/s

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A thin rod of length L and total charge Q has the nonuniform linear charge distribution λ(x)=λ0x/L, where x is measured from the

marissa [1.9K]

Answer:

Explanation:

λ(x) = λo x/ L

(a) The total charge is Q.

$Q=\int_{0}^{L}dq$

$Q=\int_{0}^{L}\frac{\lambda _{0}x}{L}dx$

$Q=\frac{\lambda _{0}}{2L}\left ( x^{2} \right )_{0}^{L}$

$Q=\frac{\lambda _{0}}{2L}$

λo = 2Q/L

(b)

Let at a distance x from the origin the charge is dq.

so, dq = (2Q/L) x/ L dx

$dq=\frac{2Qx}{L^{2}}dx$

The potential due to this small charge at a distance d to the left of origin

$dV = \frac{KdQ}{d+x}$

$\int_{0}^{V}dV = \frac{2KQ}{L^{2}}\int_{0}^{L}\frac{xdx}{d+x}$

$V = \frac{2KQ}{L^{2}}\int_{0}^{L}\left ( 1- \frac{d}{d+x}\right )dx$

$V = \frac{2KQ}{L^{2}}\times \left ( x-dln(d+x) \right )\int_{0}^{L}$

$V = \frac{2KQ}{L^{2}}\times \left ( L-dln(d+L)-0+dlnd \right )$

$V = \frac{Q}{4\pi \epsilon _{0}L^{2}}\times \left ( L+d\times ln\left (\frac{d}{d+L} \right )\right )$

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3 years ago

4786 joules fer transferred to an 89.0 gram sample of a unknown material with an initial temperature of 23.0°C what is the speci

Alika [10]

Specific heat formula is Q=M*C*ΔT
Q is amount of energy (4786 J)
M is mass (89.0 g)
C is specific heat (unknown, answer will be in J/gºC)
ΔT is the change in temp (89.5-23.0=66.5ºC)
Let's plug it in.
4786=89.0(C)(66.5)
4786=5918.5(C)
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C=.8086508406 J/gºC
with sig figs...
C=.809 J/gºC

5 0

3 years ago

Read 2 more answers

A wooden block of dimensions of 6 cm ⨯ 6 cm ⨯ 15 cm floats with the long axis oriented horizontally and the of the square faces

schepotkina [342]

Explanation:

(a) Formula to calculate volume of the submerged wooden block is as follows.

$V_{sub} = l \times w \times d$

It is given data of the wooden block is as follows.

depth = 7.96 cm, length (l) = 6 cm

width (w) = 4 cm,

So, we will calculate the volume of the submerged wooden block as follows.

$V_{sub} = l \times w \times d$

= $6 \times 6 \times 7.96$

= 286.56 $cm^{3}$

Hence, the submerged volume of the block is 286.56 $cm^{3}$ .

(b) Expression for the buoyant force acting on the wooden block is as follows.

$F_{B} = \rho_{w} g V_{sub}$

And, expression for the force of gravity of the wooden block is as follows.

$F_{g} = m_{b}g$

As the wooden block is floating on the water hence, buoyant force is balanced by the weight of the block.

$F_{g} = F_{B}$

Hence, mass of the wooden block will be calculated as follows.

$F_{g} = F_{B}$

$m_{b}g = \rho_{w}gV_{sub}$

$m_{b} = \rho_{w}V_{sub}$

= $997 kg/m^{3} \times 286.56 cm^{3}$

= $997 kg/m^{3} \times 286.56 \times 10^{-6} m^{3}$

= 0.02857 kg

Therefore, mass of the given block is 0.02857 kg

(c) Expression for the density of the block is as follows.

$\rho_{b} = \frac{m_{b}}{V_{b}}$

Now, expression for the total volume of the wooden block is as follows.

$V_{b} = l \times w \times h$

Hence, density of the given block is as follows.

$\rho_{b} = \frac{m_{b}}{V_{b}}$

= $\frac{m_{b}}{lwh}$

= $\frac{0.02857 kg}{4 \times 4 \times 15}$

= $1.19 \times 10^{-4} kg/cm^{3}$

Therefore, density of the given block is $1.19 \times 10^{-4} kg/cm^{3}$ .

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3 years ago

A gas is contained in a cylinder with a frictionless moveable piston at a pressure of 3.50 × 105 pascals and a volume of 0.02 cu

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2 years ago

A rock is thrown at an angle of 30 degrees above the horizontal with initial velocity 15m/s what is the displacement when the ro

Delvig [45]

The displacement of the rock will be the same as the total horizontal distance traveled. Here the rock's horizontal position is given by

$x=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\cos30^\circ\,t$

so to find the horizontal distance it traversed, we need to know the time it took for the rock to return to the ground. We use the rock's vertical position over time to figure that out:

$y=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\sin30^\circ\,t-\dfrac g2t^2=0$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the acceleration due to gravity. Then we find that $t\approx1.5\,\mathrm s$ , at which point we find $x\approx20\,\mathrm m$ .

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3 years ago