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3 years ago

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If each pull-up requires 300 J and Dan does a pull-up in 1.5 seconds, what is his power? 1000 watts 800 watts 400 watts 200 watt

s

1 answer:

Jlenok [28]3 years ago

8 0

300 divide 1.5=200 let me know if this was helpful

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Calculate the average net force.

MrRa [10]

Answer:

<h2><u>given</u></h2>

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<em>time</em><em>=</em><em> </em><em>1</em><em>0</em><em>.</em><em>5</em><em> </em><em>sec</em>

<h2><em>To</em><em> </em><em>find</em><em> </em></h2>

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<h2><em><u>Solution</u></em></h2>

<h3><em>☄️</em><em>Formula</em><em> </em><em>of</em><em> </em><em>force</em><em> </em></h3>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

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3 0

3 years ago

Which type of wave is described as a wave that is not confined to the space along medium?

Dmitry [639]

Traveling wave is a wave that is not confined to the space along medium.

traveling wave is a wave that travels along the medium causing disturbance. instead of vibrating at the same place, the particles actually move in the direction of wave. one common example of such type of wave is ocean waves.

4 0

4 years ago

Read 2 more answers

The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

a)

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

b)

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago

A pulley of radius 8.0 cm is connected to a motor that rotates at a rate 7000 rad s-1 and then decelerate uniformly at a rate of

zlopas [31]

Answer:

(a) α = - 1000 rad/s²

Negative sign represents deceleration.

(b) θ = 3581 rotations

(c) L = 1800 m

(d) a = - 80 m/s²

Explanation:

(a)

using First equation of motion for angular motion:

ωf = ωi + αt

where,

ωf = Final Angular Speed = 2000 rad/s

ωi = Initial Angular Speed = 7000 rad/s

α = Angular Acceleration = ?

t = time = 5 s

Therefore,

2000 rad/s = 7000 rad/s + α(5s)

α = (2000 rad/s - 7000 rad/s)/5 s

<u>α = - 1000 rad/s²</u>

<u>Negative sign represents deceleration.</u>

(b)

Using second equation of motion:

θ = ωi t + (1/2)αt²

where,

θ = No. of Rotations = ?

Therefore,

θ = (7000 rad/s)(5 s) + (1/2)(- 1000 rad/s²)(5 s)²

θ = 35000 rad - 12500 rad

θ = (22500 rad)(1 rotation/2π rad)

<u>θ = 3581 rotations</u>

(c)

Length of String = L = (Circumference of Pulley)(θ)

L = [2π(0.08 m)][3581 rotations]

<u>L = 1800 m</u>

<u></u>

(d)

Tangential Acceleration = a = rα

a = (0.08 m)(-1000 rad/s²)

<u>a = - 80 m/s²</u>

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3 years ago

2 differences between calorimeter and thermometer ?

hram777 [196]

Answer:

Calorimeter is used to measure heat in and represents that in units of joules per kelvin units J/˚C or kJ/K

A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction.

Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction. Unit measurement used by thermometer is Celsius (°C).

Explanation:

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3 years ago