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2 years ago

How do you label Time= energy • power

1 answer:

5 0

First of all, that equation is not correct, which may be the reason
that you're having trouble assigning units to the quantities.

Power is defined as [energy / time], so [Energy] = [ power x time ],
and
[Time] = [ energy / power ].

Unit-wise, these equations are correct just as they appear here,
with no proportionality constants or conversion factors, when ...

[ Power ] = watts
[ Energy ] = joules
[ Time ] = seconds .

You might be interested in

02: Describes what happens to the strength of gravity if the mass between two objects

Dmitriy789 [7]

Explanation:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. ... increases, the force of gravity decreases. If the distance is doubled, the force of gravity is one-fourth as strong as before.

5 0

3 years ago

A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th

xz_007 [3.2K]

Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

$v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}$

(b) The tension, $T$ , is given by

$v =\sqrt{\dfrac{T}{\mu}}$

where $v$ is the speed, $T$ is the tension and $\mu$ is the mass per unit length.

Hence,

$T = \mu\cdot v^{2}$

To determine $\mu$ , we need to know the mass of the cable. We use the density formula:

$\rho = \dfrac{m}{V}$

where $m$ is the mass and $V$ is the volume.

$m=\rho\cdot V$

If the length is denoted by $l$ , then

$\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}$

$T = \dfrac{\rho\cdot V}{l} v^{2}$

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

$V = \pi \dfrac{d^{2}}{4} l$

$T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2$

$T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2$

$T = 11159.4186\ldots \text{ N} = 11000 \text{ N}$

4 0

3 years ago

What is the wavelength

melamori03 [73]

The distance between two particles that are <em>in phase</em>

4 0

3 years ago

A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in

melamori03 [73]

Answers:

a) $0.5 m/s^{2}$

b) $1.5 N$

Explanation:

a) The centripetal acceleration $a_{c}$ of an object moving in a uniform circular motion is given by the following equation:

$a_{c}=\omega^{2} r$

Where:

$\omega=1 \frac{rev}{s}$ is the angular velocity of the ball

$r=0.5 m$ is the radius of the circular motion, which is equal to the length of the string

Then:

$a_{c}=(1 \frac{rev}{s})^{2} 0.5 m$

$a_{c}=0.5 m/s^{2}$ This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force $F$ ) that is directed towards the center and is equal to the tension ( $T$ ) in the string:

$F=T=m. a_{c}$

Where $m=3 kg$ is the mass of the ball

Hence:

$T=(3 kg)(0.5 m/s^{2})$

$T=1.5 N$ This is the tension in the string

7 0

3 years ago

A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10

bearhunter [10]

Answer:

605447.7066 kgm²/s

Explanation:

$m_1$ = Mass of sphere = 10000 kg

$m_2$ = Mass of rod = 10 kg

r = Radius of sphere = 2 m

l = Length of antenna = 3 m

Angular speed

$\omega=6\times 2\pi\\\Rightarrow \omega=37.69911\ rad/s$

Angular momentum is given by

$L=I\omega$

Moment of inertia of the satellite is

$I_s=\frac{2}{5}m_1r^2$

Moment of antenna of the satellite is

$I_a=\frac{1}{3}m_2l^2$

The angular momentum of the system is

$L=I_s\omega+I_a\omega\\\Rightarrow L=\left(\frac{2}{5}m_1r^2+2\times \frac{1}{3}m_2l^2\right)\omega\\\Rightarrow L=\left(\frac{2}{5}10000\times 2^2+2\times \frac{1}{3}\times 10\times 3^2\right)\times 37.69911\\\Rightarrow L=605447.7066\ kgm^2/s$

The angular momentum of the satellite is 605447.7066 kgm²/s

5 0

3 years ago