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3 years ago

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A motor does a total of 480 joules of work in 5.0 seconds to lift a 12-kilogram block to the top of a rampThe average power deve

loped by the motor is

1 answer:

Flauer [41]3 years ago

3 0

Answer:

there it is fella

we neglect the mass data

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You repeated a measurement 6 times and recorded the time using a stop watch: 5.8s, 4.6s, 4.8s, 5.1s, 4.3s, 4.6s. What average va

iVinArrow [24]

Answer and Explanation

Arranging the measured values in increasing order;

4.3s, 4.6s, 4.6s, 4.8s, 5.1s, 5.8s

The two outliers are obviously 4.3s and 5.8s; An outlier is a value in a statistical sample which does not fit a pattern that describes most other data point. Outliers make the average value complicated. So, it is usually better for data to be precise with data points spreading out around a small area.

So, the mean is the average of the four remaining data points after removing the outliers.

Mean = (4.6 + 4.6 + 4.8 + 5.1)/4

Mean = 4.775s

So, the value recorded should be 4.775s, 4.78s or 4.8s depending on the number of decimal places allowed.

QED!

8 0

3 years ago

An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6

lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

$I = \frac{P_{avg}}{A}$

$I = \frac{P_{avg}}{4\pi r^2}$

$I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m$

$I = 6.529*10^{-6}W/m^2$

The amplitude of electric field at the receiver is

$I = \frac{E_{max}^2}{2\mu_0 c}$

$E_{max}= \sqrt{2I\mu_0 c}$

The amplitude of induced emf by this signal between the ends of the receiving antenna is

$\epsilon_{max} = E_{max} d$

$\epsilon_{max} = \sqrt{2I \mu_0 cd}$

Here,

I = Current

$\mu_0$ = Permeability at free space

c = Light speed

d = Distance

Replacing,

$\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}$

$\epsilon_{max} = 0.05434V$

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0

3 years ago

Joaquin mows the lawn at his grandmother’s home during the summer months. Joaquin measured the

nordsb [41]

Velocity is define as how fast an object is moving, and in what direction, it is a vector quantity, meaning velocity has both magnitude and direction. Anything goes to the left is negative, and anything goes to the right is positive.

a. Direction from east to west, given distance 11.5 meters, and time of 7.10 s

V = displacement/time V = -11.5/7.10 S V = -1.62 m/s (going left)

b. Joaquin reaches his original position. Displacement is now zero.

Velocity of the lawnmower is equal to "zero" but if we calculate for the average speed of the lawn, you just have to add the distance covered and the time it take to go back at the original position or point of origin

7 0

3 years ago

What type of wave is sound?<br> light<br> pressure<br> electromagnetic<br> gamma

JulijaS [17]

Nu its pressure, just did this rn, 8th grade how the ear works physical science test

3 0

3 years ago

Read 2 more answers

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago