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2 years ago

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An electric motor consumes 9.00 kj of electrical energy in 1.00 min. if one-third of this energy goes into heat and other forms

of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2500 rpm?

1 answer:

liq [111]2 years ago

5 0

How do I make a question help PLZZZ

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A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho

Gnoma [55]

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

$80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)$

$80=5(50-x)$

$80=250-5x$

$5x=170$

$x=\frac{170}{5}$

$x=34 cm$

4 0

2 years ago

A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other

san4es73 [151]

Answer:

16,18,22

Or

1,3,7

Explanation:

The detailed explanation is contained in the image attached. The lengths are found using Pythagoras theorem and the two lengths reflects the two values of x yielded by the quadratic equation

8 0

2 years ago

In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a

fgiga [73]

(a) The diffraction decreases

The formula for the diffraction pattern from a single slit is given by:

$sin \theta = \frac{n \lambda}{a}$

where

$\theta$ is the angle corresponding to nth-minimum in the diffraction pattern, measured from the centre of the pattern

n is the order of the minimum

$\lambda$ is the wavelength

a is the width of the opening

As we see from the formula, the longer the wavelength, the larger the diffraction pattern (because $\theta$ increases). In this problem, since the wavelength of the signal has been decreased from 54 cm to 13 mm, the diffraction of the signal has decreased.

(b) $10.8^{\circ}$

The angular spread of the central diffraction maximum is equal to twice the distance between the centre of the pattern and the first minimum, with n=1. Therefore:

$sin \theta = \frac{(1) \lambda}{a}$

in this case we have

$\lambda=54 cm = 0.54 m$ is the wavelength

$a=5.7 m$ is the width of the opening

Solving the equation, we find

$\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.54 m}{5.7 m})=5.4^{\circ}$

So the angular spread of the central diffraction maximum is twice this angle:

$\theta = 2 \cdot 5.4^{\circ}=10.8^{\circ}$

(c) $0.26^{\circ}$

Here we can apply the same formula used before, but this time the wavelength of the signal is

$\lambda=13 mm=0.013 m$

so the angle corresponding to the first minimum is

$\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.013 m}{5.7 m})=0.13^{\circ}$

So the angular spread of the central diffraction maximum is twice this angle:

$\theta = 2 \cdot 0.13^{\circ}=0.26^{\circ}$

5 0

2 years ago

The basic function of an automobile carburetor is to atomize the gasoline and mix it with air to promote rapid combustion. Assum

tatyana61 [14]

Answer:

$A = 7.5 \times 10^6 m^2$

Explanation:

Since volume of all the liquid is always conserved

so here we know that

$V_{total} = N V$

$25 cm^3 = N(\frac{4}{3}\pir^3)$

$25 = N (\frac{4}{3} \pi (1.0 \times 10^{-5})^3)$

$N = 5.97 \times 10^{15}$

now we know that total surface area is given as

$A = N(4\pi r^2)$

$A = (5.97 \times 10^{15})(4\pi (1 \times 10^{-5})^2$

$A = 7.5 \times 10^6 m^2$

5 0

2 years ago

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

yanalaym [24]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

That would be our first equation

Now, we need to analyze its vertical motion

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Knowing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Solving for t

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

So, the magnitude of the velocity is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The <u><em>direction of the ball</em></u> is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

4 0

2 years ago