In a simple machine the energy input is 120J if the efficiency of the machine is 80% calculate the energy output

Answer:

1,860

Explanation:

[tex]{\fcolorbox{white}{lightgreen}{\bf{\textcircled{$\checkmark$}}{Verified\:answer}}}[/tex]

We know

[tex]\boxed{\sf K.E=\dfrac{1}{2}mv^2}[/tex]

[tex]\\ \sf\longmapsto 100J=\dfrac{1}{2}5\times v^2[/tex]

[tex]\\ \sf\longmapsto v^2=100\times \dfrac{2}{5}[/tex]

[tex]\\ \sf\longmapsto v^2=20(2)[/tex]

[tex]\\ \sf\longmapsto v^2=40[/tex]

[tex]\\ \sf\longmapsto v=\sqrt{40}[/tex]

[tex]\\ \sf\longmapsto v=6.2m/s[/tex]

Answer:

6.2m/s

Explanation:

Answer:

Velocity

Explanation:

You would be describing the velocity of the car.

Velocity in physics is defined as Vector quantity that describes the displacement of an object with respect to the time it takes to attain it. Displacement is the addition of direction to the speed of an object. The displacement is noted in the question, "traveling eastward". While it is stated that the car travels at 44.704 m/s. Ordinarily, it would have been tagged speed, if not for the direction added to it which makes it velocity.

I hope you understand.

Answer:

The mean is 4.5 and the standard deviation is 1.44. Step-by-step explanation: An uniform probability is a case of probability in which each outcome is equally as likely. For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b. The mean of the uniform probability distribution is: The standard deviation of the uniform probability distribution is: Uniformly distributed random variable ranging from 2 to 7. This means that . So The mean is 4.5 and the standard deviation is 1.44.

Answer:

Take up space

Explanation:

Actually we know this by the definition of matter which states that "matter is any substance that has mass and takes up space by having volume."

hope it helped you:)

Answer: C

Explanation: It's a lot of math.

The option that gives the correct value for the angle theta above the horizontal axis of the velocity of the body, v is option C.

C. 53.1°

The given parameters are;

The velocity of the body relative to the horizontal axis, vₓ = 6 m/s

The velocity of the body relative to the vertical axis, [tex]v_y[/tex] = 8 m/s

Required;

The measure of the angle theta above above the horizontal axis of the velocity of the body, v

Solution:

The velocity of the body can be represented vectorially as follows;

v = 6·i + 8·j

The vectoral form gives the legs of the right triangle that has the magnitude and direction as the hypotenuse side

Therefore;

[tex]\theta = arctan \left(\dfrac{8}{6} \right) \approx \underline{53.1^{\circ}}[/tex]

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Answer:D.Keeping track of the exact amount of wind on each plant

Answer:

9 and 4

Explanation:

The relation is:

● r = s^2 / t^2

Triplind s means multiplying it by 3. Since it's an equation we should multiply both sides by the same number

Let k be the number we should multiply by r

●k* r = (3s)^2 / t^2

●k* r = 9s^2 /t^2

We have multiplied s^2 by 9 so we should do the same for r.

k = 9

■■■■■■■■■■■■■■■■■■■■■■■■■■

Doubling t means multiplying it by 2.

Let x be the number we shoukd multiply by r.

● x* r = s^2/(2t)^2

● x*r = s^2/ 4t^2

We have multiplied t^2 by 4 so we should do the same for r.

x= 4

Answer:

The output force would decrease since exerted force is inversely proportional to contact area.

Explanation:

F ∝ [tex]\frac{1}{A}[/tex] , Where 'F' is the exerted force and 'A' is the contact area.

Answer:

W = 1320Watts

Explanation:

W = I*V

W = 12A*110V

W = 1320Watts

Answer:

0 m/s

Explanation:

Average velocity is displacement over time.

v_avg = Δx / Δt

Displacement is the distance between the start and the finish.

The circumference of the track is:

C = 2πr

C = 2π (7 m)

C ≈ 44 m

The distance covered by the object is:

d = vt

d = (11 m/s) (8 s)

d = 88 m

So the object travels 2 circumferences, meaning it ends back where it started.  Therefore, the displacement is 0 m, and the average velocity is 0 m/s.

Explanation:

SI is the international system of units

It was developed to express magnitudes and quantities

Answer:

38.024 KWh

Explanation

250 watts multiplied with 5 hours daily multiplied with 28 days monthly equals 35000 Watt hours. Divided by a 1000 to get the KWh equals 35 KWh.

= 250×5hrs× 28days

= 35000watts

= 35000/1000

= 35KWh

Four 6 watt lightbulbs equals 24 watts 4x6=24

Hence, 24×4.5hrs×30days

= 3024watts

= 3024/1000

= 3.024KWhr

The total amount of energy consumed in the month of February = 35 KWh + 3.024 KWh = 38.024 KWh

Note that I had to use 28days since we are considering the month of February.

Answer:

answer= 73.1256 [tex]i[/tex]

Explanation:

The electric charge linear density is equal to 8.8 x[tex]10^{-9}[/tex]

the length of the string is 3.1m

The magnitude of the electric field at the length of the string equal to 5.2 meters can be calculated with the formula ;

- E = λ / 4πε₀ [ [tex]l[/tex]  / α ( α +

Solution:

E =  8.8 x[tex]10^{-9}[/tex] / 4πε₀ [ 3.1/ 5.2( 5.2 + 3.1) ] [tex]i[/tex]

= 1018.0995 [0.07183] [tex]i[/tex]

=  73.1256 [tex]i[/tex]

Answer:

// EDU HACK REDDIT EDITON //

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var running = null; console.log("Program Loading...");

var iframe = document.getElementById("stageFrame");

for (i = 0; i < 1; i++) {

 setInterval(function() {

  if (running == null) { running = true; console.log('Program Loaded! Enjoy!') }

  if (iframe.contentWindow.document.getElementsByClassName('FrameRight') != null) { iframe.contentWindow.document.getElementsByClassName('FrameRight')[0].onclick(); }

  if (iframe.contentWindow.document.getElementById("invis-o-div") != null) { iframe.contentWindow.document.getElementById("invis-o-div").remove(); }

 },1000);

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} else {

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}

Explanation:

Answer:

[tex]\lambda[/tex] = 40 cm

Explanation:

given data

string length = 30 cm

solution

we take here equation of length that is

L = [tex]n \times \frac{1}{4} \lambda[/tex]     ...............1

so

total length will be here

[tex]L = \frac{\lambda}{2} + \frac{\lambda}{4}\\[/tex]

[tex]L = \frac{3 \lambda }{4}[/tex]

so [tex]\lambda[/tex]  will be

[tex]\lambda = \frac{4L}{3}\\\lambda = \frac{4\times 30}{3}[/tex]

[tex]\lambda[/tex] = 40 cm

Answer:

the photo is a bit blur

Answer:

is a force that attracts any two objects with a mass

This question is incomplete; here is the complete question:

Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the frequency as 3 hertz, which statement about the wave is accurate?

A. The wave has traveled 32.4 cm in 3 seconds.

B. The wave has traveled 32.4 cm in 9 seconds.

C. The wave has traveled 97.2 cm in 3 seconds.

D. The wave has traveled 97.2 cm in 1 second.

The answer to this question is D. The wave has traveled 97.2 cm in 1 second.

Explanation:

The frequency of a wave, which is in this case 3 hertz, represents the number of waves that go through a point during 1 second. According to this, if the frequency of the wave is 3 hertz this means in 1 second there were 3 waves. Moreover, if you multiply the wavelength (32.4cm) by the frequency (3) you will know the distance the wave traveled in 1 second: 32.4 x 3 =  97.2 cm. This makes option D the correct one as the distance in 1 second was 97.2 cm.

Answer:Is D!

Explanation:TEAT(Sorry) -_-*

Answer:

First, as you may know, the light travels at a given velocity.

In vaccum, this velocity is c = 3x10^8 m/s.

And we know that:

distance = velocity*time

Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.

This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)

Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.

The astronomers could know what was happening inside galaxies way back then by the fact that;

they examine the spectra of galaxies (or the overall colors of galaxies) with the highest redshifts they can find

Astronomers Measure the wavelength of the light that is stretched, so the light is seen as 'shifted' towards the red part of the spectrum by using spectroscopy. This measure is also called redshift.

This invokes a ray of light through a triangular prism that splits the light into various components known as spectrum.

The way the astronomers could use this concept to know what was happening in the galaxies before is by examining the spectra of galaxies that have the highest redshifts.

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Answer:

(I). The Schwarzschild radius is [tex]2.94\times10^{8}\ km[/tex]

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is [tex]1.1\times10^{-7}\ km[/tex]

(IV). The Schwarzschild radius is [tex]7.4\times10^{-29}\ km[/tex]

Explanation:

Given that,

Mass of black hole [tex]m= 1\times10^{8} M_{sun}[/tex]

(I). We need to calculate the Schwarzschild radius

Using formula of radius

[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}[/tex]

[tex]R_{g}=2.94\times10^{8}\ km[/tex]

(II). Mass of block hole [tex]m= 6 M_{sun}[/tex]

We need to calculate the Schwarzschild radius

Using formula of radius

[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]

Put the value into the formula

[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}[/tex]

[tex]R_{g}=17.7\ km[/tex]

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]

Put the value into the formula

[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}[/tex]

[tex]R_{g}=1.1\times10^{-7}\ km[/tex]

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]

Put the value into the formula

[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}[/tex]

[tex]R_{g}=7.4\times10^{-29}\ km[/tex]

Hence, (I). The Schwarzschild radius is [tex]2.94\times10^{8}\ km[/tex]

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is [tex]1.1\times10^{-7}\ km[/tex]

(IV). The Schwarzschild radius is [tex]7.4\times10^{-29}\ km[/tex]

The energy conservation allows to find the Schwarschild radius for several bodies of different masses are:

  1)  Black hole quasar is:  r = 2.9 10⁸ km

  2) Blsck hole supernove is:  r = 17.7 km

  3)  Mini black hole is:   r = 1.1 10⁻⁷ km

  4) Human body is:  r=  7 10⁻²⁹ km

The schwarschild radius is defined as the distance from a black hole center at radius  which the escape velocity is equal to the light speed, in some cases it is also called the event horizon.

Let's use Newton's second law where force is the universal law of attraction and acceleration is centripetal.

        F = ma

        F = [tex]G \frac{Mm}{r^2}[/tex]        

Where F is the force, M the mass of the black hole, m the handle of the body, r the radius and v the speed of the body.

The energy of the gravitational field is

        F = [tex]- \frac{dU}{dr }[/tex]  

         U = [tex]-G \frac{Mm}{r}[/tex]  

Let's use conservation of energy

        Em₀ = K + U = ½ m v² -  [tex]G \frac{Mm}{r}[/tex]  

In infinity the energy

        Em_f = 0

energy is conserved

       Em₀ = Em_f  

       ½ m v² - [tex]G \frac{Mm }{r}[/tex]  = 0

       r = [tex]\frac{2GM}{v^2}[/tex]

From the definition of the Schwarschild radius this speed is equal to the light speed

        v = c

        r = [tex]\frac{2GM}{c^2 }[/tex]  

They ask to calculate the radius for several cases of different mass, claculate the constant value

      V = [tex]\frac{2 \ 6.67 \ 10^{-11} }{(3 \ 10^8) ^2 }[/tex]  

      V =  1.482 10⁻²⁷

1) A black hole of mass M = 1 10⁸ [tex]M_{sum}[/tex]

The tabulated mass of the sun is [tex]M_{sum}[/tex] = 1.989 10³⁰ kg

Let's  substitute

        r =  1.482 10⁻²⁷   1 10⁸ 1.989 10³⁰

        r = 2.94 10⁸ km

With two significant figures

        r = 2.9 10⁸ km

2) A black hole of mass M = 6 [tex]M_{sum}[/tex]

        r =  1.482 10⁻²⁷ 6 1.989 10-30

        r = 17.7 km

3) a mini black hole with the mass of the moon

    Tabulated mass of the moon M = 7.35 10²² kg

       r =  1.482 10⁻²⁷  7.35 10²²    

       r = 1.1 10⁻⁷ km

4) A person of M = 50 kg

    r =  1.482 10⁻²⁷ 50

    r=  7 10-29 km

In conclusion using the conservation of energy we can find the Schwarschild radius for several bodies of different masses are:

  1)  Black hole quasar is:  r = 2.9 10⁸ km

  2) Blsck hole supernove is:  r = 17.7 km

  3)  Mini black hole is:   r = 1.1 10⁻⁷ km

  4) Human body is:  r=  7 10⁻²⁹ km

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Answer:

The answers to your questions are given below

Explanation:

A. Definition of momentum.

Momentum of an object can be defined as the product of the mass of the object and its velocity. Mathematically, it expreessed as:

Momentum = mass x Velocity

From the above equation, we can derive the SI unit of momentum as follow:

Mass is measured in Kilogram (Kg)

Velocity is measured in meter per second (ms¯¹).

Momentum = mass x Velocity

Momentum = Kg x ms¯¹

Momentum = Kg•ms¯¹

Therefore, the SI unit of momentum is Kg•ms¯¹.

Bi. Determination of the force of the body from O to A.

Mass (m) = 5 kg

Velocity (v) = 40 ms¯¹

Time (t) = 2 secs.

Force (F) =?

Next, we shall determine the acceleration of the body.

Acceleration (a) = Velocity (v) /Time (t)

a = v /t

Velocity (v) = 40 ms¯¹

Time (t) = 2 secs.

Acceleration (a) =.?

a = v/t

a = 40/2

a = 20 ms¯²

Now, we can obtain the force as follow:

Mass (m) = 5 kg

Acceleration (a) = 20 ms¯²

Force (F) =?

Force (F) = mass (m) x Acceleration (a)

F = ma

F = 5 x 20

F = 100 N

Therefore, the force of the body from O to A is 100 N.

Bii. Determination of the force of the body from B to C.

Mass (m) = 5 kg

Velocity (v) = 40 ms¯¹

Time (t) = 10 – 6 = 4 secs.

Force (F) =?

Next, we shall determine the acceleration of the body.

Acceleration (a) = Velocity (v) /Time (t)

a = v /t

Velocity (v) = 40 ms¯¹

Time (t) = 4 secs.

Acceleration (a) =.?

a = v/t

a = 40/4

a = 10 ms¯²

Now, we can obtain the force as follow:

Mass (m) = 5 kg

Acceleration (a) = 10 ms¯²

Force (F) =?

Force (F) = mass (m) x Acceleration (a)

F = ma

F = 5 x 10

F = 50 N

Therefore, the force of the body from B to C is 50 N.

We have that the time, to the nearest minute, when the water level is at 1.125 m for the second time after midnight is

[tex]t=10.0hours[/tex]

From the Question we are told that

Maximum height [tex]h_{max}=3m[/tex]

Minimum height  [tex]H_{min}=0.5m[/tex]

Time for  next high tide will occur[tex]T=12 hours =>720 min[/tex]

Generally Average Height

[tex]h_{avg}=\frac{3+0.5}{2}\\\\h_{avg}=1.75[/tex]

Therefore determine Amplitude to be

[tex]A=h_{max}=j_{avg}\\\\A=3-1.75\\\\A=1.25[/tex]

Generally, the equation for Time is mathematically given by

At t=0

[tex]h(x)=Acos(Bx)+h_{avg}[/tex]

Where

[tex]B=\frac{2\pi}{P}\\\\B=\frac{2\pi}{720}\\\\B=8.73*10^{-3}[/tex]

Therefore

[tex]h(t)=Acos8.73*10^{-3}(t)+h_{avg}[/tex]

Hence the Time at [tex]T=1.125[/tex] is

[tex]1.125(t)=1.25cos(8.73*10^{-3})(t)+1.75[/tex]

[tex]-0.1249t=1.75[/tex]

[tex]t=10.0hours[/tex]

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1. Which example best describes a restoring force?

B) the force applied to restore a spring to its original length

2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?

C) The spring exerts a restoring force to the left and returns to its equilibrium position.

3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?

D) 1 m

4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?

D)It is a vector quantity.

5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?

A) It decreases in magnatude.

Answer:

because us equal to the anticlockwise moment about that pivot.If the object are balanced:clockwise.

Explanation:

This is when an object placed on something narrow object that can act as a pivot too fir example a plank placed on a brick.Finally the edges or the curve

Answer:

Moment about a point= force × perpendicular distance

When an object is further from the pivot, the perpendicular distance of the object from the pivot increases. Since moment is the product of the force and the perpendicular distance, when the perpendicular distance increase, the moment increases too. Thus, objects located further away from the pivot have a greater moment.

Notice that

B = 4i + 6j - 2k = 2 (2i + 3j - k) = 2A

so both vectors point in the same direction and the angle between them is (A) 0°.

check the pictures(2 pictures)check the pictures(2 pictures)check the pictures(2 pictures)

The vector sum of the two forces is 20  and the magnitude of the resultant is 20 towards positive x-axis.

A vector is a quantity or phenomena with magnitude and direction that are independent of one another. The phrase also refers to a quantity's mathematical or geometrical representation.

If no vector can be written as a linear combination of the others, a set of vectors is said to be linearly independent.

The vector representation for the forces F and F are:

[tex]\rm \vec F_1 = 20 COS 60^0 \vec i + 20 SIN 60^0 \vec j\\\\ F_1 =20 \times \frac{1}{2} \vec i+20 \times \frac{\sqrt{3} }{2} \vec j \\\\ \vec F_1 = 10 \vec i+ 10 \sqrt{3} \vec J[/tex]

[tex]\rm F_2 = 20 cos 60^0 \vec i+20 sin(-60) \vec j \\\\ F_2 = 20 \times \frac{1}{2} \times \vec i+20 \times \frac{\sqrt{3} }{2} \vec j \\\\ \vec F_2 = 10 \vec I -10\sqrt{3} \vec J[/tex]

The vector sum of the two forces are;

[tex]\rm \vec R = \vec F_1+\vec F_2\\\\ \vec R = 10 \vec i+ 10 \sqrt{3} \vec J+10 \vec i-10\sqrt{3} \vec J\\\\ \vec R =20 i[/tex]

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Answer:

(a) The convex mirror image, is always upright at all positions, while images formed by concave mirrors are always inverted when the object distance from the mirror is more than the mirrors focal length.

(b) An upright image is not seen for object at a distance from a concave mirror further than the focal length of the mirror, which is the spoon in the question

Therefore, the location of her eyes of approximately, 30 cm,  from the mirror is more than the mirror's focal length

Explanation:

First find Acceleration

[tex]\boxed{\sf Acceleration=\dfrac{v-u}{t}}[/tex]

[tex]\\ \sf\longmapsto Acceleration=\dfrac{42-0}{2}[/tex]

[tex]\\ \sf\longmapsto Acceleration=\dfrac{42}{2}[/tex]

[tex]\\ \sf\longmapsto Acceleration=21m/s^2[/tex]

Using second equation of kinematics

[tex]\boxed{\sf s=ut+\dfrac{1}{2}at^2}[/tex]

[tex]\\ \sf\longmapsto s=0(2)+\dfrac{1}{2}(21)(2)^2[/tex]

[tex]\\ \sf\longmapsto s=21(2)[/tex]

[tex]\\ \sf\longmapsto s=42m[/tex]

Using 1st equation of kinematics

[tex]\\ \bull\sf\longmapsto v=u+at[/tex]

[tex]\\ \bull\sf\longmapsto v=0+0.25(180)[/tex]

[tex]\\ \bull\sf\longmapsto v=45m/s[/tex]

⇛Initial velocity (u) = 0 m/s

⇛Acceleration (a) = 0.25 m/s²

⇛Time (t) = 3 minutes

⇛Final velocity (v) = v

Converting 3 minutes into seconds

[tex] \Large\begin{gathered} {\underline{\boxed{ \rm {\red{v \: = \: u \: + \: at}}}}}\end{gathered}[/tex]

[tex] \bf \large \longrightarrow \: \: v \: = \: 0 \: + \: 0.25 \: \: (180)[/tex]

[tex]\bf \large \longrightarrow \: \: v \: = \:45 \: m/s[/tex]