calculate the work done by the person who applies force of 5000N to lift a load up to a height of 2m

The speed of the plane wave defined by sin(ωt−kx) is 2.864 × 10⁸ m/s

the speed of the plane wave defined by sin(ωt−kx) is v = ω/k where

Now ω = 3376.7 Thz = 3376.7 × 10¹² Hz and k = 11.79μm⁻¹ = 11.79 × 10⁶ m⁻¹

So, substituting the values of the variables into v, we have

v = ω/k

v = 3376.7 × 10¹² Hz/11.79 × 10⁶ m⁻¹

v = 286.4 × 10⁶ m/s

v = 2.864 × 10⁸ m/s

The speed of the plane wave defined by sin(ωt−kx) is 2.864 × 10⁸ m/s

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Answer:

150 watts

Explanation:

300/2 = 150 watts

Answer:

150 watts

Explanation:

The explanation is in the picture

PLEASE MARK BRAINLIEST

Answer:

5,920 J

Explanation:

Lowest temperature T₂ = 293 K

Highest temperature T₁ = 495 K

Highest possible efficiency e = 1 - T₂/T₁ = 1 - 293/495 = 0.408

Input heat Q₁ = 10,000 J

Find output heat Q₂

e = 1 - Q₂/Q₁

0.408 = 1 - Q₂/10,000

Q₂/10,000 = 1 - 0.408 = 0.592

so

Q₂  = 10,000 x 0.592 = 5,920 J

Answer:

A friction is very much desirable to stop the body from its moving condition. ... If the friction is more between the block and inclined surface then a large force is required to push the block on the plane.

Explanation:

hope it will help you have a great day bye and Mark brainlist if the answer is correct

[tex]kai6417[/tex]

Answer:

10m-2m = 8m

8

Hope this helps!

Peak expiratory flow readings are higher when patients are well, but lower when a  patient's airway is constricted.

This term is used in medicine to refer to the maximum rate of a person's exhalation or to how fast and how long can they exhale.

This rate is measured by using the peak expiratory flow test that requires you to deeply inhale and then exhale as fast as hard as possible.

This rate is closely related to the amount of air flowing out of the lungs. This implies the peak expiratory flow is lower if there is a disease such as asthma that can constrict the airways.

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Answer:

B

Explanation:

Answer:

average speed= total distance/total time

Explanation:

Related to the distance traveled by the object which means, it is only represented by the magnitude and direction of travel is not important. so average speed= total distance/total time

Answer:

so you have your boxes go by tens on the x-axis okay which is the one going straight up then the second number goes by 5's is your y-axis. 30,6 so 30 is your x and 6 is your y. start at 0,0 which is the little corner piece. find 30 on x and then go over 6 and that's it!

Explanation:

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: [tex]\frac{∆v}{t}[/tex]

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 [tex]\frac{feet}{second}[/tex] then sped up to 5 [tex]\frac{feet}{second}[/tex], then the change in my speed would be 2 [tex]\frac{feet}{second}[/tex], because I started walking 2 [tex]\frac{feet}{second}[/tex] faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 [tex]\frac{meters}{second}[/tex] to 16 [tex]\frac{meters}{second}[/tex]  in 4 seconds. What would the acceleration be? Let's set up an equation:

a = [tex]\frac{∆v}{t}[/tex]

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 [tex]\frac{meters}{second}[/tex] by 4 [tex]\frac{meters}{second}[/tex]. That makes sense, right? Back to the equation.

a = [tex]\frac{∆v}{t}[/tex]
a = [tex]\frac{16-4}{4}[/tex]

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = [tex]\frac{12}{4}[/tex]

a = 3 ([tex]\frac{meters}{second^{2}}[/tex])

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit [tex]\frac{\frac{meters}{second}}{second}[/tex], which can be simplified down to [tex]\frac{meters}{second^{2} }[/tex])

Let me know if you need clarification on anything I explained here!
- breezyツ

Answer:

The noble gases are located in the 18th group of the periodic table.

Answer: They’re the most reactive metals

Explanation: They have larger ionic radii and low ionization energies

Answer:

48,000 J

Explanation:

m = 60.0 kg

v = 40.0 m/s

kinetic energy = mv^2/2 = 60*40^2/2 = 48,000 J

Answer:

Force can be defined as the product of mass and acceleration.

Explanation:

~Cornasha_Weeb

Answer:

What is the effect of clouds on the "output" of a Solar PV module?

Clouds do impact photovoltaic panels. The quantity of power your photovoltaic panels can produce is directly dependent on the level of light they receive. ... They will see complete direct sunlight “plus” reflected light from the clouds! They will drink in more energy than they could on a cloudless day!

Explanation:

I hope it will help you...

#carryonleraning

Hi there!

We can use the following equation for the period of a pendulum:

[tex]\large\boxed{T = 2\pi \sqrt{\frac{L}{g}}}[/tex]

T = Period (s)
L = length of simple pendulum (.75 m)

g = acceleration due to gravity (9.8 m/s²)

Plug in the values:

[tex]T = 2\pi \sqrt{\frac{.75}{9.8}} = \boxed{1.738 s}[/tex]

Answer:

because they have opposite charges

Explanation:

If you're talking about magnetism and electricity, the laws are that like charges attact and and unlike charges repel.

Charged particles interact with each other because they have opposite charges.

When charged particles are close to each other, their electric fields interact. So particles with opposite charges attract each other whereas particles with similar charges repel each other.

So we can conclude that Charged particles interact with each other because they have opposite charges.

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Answer:

429.424

Explanation:

Trust :)

The wavelength of the light passing through the single slit is approximately 429 nanometers.

To calculate the wavelength of the light passing through a single slit, we can use the formula for the diffraction of light:

w × sin(θ) = m ×λ

Where:

w is the width of the slit

θ is the angle of the diffraction minimum

m is the order of the diffraction minimum

λ is the wavelength of the light

Given:

Width of the slit (w) = 8.77 × 10^(-6) m

Angle of the diffraction minimum (θ) = 5.62º = 5.62° = 5.62 × π/180 radians

Order of the diffraction minimum (m) = 2

To solve for the wavelength (λ),  rearrange the formula:

λ = (w × sin(θ)) / m

Substituting the given values:

λ = (8.77 × 10^(-6) m × sin(5.62° × π/180)) / 2

Calculating this expression:

λ ≈ (8.77 × [tex]10^{(-6)[/tex] m × sin(0.098)) / 2

λ ≈ (8.77 ×[tex]10^{(-6)[/tex] m × 0.098) / 2

λ ≈ 4.29 ×[tex]10^{(-7)[/tex] m

To convert the wavelength to nanometers, we multiply by[tex]10^9[/tex]:

λ = 4.29 × [tex]10^{(-7)[/tex] m × 10^9 nm/m

λ ≈ 429 nm

Therefore, the wavelength of the light passing through the single slit is approximately 429 nanometers.

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Answer:

354875 J

Explanation:

The equation for Work is:

[tex]Work=Force*Distance[/tex]

We can plug in the given values into the equation:

[tex]Work=4250N*83.5m=354875J[/tex]

Answer:

5.3 hours

Explanation:

(3180 km) / (600 km/h) = 5.3 hours

Answer:

B

Explanation:

From Kepler's Third Law, T^2 is proportional to r^3.

Now r of B is larger than r of A,

so T of B is larger than T of A.

Then B takes the longest time.

Dwarf Galaxy B takes the longest time to orbit the large galaxy. This can be explained using Kepler's Third Law of Planetary Motion, also known as the Law of Periods.

Kepler's Third Law states that the square of the orbital period (T) of a celestial body is directly proportional to the cube of the semi-major axis (a) of its elliptical orbit. Mathematically, it can be expressed as:

[tex]\[ T^2 \propto a^3 \][/tex]

Where:

T is the orbital period (time taken for one complete orbit), and

a is the semi-major axis of the elliptical orbit.

In the case of dwarf galaxies A and B revolving around the large galaxy, we can use the distances from the large galaxy to each of the dwarf galaxies as the semi-major axes (a) of their orbits.

Let's compare the values of [tex]\( a^3 \)[/tex] for the two dwarf galaxies:

For Dwarf Galaxy A:

[tex]\( a_{\text{A}} = 70,000 \)[/tex] light years

For Dwarf Galaxy B:

[tex]\( a_{\text{B}} = 500,000 \)[/tex] light years

Now, let's calculate [tex]\( a^3 \)[/tex] for each galaxy:

For Dwarf Galaxy A:

[tex]\( a_{\text{A}}^3 = (70,000)^3 = 343,000,000,000 \)[/tex] cubic light years

For Dwarf Galaxy B:

[tex]\( a_{\text{B}}^3 = (500,000)^3 = 125,000,000,000,000,000 \)[/tex] cubic light years

Since [tex]\( a_{\text{B}}^3 \)[/tex] is significantly larger than [tex]\( a_{\text{A}}^3 \)[/tex], it means that Dwarf Galaxy B has a larger a value and therefore a longer orbital period according to Kepler's Third Law. In other words, Dwarf Galaxy B takes the longest time to orbit the large galaxy.

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Answer:

I don't know what are you saying but I don't have any results

Explanation:

[tex]kai6417[/tex]

When you double the speed of your car, your braking distance quadruples. every time you double your speed, you multiply your braking distance by four.

The stopping distance of a vehicle is the sum of breaking distance and thinking distance. The stopping distance of a vehicle when we increase the velocity from 10mph to 30 mph, increases from30 feet to 90 feet.

Stopping distance is the distance covered by a vehicle when we stops. Distance is the product of time and velocity. As the velocity increases then, the time taken to cover the distance will be lower.

Stopping distance mathematically is the sum of breaking distance and thinking distance .Thinking distance is simply the speed and it is in fact means how long it takes on average to think for a break.

Breaking distance is means how long the driver takes to break or stop in actual. In general, the breaking distance is speed multiplied by 2.

Thus, for 10 mph, the thinking distance is 10 feet and breaking distance is 10 × 2 = 20 feet and therefore, the stopping distance is 10 + 20 =30 feet.

Similarly for 30 mph, breaking distance is 60 feet and thinking distance is 30 feet thus, having 60 + 30 = 90 feet of stopping distance.

Therefore, when the speed increased from 10 mph to 30 mph, stopping distance increases from 30 feet to 90 feet.

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In celestial mechanics, escape velocity or escape speed is the minimum speed needed for a free, non-propelled object to escape from the gravitational influence of a primary body, thus reaching an infinite distance from it.

PLEASE MARK AS BRAINLIEST

As the angle the plane makes with the horizontal is increased, the magnitude of the force of static friction between the block and the plane (2) increases

Since the static frictional force between the block and the plane is given by F = μN where

Now, F = μN

F = μWcosθ

Since μ and W are constant, we see that

F ∝ cosθ and cosθ ∝ θ

So, F ∝ θ

Since the static frictional force is directly proportional to the angle the plane makes with the horizontal, when the angle the plane makes with the horizontal is increased, the static friction between the block and the plane increases.

So, As the angle the plane makes with the horizontal is increased, the magnitude of the force of static friction between the block and the plane (2) increases

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oh ok where is it bring it

A car is moving at 16 m/s at a time t = 0. With an acceleration of 0.50t, in m/s², for t, in seconds, it slows down. At t = 8 seconds, it stops. Option D is correct.

The change of displacement with respect to time is defined as the velocity.  velocity is a vector quantity.

Given data;

t = 0 ,u=16 m/s

acceleration = −0.50t m/s²

v=0,t=?

The acceleration is found as;

[tex]\rm a= \frac{dv}{dt} \\\\\ dv=adt \\\\ dv=-0.50 \ t dt[/tex]

Intregating both sides we get;

[tex]\rm [v]_0^{16}=-0.50\frac{t^2}{2} \\\\ 16=0.25 t^2\\\\\ t^2=\frac{16}{0.25} \\\\\ t= 8 \ sec[/tex]

Hence, the car stops at t=8sec. Option D is correct.

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