The bus is heading southwest at 56 miles per hour is it veloctiy or speed

Answer:

Velocity = -20.45 i + 35.54 j km/h

Explanation:

given data

speed v = 41.0 km/h

direction = 60.08 north of west

solution

first we get here Speed in west direction that is express as

Speed in west direction = v × cos(60.08)     ..........................1

Speed in west direction = 41 × cos(60.08)

Speed in west direction = 20.45 km/h

now we get Speed in north direction that is

Speed in north direction = v × sin(60.08)        ............................2

Speed in north direction = 41 × sin(60.08)

Speed in north direction = 35.54 km/h

so Velocity will be here

Velocity = -20.45 i + 35.54 j km/h

Answer:

Power supplied to the pump is 980000 W or 980 KW

Note: The question is incomplete. The complete question is given below:

On a day with no wind, a fountain in Switzerland propels 30000 kg of water per minute to a height of 140 metres.

a)Calculate the power used in raising the water?

b)The efficiency of the pump which operates the fountain is 70%

calculate the power supplied to the pump?

Explanation:

Efficiency = Power output/Power input * 100%

Power output = work done/ time

where work done = mass * acceleration due to gravity (g) * height

time = 1 minute = 60 seconds

mass of water = 30000 kg, height = 140 meters, g = 9.8 m/s²

Then, work done = 30000 * 140 * 9.8 = 41160000 J

Power output then = 41160000 J / 60 s

Power output = 686000 W

The power input can then be derived from the formula of efficiency

Power input = (Power output * 100)/Efficiency

Power input = (100 * 686000) / 70

Power input = 980000 W

Therefore, power supplied to the pump is 980000 W or 980 KW

Answer:

the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg

Explanation:

Given the data in the question;

first we compute the total volume of mud

V = ( 1.9×1000)m × (0.55×1000)m × (2.4×1000)m

V = 2.508 × 10⁹ m³

same volume of mud spread uniformly over 1.4 × 1.4 km² area

Hence,

V = (1.4×1000)m × (1.4×1000)m × depth

2.508 × 10⁹ = 1400m × 1400m × depth

2.508 × 10⁹ m³ = 1960000m² × depth

depth = 2.508 × 10⁹ m³ / 1960000 m²

depth = 1279.59 m

so volume of the mud sitting above 3.5 m² area of the valley floor will be;

⇒ 3.5 m² × 1279.59 m

⇒ 4478.565 m³

so mass will be

m = 4478.565 × 1900 kg

m = 8509273.5 kg

Therefore, the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg

Answer:

iron/copper

iron faster because it in the core so iron is the answer with would be 15C because its the hottest in the core and is very quick there to being moved by the core so iron being 15C is the answer.

Explanation:

Answer:

convergence and trench

Explanation:

LOL

Answer:

Assumption: there is no object between this point charge and the observer.

Explanation:

The electric field of a point charge is inversely proportional to the square of the distance from that point charge.

Let [tex]k[/tex] denote Coulomb's constant ([tex]k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-1}[/tex].) Let the magnitude of that point charge be [tex]q[/tex]. At a distance of [tex]r[/tex] from this charge, the electric field due to this charge would be:

[tex]\displaystyle E = \frac{k \cdot q}{r^{2}}[/tex].

Convert the magnitude of the point charge in this question to standard units:

[tex]q = 1\; \rm nC = 10^{-9}\; \rm C[/tex].

Apply that equation to find the magnitude of the electric field due to this point charge:

[tex]r = 1\; \rm m[/tex]:

[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(1\; \rm m)^{2}} \\ &\approx 9.0\; \rm N \cdot C^{-1}\end{aligned}[/tex].

[tex]r = 2\; \rm m[/tex]:

[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(2\; \rm m)^{2}} \\ &\approx 2.2\; \rm N \cdot C^{-1}\end{aligned}[/tex].

[tex]r = 3\; \rm m[/tex]:

[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(3\; \rm m)^{2}} \\ &\approx 1.0\; \rm N \cdot C^{-1}\end{aligned}[/tex].

The direction of the electric field at a point is the same as the direction of a force from this field onto a positive point charge at this point.

Because the [tex](+1\; \rm nC)[/tex] point charge here is positive, the electric field of this charge would repel other positive point charges. Hence, the electric field around this [tex](+1\; \rm nC)\![/tex] point charge at any point in the field would point away from this charge.

Answer:I think sprinting

Explanation:

Answer:

How can smoking affect the respiratory system?

Smoking and Respiratory Disease

Smoking can cause lung disease by damaging your airways and the small air sacs (alveoli) found in your lungs. Lung diseases caused by smoking include COPD, which includes emphysema and chronic bronchitis. Cigarette smoking causes most cases of lung cancer.

Explanation:

I cannot create videos. However, I can help you develop a script for your health teaching package.

Your lungs and general health can be significantly impacted by smoking tobacco and other substances. It's crucial to comprehend the actual scientific data and take action to manage and avoid respiratory illnesses. Do not hesitate to ask if you have any queries or concerns.

Hello, and thank you for coming to me for health teaching. I understand that you smoke tobacco, and you may have doubts about the impact it can have on your lungs and overall health. I'm here to show you the true scientific facts around smoking and lung disease.

Let's start with the respiratory system. The respiratory system is made up of many intricate structures that work together to deliver oxygen to your body and remove carbon dioxide. When you inhale smoke from tobacco or other substances, you introduce harmful chemicals into your lungs that can damage these structures.

Over time, smoking can lead to a range of respiratory diseases, including chronic bronchitis, emphysema, and lung cancer. Symptoms of these diseases can include coughing, wheezing, shortness of breath, and chest pain. These symptoms can significantly impact your quality of life and even be life-threatening.

Prevention is key when it comes to respiratory diseases caused by smoking. The best way to prevent these diseases is to quit smoking altogether. If you're struggling to quit, there are many resources available to help you, including nicotine replacement therapy and counseling.

If you're experiencing symptoms of a respiratory disease, it's important to seek medical attention. Your doctor can diagnose the disease and recommend a treatment plan that may include medication, lifestyle changes, and possibly surgery. Early diagnosis and treatment can greatly improve your chances of recovery and help you manage your symptoms.

Now, let me show you a diagram of the respiratory system so you can see how smoking affects it. [At this point, you can show the diagram and point out the different structures affected by smoking, such as the bronchi and alveoli.]

Therefore, smoking tobacco and other substances can have a significant impact on your lungs and overall health. It's important to understand the true scientific facts and take steps to prevent and manage respiratory diseases. If you have any questions or concerns, please don't hesitate to ask.

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That depends on the circle's radius, which you neglected to mention.

If the radius is 1 meter, then an arc of 3.05 degrees is 5.32 centimeters long.

If the radius is 1 kilometer, then an arc of 3.05 degrees is 53.2 meters long.

Explanation:

Q= mc∆T

∆T= 5-24=- 19

Q= 0.5*4186*-19

Q= -39767 J

negative sign show heat releases

Answer:

south

Explanation:

Answer:

The north and south pole of a solenoid depends on two factors. One, the direction of the current flow and two, the direction of the winding (clockwise or counter-clockwise). ... If clockwise in relation to the positive wire then is the south pole, if anti-clockwise then is the north pole.

Explanation:

The north and south pole of a solenoid depends on two factors. One, the direction of the current flow and two, the direction of the winding (clockwise or counter-clockwise). ... If clockwise in relation to the positive wire then is the south pole, if anti-clockwise then is the north pole.

Answer:

4.9 x 10-23 N

Explanation:

I took the test

Answer:

Factors Affecting the Strength of the Magnetic Field of an Electromagnet: Factors that affect the strength of electromagnets are the nature of the core material, strength of the current passing through the core, the number of turns of wire on the core and the shape and size of the core

Explanation:

Follow me for more

The following factors affect the strength of an electromagnet; number of turns, wire size, current and the presence of iron core.

This is a type of magnet  in which the magnetic field is produced by an electric current.

The following factors affect the strength of an electromagnet;

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Answer:

The answer is B

Explanation:

keep it up

Answer:

and yes, sound travels at the speed of entity expressions is whhy you need math for a job in mc donalds :

293 x gj -2 = 23r = velocity of rocket ship is 2.74

Answer:

it will take 4 seconds to travel 2 meters.

Answer:

Explanation:

Given that:

The area of the metal tube = 1.60 cm² = 1.60 × 10⁻⁴ m²

Length of cube L = 0.150 m

The exerted force by water = 1.00 × 10³ N

At the top of the cube;

The force on water is F = pgL

F = 9.8 m/s² × (1000 kg/m³) (0.150 m)(0.150 m)²

F = 33.075 N

∴

[tex]pgh = \dfrac{F}{A}[/tex]

[tex]h = \dfrac{F}{Apg}[/tex]

[tex]h = \dfrac{33.075}{(1.60 \times 10^{-4} \ m^2) \times (1000 \ kg/m^3)(9.8)}[/tex]

h = 21.09 m

The volume of the water V = L³

V = (0.150 m)³

V = 0.003375 m³

weight of the water w = p(V + Ah)

w = 1000 (0.003375 + (1.60 × 10⁻⁴ × 21.09))

w = 6.75 kg

Answer:

1 - porch, 2 - stay, 3 - electronic

Answer:

change is imperceptible

          w_f = 7.272  10⁻⁵ rad / s

Explanation:

For this exercise we can use the conservation of angular momentum.

Initial. Before the crash

        L₀ = I w₀

final. After the crash

         L_f = I w_f + p r

where the moment is

        p = mv

As the system is formed by the two bodies, the forces during the impact are internal, therefore the angular momentum is conserved

        L₀ = L_f

        I w₀ = I w_f + m v r

        w_f = w₀ - [tex]m \frac{ v \ r }{I}[/tex]

We can approximate the Earth to a sphere, so its angular momentum is

       I = 2/5 M r²

we substitute

         w_f = w₀ - [tex]\frac{5 \ m \ v}{2 \ M \ r}[/tex]

We can find the angular velocity of the Earth with the duration of a spin which is the period of one day

        w₀ = 2π / T

        T = 24 h (3600 s / 1h) = 86 400 s

         w₀ = 2π / 86400

         w₀ = 7.272 10⁻⁵ rad / s

let's calculate

         w_f = 7.27 10⁻⁵ - [tex]\frac{5 \ 250 \ 10^{13} \ 33.0 \ 10^{3} }{ 2 \ 5.98 \ 10^{24} \ 6.37 10^{6} }[/tex]

         w_f = 7.272 10⁻⁵ - 1.0829 10⁻¹³

         w_f = 7.27199999 10⁻⁵

this change is imperceptible

          w_f = 7.272  10⁻⁵ rad / s

Answer:

It is kinetic energy.

Explanation:

The kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate the volleyball from rest to its stated velocity. Having gained this energy from the player spiking it during its acceleration, the body maintains this kinetic energy unless its speed changes.

Answer:

h = 18.4 cm

Explanation:

Given that,

The speed of the bob at the bottom of the swing is 1.9m/s.

We need to find the initial height of the bob. Let it is h.

We can find it using the conservation of energy i.e.

[tex]mgh=\dfrac{1}{2}mv^2[/tex]

Where

v is speed of the bob

So,

[tex]h=\dfrac{v^2}{2g}\\\\h=\dfrac{(1.9)^2}{2\times 9.8}\\\\h= 0.184\ m[/tex]

or

h = 18.4 cm

So, the initial height of the bob is 18.4 cm.

The initial height of bob will be "18.4 cm".

Speed would be the scalar quantity that describes "how quickly an attribute moves." This could sometimes be defined as this same rate whereby an object travels a given distance.

According to the question,

The speed of the bob = 1.9 m/s

Let,

The initial height be "h".

By using the conservation of energy, we get

→ mgh = [tex]\frac{1}{2}[/tex]mv²

or,

→      h = [tex]\frac{v^2}{2g}[/tex]

By substituting the values, we get

          = [tex]\frac{(1.9)^2}{2\times 9.8}[/tex]

          = 0.184 m or,

          = 18.4 cm

Thus the above answer is correct.

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Answer:

A.

Explanation:

A large fraction of astronomical gamma rays are screened by Earth's atmosphere. Gamma rays are ionizing radiation and are thus biologically hazardous. Due to their high penetration power, they can damage bone marrow and internal organs.

The extremely high energy of gamma rays allows them to penetrate just about anything. They can even pass through bones and teeth. This makes gamma rays very dangerous. They can destroy living cells, produce gene mutations, and cause cancer.

Answer:

a) the markers are 29.76 m far apart in the laboratory

b) According to B, the markers are 23.21 m far apart in the laboratory

Explanation:

Given that;

speed of particle = 0.624c = (0.624 × 3×10⁸m/s)

{ where c is the speed of light}    

lifetime of the particle t = 159ns  = 1.59 × 10⁻⁷

a) Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory

lets apply the formula distance = velocity × time

lets substitute

distance = (0.624×3×10⁸ m/s)  × (1.59 × 10⁻⁷ s)

distance = 29.76 m

Therefore, the markers are 29.76 m far apart in the laboratory

b) Observer B, who observes the particle to be at rest and measures its lifetime to be 124ns. According to B, how far apart are the two markers in the laboratory?

lifetime of the particle t = 124ns  = 1.24 × 10⁻⁷

distance = (0.624×3×10⁸ m/s)  × (1.24 × 10⁻⁷ s)

distance = 23.21 m

Therefore,  According to B, the markers are 23.21 m far apart in the laboratory

Answer:

Gigantic towers, monuments and tall structures are always supported by a very thick layer of concrete that’s blended into their foundation. The purpose of this is quite straightforward: nothing provides greater support to a super heavy (and stationary) structure than a layer of concrete.  There may be other things that, in theory, could provide a stronger foundation, but they wouldn’t be nearly as cost effective as concrete.

Answer:

Because it has essentially zero strength in tension, it is almost always used as reinforced concrete, a composite material. It is a mixture of sand, aggregate, cement and water.

Answer:

Potential difference [tex]V=1.6\times 10^{-10}\ V[/tex]

Explanation:

Capacitance of capacitor, C = 0.9 MF

Charge, [tex]Q=1.4\times 10^{-4}\ C[/tex]

We need to find the amount of potential difference. We know that,

[tex]Q=CV[/tex]

V is potential difference

So,

[tex]V=\dfrac{Q}{C}\\\\V=\dfrac{1.4\times 10^{-4}}{0.9\times 10^6}\\\\V=1.6\times 10^{-10}\ V[/tex]

So, the required potential difference is [tex]1.6\times 10^{-10}\ V[/tex].

Answer:

100

w=f*s

20*5=100....

100 Joules of work is done on the cart.

Work done by a force is defined as the product of the displacement and the component of the applied force on the object in the direction of displacement. When we push a block with some force, the body moves with some acceleration, so it is called work done.

Work done is expressed as W=Fd and its unit is joules which can be defined as the amount of work done by a force in Newton is applied to an object, as a result of which it is displaced in meter.

For above given information,

Force= 20 N

Distance= 5 m

So, work done= 20*5= 100 Joules

Thus, 100 Joules of work is done on the cart.

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Answer: look at the file

yes it is boiling creating bubbles that aren't their originally

Explanation:

When cooking oils are subjected to heat in the presence of air and water (from food), such as in deep-fat frying and sautéing (pan frying), they can undergo at least three chemical changes: 1) oxidation of the fatty acids, 2) polymerization of the fatty acids, and 3) breaking apart of the triglyceride molecules into free fatty acids and glycerol by hydrolysis (reaction with water from the food being cooked) (Choe and Min 2007)