A ball is released freely from the top of a building 80m high. Find its speed.​

Answer:

Black hole accretion disk emit in the X-rays but neutron star disks are cooler

Hope this Helps

we know 1kg=9.8N​ so 30 kg= 30 x 9.8 = 294 N

work is done when a 30 kg mass is to be lifted through a height 6m :

A = 294 x 6 = 1764 J

ok done. Thank to me :>

Answer:

This demonstration is often done following a discussion of the ideal gas equation of state, PV=nRT.

We begin by weighing a balloon, then blowing it up and weighing it again. In the photo shown on right, the mass indication increased from 3.4 to 3.5 grams. At this point, it is important to note that the scale measures force, even though it reports a conclusion about mass based on the force measurement.

One assumption made in reaching the conclusion is that the buoyant force on the object being weighed is negligible. In the case of the balloon, this is incorrect. The buoyant force on this balloon is equal to the weight of the air displaced.

Since the volume of air inside the balloon is essentially the same as the volume of air displaced, we should expect that the buoyant force would support the weight of the air inside the balloon: The reported mass should not go up at all, because the force required of the scale should not change.

The increase in reported mass of .1 gram is attributed to the higher density of the air inside the balloon: The tension in the balloon compresses the air inside, as attested by the pressure required to blow the balloon up. Evidently, for this experiment, the pressure inside is greater than atmospheric by about 2%.

In the picture at right, the balloon is being pressed into a pan of liquid nitrogen. (The pan is the styrofoam lid of a small lunch box.) The balloon floats lightly on the liquid nitrogen unless pressed down. Pressing down places more surface area in contact with the cold nitrogen and speeds the demonstration. It is interesting to note the buoyant force by this liquified constituent of air.

The balloon shrinks dramatically, as indicated below. When left in contact with the liquid nitrogen long enough (perhaps 5 minutes) the oxygen inside the balloon liquifies, and then the nitrogen liquefies also. Close observation of the photo at the upper left corner of the pan shows some liquid nitrogen bubbles may forming above the dark spot in the center of the pan. One can also make out a faint line at the upper left corner of the pan which is the liquid nitrogen surface. The balloon still floats, riding rather high on that surface. Evidently, some of the balloon contents remain in the gas phase, making the mass of the balloon less than the mass of the displaced liquid nitrogen.

Next, we take the shrunken balloon and place it back on the scale, as above. In this instance, the reported mass is 8.7 grams, an increase of 5.2 grams.

A look at the figure on the right shows a faint line near the bottom of the cold balloon. Above that line, the balloon contains gas; below the liquid. That line represents the top surface of the liquid air inside the balloon. With this evidence, the easy thing to say would be, "Of course, liquids are heavier than gases," but that would be incorrect. We assert that the amount of air inside the balloon has not changed and that the mass of that air is not dependent on temperature.

If these assertions are true, then the force of gravity on the balloon has not changed. The scale reading is determined by the force which it must exert on the balloon in order to keep it stationary. Evidently, the required force is larger when the balloon is shrunken. The reason is that the buoyant force (upward) has decreased to practically zero, leaving the scale alone to balance the downward force by gravity.

From the data, we can say that the change in the buoyant force is equal to the weight associated with the apparent change in mass. The weight of 5.2 grams is about .052 newtons. The buoyant force is less now because the balloon displaces less air. If we could measure the change in volume of the balloon as DV, then the buoyant force would be (r g DV) upwards, where r is the density of air that was displaced by the balloon, and g is the gravitational field strength, 9.8 Newton/kg.

Note that the .052 newton force is not the weight of the air inside the balloon. Rather, it is the weight of the air that was displaced by the balloon. If we ignore the compression of air inside the balloon, the two numbers are the same. However, the two samples are completely different.

We can estimate the volume of the balloon by assuming that the hand in the photograph is about .1meters across. For purposes of estimation, we say that the volume shrank to almost zero when the balloon was cold so that the change in volume was nearly equal to the original volume. Plugging in numbers gives fair agreement with the book value of 1kg/cubic meter for the density of air.

The value for the density of air is secondary to two main features of this demonstration:

Large changes in temperature produce the large changes in volume that are indicated by the ideal gas equation.

The mass of air in a volume equal to the volume of a balloon can be determined provided that the buoyant force is understood.

Answer: about 29.4 Earth years

Orbit and Rotation

One day on Saturn takes only 10.7 hours (the time it takes for Saturn to rotate or spin around once), and Saturn makes a complete orbit around the Sun (a year in Saturnian time) in about 29.4 Earth years (10,756 Earth days).

Answer:

Choice C: Approximately [tex]1.64[/tex].

(Assuming that light entered into this glass block from air.)  

Explanation:

Let [tex]n_{1}[/tex] denote the refractive index of the first medium (in this example, air.) Let [tex]\theta_{1}[/tex] denote the angle of incidence.

Let [tex]n_{2}[/tex] denote the refractive index of the second medium (in this example, the glass block.) Let [tex]\theta_{2}[/tex] denote the angle of refraction.

By Snell's Law:

[tex]n_{1} \, \sin(\theta_{1}) = n_{2}\, \sin(\theta_{2})[/tex].

Rearrange the equation to find an expression for [tex]n_{2}[/tex], the refractive index of the second medium (the glass block.)

[tex]\begin{aligned}n_{2} = \left(\frac{\sin(\theta_{1})}{\sin(\theta_{2})}\right)\, n_{1}\end{aligned}[/tex].

The refractive index of air is approximately [tex]1.00[/tex]. Substitute in the values and solve for [tex]n_{2}[/tex], the refractive index of the glass block:

[tex]\begin{aligned}n_{2} &= \left(\frac{\sin(\theta_{1})}{\sin(\theta_{2})}\right)\, n_{1} \\ &= \left(\frac{\sin(46^{\circ})}{\sin(26^{\circ})}\right)\times 1.00 \\ &\approx 1.64\end{aligned}[/tex].

Explanation:

hope this is right!!!!!!!!!!!!!

1. a. Electron 1 -1

b. Proton 2000 +1

c. Neutron 2000 0

2. a. An uncharged nucleon is neutron. It is a chargeless subatomic particle.

b. The particle with the least mass is electron. The mass of an electron is so small that it is considered to be negligible while calculating the atomic mass of an element.

c. The particle with the same mass as a neutron is proton. However if we see the absolute mass of neutron and proton then it is seen that neutron has a greater mass than proton.

d. The particle with the same amount of charge as an electron is proton. Note that the amount of charge is only same. Electron is negatively charged while proton is positively charged.

e. A particle that is negatively charge is electron. The relative charge of electron is -1.

Hope you could understand.

If you have any query, feel free to ask.

Answer:

False

Explanation:

To find how many liters 251mL is you would need to divide 251 by 100 which is 0.251.

Answer:

She took it as priority and got to work

Answer:

stored Energy = Power x time = 65 x 25x 60 = 105.6 Kjoules

Temperature inversion leads to an increase in temperature as altitude increases.

The term temperature inversion refers to a situation in which a layer of warm air lies over a layer of cool air. This is also referred to as thermal inversion. This occurs when the air below to loose heat rapidly.

One of the effects of temperature inversion is reduction in visibility. So, thermal inversion leads to an increase in temperature as altitude increases.

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If the spring constant is known, the extension or compression of the spring can be used to calculate the applied force (pull or push) on the spring.

According to Hook's law, the force applied to an elastic material is directly proportional to the extension of the material.

The force applied;

F = kx

where;

The applied force can be in form of push or pull. A push force can result in compression of the spring while a pull force can result in extension of the spring.

If the spring constant is known, the extension or compression of the spring can be used to calculate the applied force (pull or push) on the spring.

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Answer:

The thermodynamic processes that occur in nature convert thermal energy into mechanical energy.

The thermodynamic processes that occur in nature are according to the law of thermodynamics.

The thermodynamics processes seen in nature are guided by the law of thermodynamics. The role of energy and heat transmission in physical systems is guided by these laws.

The laws are:

(1) Energy cannot be created nor destroyed in a system, according to the first law of thermodynamics, Also, referred to as the law of energy conservation. Only from one form to another, the energy can be transmitted or altered. Energy is conserved through natural processes, keeping the overall amount constant.

(2) Second law of thermodynamics: states that the overall entropy of an isolated system tends to increase over time in any natural process. This suggests that processes progress towards a state of increasing disorder.

Hence, the thermodynamic processes are guided by the law of thermodynamics.

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Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 50 × 9.8 × 600 = 294,000

We have the final answer as

Hope this helps you

Answer:

it is 1 / square root of the change in distance

Explanation:

since the equation of the gravitational force is

[tex]F=G{\frac{m_1m_2}{r^2}}[/tex]

then F and r are inversely related, by the equation

[tex]F \alpha \frac{1}{r^2}[/tex]

this means when F is multiplied by 2 then r^2 is multiplied by half

so when r is multiples by half we first square the number (0.5) ^ 2 = 0.25

so

[tex]F \alpha \frac{1}{(0.25)r^2}[/tex]

which means F is multiples by 1/0.25 or 4 when the radius is halfed

Answer:

It is defined as an astronomical phenomenon which occurs when one spatial object comes within the shadow of another spatial object.

Answer:

5.7km/h north and 5.8km/h west are instantaneous speeds, while 8.1km/h Northwest is the average speed.

Hi there!

Since the ball is thrown with an initial HORIZONTAL velocity, we can treat this as a free-fall situation since the horizontal motion does NOT impact the ball's vertical motion.

We can use the derived kinematic equation:

[tex]t = \sqrt{\frac{2h}{g}}[/tex]

Plug in the given values:

[tex]t = \sqrt{\frac{2(150)}{9.8}} = \large\boxed{5.53 s}}[/tex]

The angle of projection is 0° because there is no vertical component to the velocity as the initial velocity is purely horizontal.

Range:

We can use the equation:

dₓ = vₓt

displacement (x direction) = velocity (x direction) · time

Used the solved-for time and given velocity:

dₓ = 100 · 5.53 = 553 m

Answer: 540 g m/s.

Explanation:

Answer: A.

A momentum of 540 gram meters per second!

Explanation: On Edge!!

Answer:

100 km /hour

Explanation:

/ or per = every once so every hour you are traveling 100 kilometera

Answer:

100kmh-1

Explanation:

100km÷1hour

100kmh-1

Answer:

Explanation:

Ignoring air resistance

1) from vertical rest

s = ½at²

t = √(2s/a)

t = √(2(80)/9.8)

t = 4.040610...

t = 4.0 s

2) d = vt

   d = 25(4.0)

   d = 100 m

3) v² = u² + 2as

   v² = 25² + 2(9.8)(80)

   v² = 2193

   v = 46.8294...

    v = 47 m/s

θ = - 90 + arcsin(25/46.8294)

θ = -57.7339...

θ = 58° below the horizontal

Answer:

Od- Wooden

Explanation:

It would not make sense if it was non-wood because that make it no a wood base product, OC is not right because its not a a plant that produces wood as its structural tissue and thus has a hard stem.   An OA justT o take or get a supply of wood. so its Od

Answer:

"work against the force of the air in the tire"

The air in the tire provides a force opposing the force of the air provided by the plunger.

Answer:

Pushing a car and a truck

Explanation:

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

[tex] E_{t} = KE_{i} + PE_{i} [/tex]

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

[tex] E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J [/tex]

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

[tex] PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J [/tex]

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

[tex] KE_{A} + PE_{A} = KE_{B} + PE_{B} [/tex]

[tex] KE_{B} = KE_{A} + PE_{A} - PE_{B} [/tex]

Since

[tex] KE_{A} + PE_{A} = KE_{i} + PE_{i} [/tex]

we have:

[tex] KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J [/tex]

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

[tex] PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J [/tex]

5) The kinetic energy of the roller coaster at point C is:

[tex] KE_{i} + PE_{i} = KE_{C} + PE_{C} [/tex]            

[tex] KE_{C} = KE_{i} + PE_{i} = 19820 J [/tex]      

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

[tex] KE_{C} = \frac{1}{2}mv_{C}^{2} [/tex]

[tex] v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s [/tex]

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

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Answer:

Ink

mud

hcl

ocean water

co2

h20

The slope of the graph is acceleration of the car with a value of 3.2 m/s².

The given parameters;

The initial velocity of the car in m/s is calculated as follows;

[tex]60 \ \frac{km}{hr} \times \frac{1 \ hr}{3600 \ s} \times \frac{1000 \ m}{1 \ km} = 16.67 \ m/s[/tex]

The acceleration of the car is calculated as follows;

[tex]a = \frac{v- u}{t} \\\\a = \frac{80 - 16.67 }{20} \\\\a = 3.2 \ m/s^2[/tex]

The graph of the velocity against time and acceleration against time is in the image uploaded.

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