What is the equation for density?
O D= m
V
O D= V
-
m
O D= m
O D= { mu?
2

Answer:

a)object is in uniform motion

Explanation:

velocity is constant.and a=0

body is in uniform motion.

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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:

Explanation:

The kinetic energy of an object can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass in kg

v is the velocity in m/s

From the question

m = 20 kg

v = 2.5 m/s

From the question we have

[tex]k = \frac{1}{2} \times 20 \times {2.5}^{2} \\ = 10 \times 6.25 \\ = 62.5\: \: \: \: \: \: \: \: \: \: [/tex]

We have the final answer as

Hope this helps you

Answer:it equals 0

Explanation:hope this is a good answer

Angular velocity is the ratio of angular displacement to time. Then the number of turns by the shaft is zero.

The rate of change of angular displacement is called angular velocity. It is a vector quantity.

We know that the angular velocity formula

[tex]\rm Angular \ velocity = \dfrac{Angular \ displacement}{time}[/tex]

The motor shaft turns through between the time when the current is reversed and the instant when the angular velocity is zero.

Then the number of turns will be

[tex]\rm Number \ of\ turns = Angular \ velocty \times time \\\\Number \ of\ turns = 0 *time \\\\Number \ of\ turns = 0[/tex]

The number of turns by the shaft is zero.

More about the Angular velocity link is given below.

https://brainly.com/question/7359669

Answer:

this is an example of Newton's Second Law, acceleration happens when a force acts on a mass.

the force that makes the bicycle move is your foot pushing against the pedal

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: friction :)

Explanation:

I took the test and got 100%

Please find attached photograph for your answer.

Hope it helps.

Do comment if you have any query.

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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Answer: beta particles

Explanation:

that's the answer

Answer:

Ink

mud

hcl

ocean water

co2

h20

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.

Read more here:

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I hope it helps you!

Explanation:

Shorter wavelength since UV light has more energy than visible light

365 nm - What is the peak energy wavelength of UV light?  It allows both infrared daylight and ultraviolet night-time communications by being transparent between 320 nm and 400 nm and also the longer infrared and just-barely-visible red wavelengths. Its maximum UV transmission is at 365 nm

Explanation:

u=0, v=?, a=5m/s², t=8sec

So, by the formula,

v=u+at

v=0+5×8

v=40m/s.

hope this helps you .

Answer:

Pushing a car and a truck

Explanation:

Answer:

It should be mentioned that an efficient way to work this vector addition problem is with the cosine law for general triangles (and since  

a

,

b

 and  

r

 from an isosecles triangle, the angles are easy to figure). However, in the interest of reinforcing the usual systematic approach to vector addition, we note that the angle  

b

 makes with the +x axis  is 30

0

+105

0

=135

0

(a) The x component of  

r

x

​

​

=(10.0m)cos30

0

+(10.0m)cos135

0

=1.59m

(b)  the y component of  

r

 is r

y

​

=(10.0m)sin135

0

 = 12.1 m.

(c) the magnitude of  

r

 is r =  ∣

r

∣=

(1.59m)

2

+(12.1m)

2

​

=12.2m

(d) The angle between  

r

 and the +x direction is tan

−1

[(12.1m)/(1.59m)]=82.5

0

Explanation:

Answer:

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

39°C=1.28L

8°C=? Less

=8°C/39°C×1.28L

=8/39×1.28L

=10.24/39

=512/195

=2.6L

Answer:

[tex]4 ms^{-2}[/tex]

Explanation:

Normal force - should be the reaction from the surface it's on - nullifies the weight of the object. At this point the only force is of [tex]22-14 = 8N[/tex] from whoever is pushing the square and the friction, towards the right. At this point we divide the force by the mass of the object to obtain an acceleration of [tex]8/2 = 4 ms^{-2}[/tex]

Hi there!

We can use the following:

U (Gravitational Potential Energy) = mgh

m = mass of object (kg)

g = acceleration due to gravity OR gravitational field strength

h = height, or displacement (m)

ΔU = Uf - Ui, so:

Uf = 3 × 1.5 × 10 = 45 J

Ui = 3 × 0 × 10 = 0

ΔU = Uf - Ui = 45 - 0 = 45 J

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:

Explanation:

Givens

m = 15000 kg

t = 12.1 seconds

vi = 86 km / hr

vf =0

Formula

F = m*a

a = (vf - vi)/ t

vi has to be converted to m/s

86 km/hr [1000 m / 1 km] * [1 hour / 3600 seconds]

86 * 1000 / 3600 = 23.89 m/s

Solution

a = (vf - vi) / t

a = (0 - 23.89)/12.1

a = - 1.97 m/s^2 The minus means that the trains is slowing down.

F = 15000 kg * - 1.97 m/s^2

F = -29615.7 Newtons.

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:

whats the question

Explanation:

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].

Answer:

D

Unbalanced or net forces acting on the car will cause it to move

The correct answer is option D. The net force on the car is zero​.

Since the car is not accelerating, the force of air resistance and the friction force must add to give the exact opposite of the applied force. The net force here would be zero, as well.

Newton's First Law is the law of inertia. An object with no net forces acting on it which is initially at rest will remain at rest. If it is moving, it will continue to move in a straight line with constant velocity.

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Answer:

a balance and a beaker of water

Explanation:

The mass can easily be measured with balance. If you drop the key in a beaker of water, the water inside the beaker will increase and this increases the volume of water that will be equal to volume of key.

Answer:

physical science

Explanation:

i guess??