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3 years ago

You throw a ball upwards at 22 m/s. How high will it go?

Physics

2 answers:

Anna11 [10]3 years ago

7 0

Answer:

24.69 meters

Explanation:

sorry if it's not right.

NNADVOKAT [17]3 years ago

6 0

answer:

$h=24.69m$

step-by-step explanation:

$eg=mgh \\ek=\frac{1}{2} mv^2$

eg= gravitational energy

ek= kinetic energy

now, since no mass is given of the ball, both equations on their own do nothing for us, except leave us scratching our heads wondering how to figure out the problem. but, since the question states, “and no air resistance,” we now know, according to the law of conservation of energy, that the energy of the two equations will equal each other because none of the energy has dissipated or left the system.

the amount of energy present during the initial phase of the woman about to throw the ball will be present in the final phase, which will be at its highest point (according to this problem).

so now $eg=ek$

knowing this, we can now set the equations equal

$eg=ek\\mgh=\frac{1}{2} mv^2$

the two m’s cancel out, making the mass of the ball insignificant and not influential; next, substitute the values we are given in the problem

$(22m/s),(9.8m/s^2)\\m(9.8m/s^2)h=\frac{1}{2} m(22m/s)^2\\(9.8 m/s^2)h=\frac{1}{2} (22m/s)^2\\(9.8m/s^2)h=1/2 (484m^2/s^2)\\(9.8m/s^2)h=1/2 (242m^2/s^2)\\\\h= (242m^2/s^2)/(9.8m/s^2)$

as you can see, all units that need to be canceled out are indeed canceled, leaving us with just m, meters, which is what height is measured in

therefore, $h=24.69m$

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Romashka [77]

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3 years ago

When light propagates through two adjacent materials that have different optical properties, some interesting phenomena occur at

Veseljchak [2.6K]

Answer:

First, the different indices of refraction must be taken into account (in different media): for example, the refractive index of light in a vacuum is 1 (since vacuum = c). The value of the refractive index of the medium is a measure of its "optical density": Light spreads at maximum speed in a vacuum but slower in others transparent media; therefore in all of them n> 1. Examples of typical values of are those of air (1,0003), water (1.33), glass (1.46 - 1.66) or diamond (2.42).

The refractive index has a maximum value and a minimum value, which we can calculate the minimum value by means of the following explanation:

The limit or minimum angle, α lim, is defined as the angle of refraction from which the refracted ray disappears and all the light is reflected. As in the maximum value of angle of refraction, from which everything is reflected, is βmax = 90º, we can know the limit angle (the minimum angle that we would have to have to know the minimum index of refraction) by Snell's law:

βmax = 90º ⇒ n 1x sin α (lim) = n 2 ⇒ sin α lim = n 2 / n 1

Explanation:

When a light ray strikes the separation surface between two media different, the incident beam is divided into three: the most intense penetrates the second half forming the refracted ray, another is reflected on the surface and the third is breaks down into numerous weak beams emerging from the point of incidence in all directions, forming a set of stray light beams.

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3 years ago

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

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Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

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3 years ago

A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig

Gre4nikov [31]

Answer:

The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame

Explanation:

From the question we are told that

The distance between earth and Retah is $d = 20 \ light \ hours = 20 * 3600 * c = 72000c \ m$

Here c is the peed of light with value $c = 3.0*10^8 m/s$

The time taken to reach Retah from earth is $t = 25 \ hours = 25 * 3600 =90000 \ sec$

The velocity of the spacecraft is mathematically evaluated as

$v_s = \frac{d }{t}$

substituting values

$v_s = \frac{72000 * 3.0*10^{8} }{90000}$

$v_s = 2.40*10^{8} \ m/s$

The time elapsed in the spacecraft’s frame is mathematically evaluated as

$T = t * \sqrt{ 1 - \frac{v^2}{c^2} }$

substituting value

$T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }$

$T = 54000 \ s$

=> $T = 15 \ hours$

So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame

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3 years ago

The melting point of a solid is 90.0C. What is the heat required to change 2.5 kg of this solid at 30.0C to a liquid? The specif

Neko [114]

Hey again!

Ok..

Now... The melting Point of this solid is 90°C.

Meaning That as soon as it gets to this temp... It STARTS Melting.

So at that temp... It still has some solid parts in it.

You can say its a Solid Liquid Mixture.

Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.

After Fusion...It'd then Be a Pure Liquid with no solids in it.

Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c

Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.

Q= Q' + Q"

Q= mc∆0 + ml

∆0 = 90°c - 30°c = 60°c

Q= 2.5(390)(60) + (2.5)(4000)

Q=6.9 x 10⁴Joules

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3 years ago