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Irina-Kira [14]
3 years ago
10

How do you know if an experimental result is acceptable or trustworthy? What gives you confidence that your data are trustworthy

?
Physics
1 answer:
Virty [35]3 years ago
6 0

Explanation:

For an experimental result to be considered acceptable, all relevant variables involved in the experiment must be taken into account, by isolating it, performing it under controlled conditions and modifying the conditions under which it takes place. This, with the objective of excluding alternative explanations in the analisis of the experimental data. Therefore, if these steps are followed appropriately, experimental data are trustworthy. The reliability of the experiment increases when it is replicated by other researchers and the same results are obtained.

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A submarine is 2.99 ✕ 102 m horizontally from shore and 1.00 ✕ 102 m beneath the surface of the water. A laser beam is sent from
Vsevolod [243]

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of  incidence  beam striking the water is \theta = 49.63^o

c  the angle of  refraction  beam striking the water is r = 59.7^o

d The angle the refracted beam make with respect to the horizontal is = 30.3^o

e The height of the target above sea level is  

                   h= 125.05m

Explanation:

From the diagram we see that the angle of the beam striking the water is

                   tan \theta = \frac{100}{85}

                        \theta = tan^{-1}(\frac{100}{85} )

                           = 49.63^o

According to Snell's law

                   \mu_{water} *sin(i) = \mu_{air}  *sin(r)

Where \mu_{water } is the refractive index of water =  1.333

           i is the angle of incidence

          \mu_{air} is the refractive index of air  = 1

            r is the angle of refraction

 Substituting values accordingly

          1.33 * sin (40.37) = 1 * sin(r)

    Making r the subject of the formula

                       r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})

                          = 59.7^o

looking at the diagram we can see that to  obtain the angle the refraction beam makes with the horizontal   by subtracting the angle refraction from 90°

                 i.e  90 -59.7 = 30.3 °

From the diagram we see that the height  target above sea level can be obtained by this relation

                   tan \theta = \frac{h}{214}\\

Where h is the is the height

                   tan(30.3) = \frac{h}{214}

                         h = 214 * tan (30.3)

                            =125.05m

           

4 0
3 years ago
Which of the following is not in the form of waves?
Nady [450]
C. Seismic energy
This is energy that is released in earthquakes.
6 0
3 years ago
Read 2 more answers
6. A football is kicked with an initial speed of 10.2 m/s at an angle of 40.00 above the horizontal. It lands on the ground 2.12
igor_vitrenko [27]

Answer:

7.81 m/s

Explanation:

Given,

initial speed, u = 10.2 m/s

angle of inclination, θ = 40°

time, t = 2.12 s

Horizontal component of the velocity:

u_x = u cos \theta

u_x = 10.2\times cos 40^0

u_x = 7.81 m/s

In projectile motion horizontal component of the velocity remain same at every point because there is no acceleration.

So, Velocity at the Pinnacle is equal to 7.81 m/s

7 0
3 years ago
How much energy is needed to heat and melt 3.0 kg of copper initially at 83°C?
Ne4ueva [31]

As we know that in order to melt the copper we need to take the temperature of copper to its melting point

So here heat required to raise the temperature of copper is given as

Q = ms\Delta T

We know that

melting temperature of copper = 1085 degree C

Specific heat capacity of copper = 385 J/kg C

now we have

Q = 3(385)(1085 - 83)

Q = 1157310 J

Q = 1157.3 kJ

now in order to melt the copper we know the heat required is

Q = mL

here we know that

L = 205 kJ/kg

now from above formula

Q = 3(205) kJ

Q = 615 kJ

now total heat required will be

Q = 1157.3 kJ + 615 kJ

Q = 1772.3 kJ

As we know that

1 Cal = 4.18 kJ

now we have

Q = \frac{1772.3}{4.18} = 430 KCal

6 0
3 years ago
Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees res
damaskus [11]

Answer:

Same, the electric fields point in opposite directions and therefore cancel at some midpoint.

Explanation:

The Electric field net at some point between them is zero, only if they point in opposite direction (they cancel to the each other). In order the electric fields  have opposite direction, at some point between the bees , the bees must have the same sign of electric charge

3 0
3 years ago
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