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3 years ago

What is an example for outcome

Physics

2 answers:

Jet001 [13]3 years ago

4 0

Answer:

An example of outcome: It was observed that a higher concentration of nitrogen increases the growth of bean seedlings.

Explanation:

Outcome is the phase of a scientific experiment where the researcher analyzes each of the results to see if they are sufficient to explain each of the problems raised and whether they are in accordance with the hypotheses.

If the results are not satisfactory, new hypotheses can be raised for new experiments to occur. If the results of the experiment are satisfactory, the scientist moves on to the conclusion stage.

SCORPION-xisa [38]3 years ago

3 0

Outcome- possible results of an experiment is called an outcome.

ex- 2 is a possible outcome when a number cube is rolled

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A straight vertical wire carries a current of 1.0 A into the page in a region where the magnetic field strength is 0.60 T. What

Advocard [28]

Answer:

force on the wire of section cm will be $6\times 10^{-3}N$

Direction of force on the wire will be in south direction

Explanation:

We have given current in the wire i = 1 A

Magnetic field strength B = 0.6 T

We have to find the force on 1 cm section of the wire so l = 1 cm = 0.01 m

Force on the wire containing current is equal to

$F=iBL$

$F=1\times 0.6\times 0.01=6\times 10^{-3}N$

So force on the wire of section cm will be $6\times 10^{-3}N$

Direction of force on the wire will be in south direction

5 0

3 years ago

Multiple-Concept Example 7 and Interactive LearningWare 26.1 provide some helpful background for this problem. The drawing shows

Vlada [557]

Answer:

n = 1.4266

Explanation:

Given that:

refractive index of crystalline slab n = 1.665

let refractive index of fluid is n.

angle of incidence θ₁ = 37.0°

Critical angle $\theta _c = sin^{-1} (\frac{n}{n_{slab}} )$

$sin \theta _ c =\frac{n}{n_{slab}}$

According to Snell's law of refraction:

$n sin \theta _1 = n_{slab} \ sin \ (90- \theta_c)$

At point P ; $90 - \theta _2 \leq \theta _c$

$\theta _2 = 90 - \theta _c$

Therefore:

$n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)} \\ \\ n \ sin \theta_1 = n_{slab} \sqrt{(1- \frac{n}{n_{slab}} )}$

Then maximum value of refractive index n of the fluid is:

$n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }$

$n = \frac{1.665}{\sqrt{1+ sin^2 \ 37} }$

n = 1.4266

3 0

3 years ago

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.60 m/s. How

kirill115 [55]

Answer:

Efriction = 768.23 [kJ]

Explanation:

In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.

We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.

The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.

$E_{pot}-E_{friction}=E_{kin}$

where:

$E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}$

m = mass = 56 [kg]

h = elevation = 1400 [m]

v = velocity = 5.6 [m/s]

$(56*9.81*1400)-E_{friction}=\frac{1}{2}*56*(5.6)^{2}\\769104 -E_{friction}= 878.08 \\E_{friction}=769104-878.08\\E_{friction}=768226[J] = 768.23 [kJ]$

4 0

3 years ago

At a particular instant the magnitude of the momentum of a planet is 2.60 × 10^29 kg·m/s, and the force exerted on it by the sta

aleksley [76]

Answer:

F=(-4.8*10^22,0,0) N

Explanation:

<u>Given :</u>

We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°

Solution :

We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.

The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next

F=|F|cosФp (1)

Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet

F=|F|cosФp

=-4.8*10^22 N*p

<em>As this force is in one direction, we could get its vector as next </em>

F=(-4.8*10^22,0,0) N

F=(0,-4.8*10^22,0) N

F=(0,0-4.8*10^22) N

The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.

8 0

3 years ago

The mass of an object is 10 kg and the velocity is 4 m/s, what is the momentum?

eduard

The answer is 40 kg. m/s.

Formula for momentum:

p=mv

p=(10 kg.)(4 m/s)

So, therefore, the final answer is p=40 kg. m/s.

I hope this helped answer your question. Enjoy your day, and take care!

3 0

3 years ago