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3 years ago

” Why is it important to identify a control group?

Physics

1 answer:

Gennadij [26K]3 years ago

5 0

Answer:

because if not then you wont be able to properly document the change happening in your experiment

Explanation:

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A particle moves along a straight path through displacement while force acts on it. (Other forces also act on the particle.) Wha

allsm [11]

a) c = 1.85

b) c = 0.8

c) c = 2.33

Explanation:

The displacement of the particle is given by

$d=2.2i+cj$

While the force applied on the particle is

$F=3.2i-3.8 j$

So we have a problem in 2-dimensions.

The work done on the particle is given by the scalar product between force and displacement:

$W=F\cdot d$ (1)

Here the work done on the particle is zero, so

W = 0

Therefore from eq(1) we find:

$0=(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=7.04\\c=\frac{7.04}{3.8}=1.85$

In this problem, the work done on the particle is

$W=4.0 J$

The force and displacement are still

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

Therefore, by calculting the scalar product between force and displacement and equating it to the work done (4.0 J), we find:

$W=F\cdot d$

$4.0 =(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=3.04\\c=\frac{3.04}{3.8}=0.8$

In this problem instead, the work done on the particle is negative:

$W=-1.8 J$

As before, the force and displacement are

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

And so again, we calculate the scalar product between force and displacement and we equate it to the work done on the particle, -1.8 J.

Doing so, we find:

$W=F\cdot d$

$-1.8=(3.2i-3.8j)\cdot (2.2i+c)=7.04-3.8c\\3.8c=8.84\\c=\frac{8.84}{3.8}=2.33$

7 0

3 years ago

When 23Na is bombarded with protons, the products are 20Ne and A. a neutron B. an alpha particle C. a deuteron D. a gamma ray pa

Vikki [24]

Answer:

The correct option is B. an alpha particle.

Explanation:

When ²³Na is bombarded with protons $_{1}^{1}\textrm{H}$ , ²⁰Ne and one alpha particle $_{2}^{4}\textrm{He}$ is released.

The reaction is as follows:

$_{11}^{23}\textrm{Na} + _{1}^{1}\textrm{H} \rightarrow _{10}^{20}\textrm{Ne} + _{2}^{4}\textrm{He}$

Therefore, an alpha particle and ²⁰Ne are released when ²³Na is bombarded with protons.

5 0

4 years ago

check the file.........................................................................................................

White raven [17]

What is meant by the number of complete oscillation made by an oscillating body in 10 seconds is 500 complete oscillations is that the frequency of the oscillating body is 50Hz

<h3>What is frequency?</h3>

Frequency of an oscillating body can be defined as the number of complete oscillations per unit time

Frequency is measured in hertz (Hz).

Also, the frequency of 1Hz is one oscillation per second.

frequency = number of oscillations/ time taken

frequency = 500/10 = 50 Hz

Thus, what is meant by the number of complete oscillation made by an oscillating body in 10 seconds is 500 complete oscillations is that the frequency of the oscillating body is 50Hz

Learn more about frequency here:

brainly.com/question/1199084

#SPJ1

6 0

2 years ago

Suppose a star has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 ) but is located at a distance of 300 ligh

iragen [17]

For a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as

L=2.7*10^30w

<h3> What is its luminosity?</h3>

Generally, the equation for the luminosity is mathematically given as

L=4*\pi^2*b

Therefore

L=4*\pi^2*b

L=4* \pi *(2.83*10^{18})*2.7*10^{-8}

L=2.7*10^30

In conclusion, the luminosity

L=2.7*10^30w