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In-s [12.5K]
2 years ago
7

An asteroid has acquired a net negative charge of 149 C from being bombarded by the solar wind over the years, and is currently

in equilibrium whereby it expels electrons at the same rate as it acquires them. How many more electrons does it have than protons
Physics
1 answer:
bearhunter [10]2 years ago
5 0

Answer:

93.125 × 10^(19)

Explanation:

We are told the asteroid has acquired a net negative charge of 149 C.

Thus;

Q = -149 C

charge on electron has a value of:

e = -1.6 × 10^(-19) C

Now, for us to determine the excess electrons on the asteroid, we will just divide the net charge in excess on the asteroid by the charge of a single electron.

Thus;

n = Q/e

n = -149/(-1.6 × 10^(-19))

n = 93.125 × 10^(19)

Thus, it has 93.125 × 10^(19) more electrons than protons

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Steel is made of atoms of iron and carbon. Would iron and carbon form metallic bonds? Explain your answer choice.
Hatshy [7]

Sample Response: "No, steel and carbon would not form metallic bonds because metallic bonds only form between metals. Iron is a metal, but carbon is not."

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2 years ago
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Written accounts, objects, and buildings are examples of the types of evidence that historians and archaeologists study to expla
Lady_Fox [76]
     Your answer would be true. Because if we didn't have those pieces of evidence, we wouldn't know about a lot of the ancient civilizations that we know today without that. Small pieces of evidence like that can help us to determine how they lived, or what they used to do, or even what they ate.
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3 years ago
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a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the
tatuchka [14]
The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N
the force of the wind F, acting horizontally, with intensity
F=12 N
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
T \cos \alpha -F=0
T \sin \alpha -W=
By dividing the second equation by the first one, we get
\tan \alpha =  \frac{W}{F}= \frac{24.5 N}{12 N}=2.04
From which we find
\alpha = 63.8 ^{\circ}
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N




7 0
3 years ago
A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary
Snezhnost [94]

Explanation:

Let N_p\ and\ N_s are the number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p}{N_s}=\dfrac{1}{3}

A resistor R connected to the secondary dissipates a power P_s=100\ W

For a transformer, \dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}

V_s=(\dfrac{N_s}{N_p})V_p

V_s=3V_p...............(1)

The power dissipated through the secondary coil is :

P_s=\dfrac{V_s^2}{R}

100\ W=\dfrac{V_s^2}{R}

V_p^2=\dfrac{100R}{9}.............(2)

Let N_p'\ and\ N_s' are the new number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p'}{N_s'}=\dfrac{1}{24}

New voltage is :

V_s'=(\dfrac{N_s'}{N_p'})V_p'

V_s'=24V_p...............(3)

So, new power dissipated is P_s'

P_s'=\dfrac{V_s'^2}{R}

P_s'=\dfrac{(24V_p)^2}{R}

P_s'=24^2\times \dfrac{(V_p)^2}{R}

P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}

P_s'=6400\ Watts

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0
3 years ago
It takes you 8.3 min to walk with an average velocity of 1.6 m/s to the north from the bus stop to the museum entrance. How far
Anarel [89]

solution:

1.6 m/s = 96 m/min (in other words, 1.6 m/s x 60 s/min)  

96 m/min x 8.3 min = 796.8 m

s=ut +\frac{1}{2}at^2\\there is no accleration mentioned so,\\s= uv\\8.3\times60=498(s)\\510\times1.6=816(m)


3 0
3 years ago
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