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1 year ago

12

"I know how many electrons the atom has, and I know how many protons it has, but I don't know whether or not it is neutral," a f

riend says. How would you respond?

1 answer:

finlep [7]1 year ago

4 0

The atom is neutral (no electric charge)

Explanation:

An atom consists of:

- A nucleus, containing protons (positively charged) and neutrons (no electric charge)

- Electrons (negatively charged), orbiting around the nucleus

The charge of one proton is equal in magnitude (but opposite in sign) to that of an electron, and it is

$e=1.6\cdot 10^{-19} C$

Moreover, for a normal atom, the number of protons in the nucleus is equal to the number of electrons around it.

This means that for a normal atom, the net charge of it is zero, since the total charge of the protons balance that of the electrons.

So, the answer to the question is that the atom is neutral.

Learn more about atoms here:

brainly.com/question/2757829

#LearnwithBrainly

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You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it at position y0. Imagine that you

olga nikolaevna [1]

Answer:

a) Δy = 0.144 m

b) W = 0.145 J

c) Us = 0.32 J

d) ymax = 0.144 m

Explanation:

a) First let's find the spring constant using Hooke's Law

F = k*Δy ⇒ k = F/Δy

where

F = m*g = 0.1 kg*9.81 m/s² = 0.981 N

and Δy = 0.07 m. Hence

k = 0.981 N/0.07 m = 14.014 N/m ≈ 14 N/m

In order to find the position of the block when we let it go, we need to find the force that caused this expansion in the spring, we know that the reading of the scale was 3 N and this reading includes the force we want to find and the weight of the block, therefore:

f = 3 N - F = 3 N - 0.981 N = 2.019 N

Now that we have found the force we can use Hooke's Law in order to find the position of the block

f = k*Δy ⇒ Δy = f/k

⇒ Δy = 2.019 N/14 N/m

⇒ Δy = 0.144 m

b) First, notice that there are two kind of potential energy: the potential energy in the spring and the potential energy due to the gravitational field:

W = ΔU

W = ΔUs + ΔUg

W = (Usf - Usi) + (Ugf - Ugi)

Notice that

Us = 0.5*k*y²

where

yf = 0.07 m + 0.144 m = 0.214 m and

yi = 0.07 m

and we will take the zero level to be the equilibrium position where the block was hanging at rest. Hence

W = 0.5*k*(yf² - yi²) + m*g*(0 - Δy)

⇒ W = 0.5*14 N/m*((0.214 m)² - (0.07 m)²) + (0.1 kg)*(9.81 m/s²)*(0 - 0.144 m)

⇒ W = 0.145 J

c) When we let the block go the spring was stretched by

y = 0.07 m + 0.144 m = 0.214 m

Therefore:

Us = 0.5*k*y²

⇒ Us = 0.5*14 N/m*(0.214 m)²

⇒ Us = 0.32 J

d) Because the position that we pulled the block to it is considered as the amplitude for the vibrational motion that will happen after we release the block, then the maximum height the particle will reach above the equilibrium position is

ymax = Δy = 0.144 m

3 0

2 years ago

A worker kicks a flat object lying on a roof. The object slides up the incline 10.0 m to the apex of the roof, and flies off the

11111nata11111 [884]

Answer:15.20 m

Explanation:

Given

initial velocity $(v_i)=15 m/s$

inclined length=10 m

$\mu _k=0.435$

inclination $\theta =43.5^{\circ}$

$F_{net}=f_r+mg\sin \theta$

$a_{net}=\mu _kg\cos \theta +g\sin \theta$

$a_{net}=0.435\times 9.8\times \cos 43.5+9.8\times \sin 43.5$

$a_{net}=3.09+6.74=9.83 m/s^2$

$v^2-u^2=2as$

$v^2=15^2-2\times (9.83)10$

$v=5.31 m/s$

So Particle launches with a speed of 5.31 m/s at an angle of \theta $=43.5$

$h_{max}=\frac{u^2\sin^2\theta }{2g}$

$h_{max}=\frac{(5.31)^2\sin ^2(43.5)}{2\times 9.81}$

$h_{max}=0.679$

Total height raised is $0.679+\frac{10}{\sin 43.5} =15.20 m$

6 0

1 year ago

Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. Wha

aliina [53]

Answer:

P=627.47W

Explanation:

To solve this problem we have to take into account, that the work done by the winch is

$W=Fh$

the force, at least must equal the gravitational force

$F=Mg=(156kg)(9.8\frac{m}{s^2})=1258.8N$

with force the tension in the cable makes the winch go up.

The work done is

$W=(1258.8N)(58.0m)=73010.4J$

To calculate the power we need to know what is the time t. But first we have to compute the acceleration

The acceleration will be

$v_f^2=v_0+2ah\\a=\frac{v_f^2}{2h}=\frac{(24.9\frac{m}{s})}{2(58.0m)}=0.214\frac{m}{s^2}$

and the time t

$v_f=v_0+at\\t=\frac{v_f}{a}=116.35s$

The power will be

$P=\frac{W}{t}=\frac{73010.4J}{116.35s}=627.47W$

HOPE THIS HELPS!!

6 0

2 years ago

The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a

Amanda [17]

Answer:

amount of energy = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

so

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = $\frac{excess initial cost}{\Delta M}$

payback period = $\frac{32 + 66}{520.34}$

payback period = 0.188 year

8 0

2 years ago

A boy does 465 J of work pulling an empty wagon along level ground with a force of 111 N [31o below the horizontal]. A frictiona

vovikov84 [41]

Can you list the answer's

6 0

1 year ago