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3 years ago

What is the charge of an atom with 20 protons and 18 electrons?

Physics

2 answers:

aev [14]3 years ago

8 0

The answer is 2+ because I just know it’s easy

aleksandr82 [10.1K]3 years ago

3 0

The charge is +2

20-18=2

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Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

nirvana33 [79]

Answer:

1.327363 m/s

0.00090243 m

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of flea

Energy

$E=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 3.7\times 10^{-4}=2.1\times 10^{-4}(v^2-0)\\\Rightarrow v=\sqrt{\frac{3.7\times 10^{-4}}{2.1\times 10^{-4}}}\\\Rightarrow v=1.32736\ m/s$

The velocity of the flea when leaving the ground is 1.327363 m/s

$W=F\times s\\\Rightarrow s=\frac{W}{F}\\\Rightarrow s=\frac{3.7\times 10^{-4}}{0.41}\\\Rightarrow s=0.00090243\ m$

The flea will travel 0.00090243 m upward

8 0

3 years ago

Choose the correct statement regarding a compound microscope being used to look at a mite. The eyepiece lens forms a virtual ima

Nata [24]

Answer:

a) True. The image of the mite is virtual

e) True. The image must be within the focal length of the eyepiece len

Explanation:

Let's review the general characteristics of compound microscopes

Formed by two converging lenses

Magnification is

M = -L/fo 0.25/fe

Where fo is the focal length of the objective lens and fe is the focal length of the ocular lens, L is the tube length

Let's review the claims

a) True. The image of the mite is virtual

b) False. The effect is the opposite of the magnification equation

c) False. The objective lens forms a real image

d) False. As the seal distance increases the magnification decreases

e) True. The image must be within the focal length of the eyepiece len

4 0

4 years ago

Match the terms to the correct descriptions.

Ulleksa [173]

2. kinetic energy: due to it being transferred through collisions

5 0

3 years ago

Read 2 more answers

Can anyone please help me with the steps

KIM [24]

Answer:

A) v_average = - 10 km / h, B) v = 1.6 m / s, v = 17.6 m / s

Explanation:

A) the average speed is the average speed of a body, if we assume that the direction of going up the hill is positive

v₁ = 40 km / h

v₂ = - 60 km / h

the average speed is

v_average = $\frac{v_1 + v_2}{2}$

v_average = ( 40 - 60)/2

v_average = - 10 km / h

B) in this case they indicate the acceleration a = 3.2 m / s² and the velocity vo = 9.6 m / s

i) the speed for 2.5 s above

v = v₀ + a t

as the time is earlier

t = - 2.5 s

we substitute

v = 9.6 - 3.2 2.5

v = 1.6 m / s

ii) the velocity for a subsequent time of 2.5 s

t = 2.5 s

we substitute

v = 9.6 + 3.2 2.5

v = 17.6 m / s

3 0

3 years ago

A string fixed at both ends is 8.40 m long and has a mass of 0.120 kg. It is subjected to a tension of 96.0 N and set oscillatin

Luden [163]

Answer:

81.9756 m/s

16.8 m

4.8795 Hz

Explanation:

m = Mass of string = 0.12 kg

L = Length of string = 8.4 m

T = Tension on string = 96 N

Linear density is given by

$\mu=\dfrac{m}{L}\\\Rightarrow \mu=\dfrac{0.12}{8.4}$

Spee of the wave is given by

$v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow v=\sqrt{\dfrac{96}{\dfrac{0.12}{8.4}}}\\\Rightarrow v=81.9756\ m/s$

The speed of the waves on the string is 81.9756 m/s

Wavelength is given by

$\lambda=2L\\\Rightarrow \lambda=2\times 8.4\\\Rightarrow \lambda=16.8\ m$

The longest possible wavelength is 16.8 m

Frequency is given by

$f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{81.9756}{16.8}\\\Rightarrow f=4.8795\ Hz$

The frequency of the wave is 4.8795 Hz

3 0

3 years ago