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adell [148]
3 years ago
13

Just how small do you think an individual atom is

Physics
2 answers:
jok3333 [9.3K]3 years ago
8 0
An atom is probably less than 1 nano-meter in size
harina [27]3 years ago
8 0
An atom is so miniscule that you can't even see a single atom. Think of the smallest thing in the world and know that possibly millions of atoms form that object
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An electron with a charge value of 1.6 x 10-19 C is moving in the presence of an electric field of 400 N/C. What force does the
Vsevolod [243]

Answer:

4.0×10⁻¹⁷ N

Explanation:

4 0
2 years ago
28. Which of the following correctly shows the order of highest amount of friction to the lowest amount of
Bad White [126]

Answer:

\mathrm{d.\:Static,\: Sliding,\:Rolling}

Explanation:

Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.

During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.

Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.

4 0
2 years ago
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
Read 2 more answers
As a wave travels through a medium, it displaces particles in a direction parallel to the motion of the wave. We can conclude th
postnew [5]
We can conclude that it is a longitudinal wave because the wave is traveling through a medium displacing particles<span>
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8 0
3 years ago
Read 2 more answers
Suppose that two objects attract each other with a gravitational force of 50N. If the mass
vagabundo [1.1K]

Answer:

112.5 N

Explanation:

50 = GMm/r^2

Let F be the new force of attraction

F/50 = ( G(3M)(3m)/(2r)^2 ) / (GMm/r^2)

[Elimiating G,M,m,r]

F = 112.5 N

7 0
3 years ago
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