4 years ago

The radius of an atom is closest in size to a

Physics

2 answers:

ioda4 years ago

8 0

the radius of the entire atom was 0.00000001 cm.

lions [1.4K]4 years ago

3 0

From all the choices available in that question, it's closest in size to the dot over the ' i ' in the words "radius", "is" and "in".

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A force of 6.0 N pulls a box 0.40 m along a frictionless plane that is inclined at 36°. What work is being done by the pulling f

lys-0071 [83]

Answer:

Expression of work done is

$W = Fd cos\theta$

Work done to move the sled is given as 1.94 J

Explanation:

As we know that the formula of work done is given as

$W = Fd cos\theta$

here we know that

F = 6 N

d = 0.4 m

$\theta = 36 degree$

so we will have

$W = 6 \times 0.4 cos36$

$W = 1.94 J$

7 0

3 years ago

ASAP I WILL GIVE BRAINLIEST A big wheel with triple the circumference of a smaller wheel will rotate with ____ the force. Group

Marianna [84]

Answer:

A big wheel with triple the circumference of a small wheel will rotate with triple the force.

Explanation:

when this circumference is tripled, it means triple the force (they are directly correlated)

4 0

3 years ago

What is the difference between sounds that have the same pitch and loudness?

eduard

Sounds that have the same pitch and loudness can be differentiated by intensity and tone.

8 0

3 years ago

Read 2 more answers

Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the sp

maxonik [38]

Answer:

kg m/s

Explanation:

e = Charge = C

V = Voltage = $\dfrac{N}{C}m$

c = Speed of light = m/s

Momentum is given by

$\dfrac{MeV}{c}=\dfrac{e\times V}{c}\\\Rightarrow \dfrac{MeV}{c}=\dfrac{C\times \dfrac{N}{C}\times m}{m/s}\\\Rightarrow \dfrac{MeV}{c}=Ns\\\Rightarrow \dfrac{MeV}{c}=kg\times \dfrac{m}{s}\times s\\\Rightarrow \dfrac{MeV}{c}=kg\cdot m/s$

The unit of MeV/c in SI fundamental units is kg m/s

5 0

3 years ago

A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 5.

Kobotan [32]

Explanation:

The given data is as follows.

F = $5.75 \times 10^{-16} N$

q = $1.6 \times 10^{-19} C$

v = 385 m/s

$sin (63.9^{o})$ = 0.876

Now, we will calculate the magnitude of magnetic field as follows.

B = $\frac{F}{qv sin (\theta)}$

= $\frac{5.75 \times 10^{-16} N}{1.6 \times 10^{-19} C \times 385 m/s \times 0.876}$

= $0.01065 \times 10^{3}$ T

= 10.65 T

Thus, we can conclude that magnitude of the magnetic field is 10.65 T.

4 0

3 years ago