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tekilochka [14]
3 years ago
10

Though atomic physicists have made significant progress at understanding how God created the atom, a there is not yet a complete

understanding of all atomic interactions that exist.
True or False
Physics
1 answer:
MakcuM [25]3 years ago
7 0
True because it’s true
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Ay tawaging bayani sa makabagong panahon?
agasfer [191]

Answer:

Explanation: Bagong bayani ang turing sa ating mga Overseas Filipino. Workers (OFWs) sapagkat ang kanilang pagpapagod at pakikipagsapalaran sa ibayong dagat ay hindi lamang ... kasigurahan ang pagkukunan ni Lito sa panahon ng kanyang.

7 0
2 years ago
Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87
marshall27 [118]

Answer:

a)  \lambda=0.935\ \textup{m}

b) f=36.19\approx 36\ \textup{Hz}

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

L= n\frac{\lambda}{2}

where,

\lambda = wavelength

on substituting the values, we get

1.87= 4\times \frac{\lambda}{2}

or

\lambda=0.935\ \textup{m}

b) Speed of the wave (v) in the string is given as:

v =f\lambda

also,

v=\sqrt\frac{T}{(\frac{m}{L})}

equating both the formula for 'v' we get,

f\lambda=\sqrt\frac{T}{(\frac{m}{L})}

on substituting the values, we get

f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}

or

f=\frac{33.84}{0.935}

or

f=36.19\approx 36\ \textup{Hz}

5 0
2 years ago
Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec
Dmitrij [34]

Answer:

(a) F_{net} = 4.19 x 10^{-5} N, and its direction is towards m_{2}.

(b) It must be placed inside a hollow shell.

Explanation:

Let, m_{1} = 165 kg, m_{2} = 465 kg, m_{3} = 60 kg, and the distance between m_{1} and m_{2} is 0.340 m.

(a) Since m_{3} is placed midway between m_{1} and m_{2}, then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = \frac{Gm_{1}m_{2}  }{r^{2} }

Where all variables have their usual meaning.

Then,

a. F_{net} = F_{23} - F_{13}

F_{13} = \frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }

     = 2.25 x 10^{-5} N

F_{23} = \frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }

     = 6.44 x 10^{-5} N

∴ F_{net} =  = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

The net force exerted by the two masses on the 60 kg object is 4.19 x 10^{-5}  N.

(ii) /F_{net}/ = /F_{23}/ - /F_{13}/

              = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

(iii) The direction of the net force is to the right i.e towards m_{2}.

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0
3 years ago
two negative charges that are both -3.0 C push each other apart with a force of 19.2 N how far apart are the charges
Hoochie [10]
The electrostatic force between two charges q1 and q2 is given by
F=k_e  \frac{q_1 q_2}{r^2}
where k_e =8.99 \cdot 10^9 N m^2 C^{-2} is the Coulomb's constant and r is the distance between the two charges.

If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the  distance between the two charges by re-arranging the previous formula:
r= \sqrt{k_e \frac{q_1 q_2}{F} }= \sqrt{ 8.99 \cdot 10^9 N m^2 C^{-2} \frac{(-3.0C)^2}{19.2 N} } =6.49 \cdot 10^4 m=64.9 km
5 0
3 years ago
What are some alike things between the ozone layer and the greenhouse effect
umka2103 [35]
The ozone layer traps heat from the sun's heat. only three-fourths are reflected back out into space by the ozone layer. the greenhouse effect traps carbon dioxide and so does the ozone layer.
6 0
2 years ago
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