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3 years ago

Can someone please give me this answe

Physics

1 answer:

Helen [10]3 years ago

4 0

Answer:

Explanation:

Generally, longer wavelengths have less energy than shorter wavelengths because the higher the frequency, the shorter the wavelength. Since radio waves have longer wavelengths than visible light, the correct answer should be B.

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PLEASE HELP

Gennadij [26K]

Answer:

3.675 m

Explanation:

$a_{x} =0$ $v_{xo}=100$ $a_{y} =-g$ $v_{yo}=0$

X-direction | Y-direction

$R=x_{o}+ v_{xo} t$ | $y=y_{o}+v_{yo}t+\frac{1}{2}a_{y}t^2$

$75=100t$ | $y=0+0+\frac{1}{2} (9.8)(0.75)$

$\frac{75}{100} =t$ | $y=3.675 m$

$0.75s=t$

Hope it helps

3 0

3 years ago

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

4 0

3 years ago

Scientists have differing views on the origin of the universe. One view suggests the universe is expanding everywhere at the sam

GalinKa [24]

Oscillating model is the right answer

3 0

3 years ago

Identify trends in physical properties based on intermolecular forces

umka21 [38]

The Boiling point,melting point, surface tension and viscousity will increase while the Vapor pressure will decrease.

<h3 /><h3>What are intermolecular forces?</h3>

Intermolecular forces are the forces that bind two molecules together. Physical properties are affected by the strength of intermolecular forces

An increase in the strength of intermolecular forces increases will lead to an increase in force applied to break the barriers posed by the strength of the molecules.

This increased intermolecular strength will cause a rise in boiling point,melting point, viscousity and surface tension.

The Vapor pressure reduces with increasing intermolecular strength. Vapor pressure is the amount of vapor that is equilibrium with its own liquid or solid. Hence,with increasing intermolecular strength the amount of vapor that is in equilibrium with its own liquid will reduce.

To know more about intermolecular forces follow

brainly.com/question/13588164

#SPJ4

6 0

2 years ago

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of

Aleonysh [2.5K]

Answer:

r = 0.0548 m

Explanation:

Given that,

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.

We need to find the radius of their circular path. The formula for the radius of path is given by :

$r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$

m is mass of Singly charged uranium-238 ion, $m=3.95\times 10^{-25}\ kg$

q is charge

So,

$r=\dfrac{1}{1.9}\times \sqrt{\dfrac{2\times 3.95\times 10^{-25}\times 2.2\times 10^3}{1.6\times 10^{-19}}}\\\\r=0.0548\ m$

So, the radius of their circular path is equal to 0.0548 m.

4 0

3 years ago