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Nostrana [21]
3 years ago
5

Imagine a bouncing ball that does not lose any energy

Physics
1 answer:
OleMash [197]3 years ago
4 0
It would never stop bouncing.
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Calculate the maximum kinetic energy of electrons ejected from this surface by electromagnetic radiation of wavelength 236 nm.
nydimaria [60]

Answer:

Explanation:

Using E= hc/wavelength

6.63 x10^-34 x3x10^8/ 236nm

19.86*10^-26/236*10^-9

=0.08*10^-35Joules

8 0
3 years ago
How do velocities combine
Anarel [89]

Answer:

They combine if they have the same units

Explanation:

I am not quite sure what you mean by this question. I will try my best to answer this.

Velocities combine only if they both have the same units. The linear velocities have the unit meters per second. For example, a linear velocity and an angular velocity cannot combine together because angular velocity has a unit of radians per second.

8 0
3 years ago
Microscopes are often equipped with blue filters. give a reason for using blue light in a microscope
timama [110]
The blue light removes red parts of the color spectrum giving it a cooler color

8 0
4 years ago
What will be the pressure exerterd by the 0bject if 4000n is acting on an area of 50msqure
Ivahew [28]
<h3>Given, </h3>

Force,F = 4000 N

Area,a = 50 m²

<h3>We know that, </h3>

Pressure = Force/Area

★ Putting the values in the above formula,we get:

\sf \rightarrow \: pressure =  \dfrac{4000}{50}

\sf \rightarrow pressure = 80 \: N {m}^{ - 2}

7 0
3 years ago
Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q
katen-ka-za [31]

Answer:

c) 4.2*10^{-5}C

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

F=\frac{kq_{1}q_{2}}{r^{2}}

where k is a proportionality constant with the value k=9*10^{9}\frac{Nm^{2}}{C^{2}}

In this case q_{1}=q_{2}=q, so we have:

F=\frac{kq^{2}}{r^{2}}

Solving the equation for q, we have:

kq^{2}=Fr^{2}

q^{2}=\frac{Fr^{2}}{k}

q=\sqrt{\frac{Fr^{2}}{k}}

Replacing the given values:

q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}

q=4.2*10^{-5}C

3 0
3 years ago
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