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fgiga [73]
3 years ago
5

Which scientist discovered the two radioactive elements radium and polonium?

Physics
1 answer:
kolezko [41]3 years ago
6 0
The scientist who discovered the two elements radium and polonium would be Marie Curie in 1898.

Hope this helps!
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Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of v=0.8c
Studentka2010 [4]

Answer:

30.96 m

Explanation:

If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:

d = v * Δt

v = 0.8*c = 0.8 * 3e8 = 2.4e8

Δt = ζ = 129 ns = 1.29e-7 s

d = 2.4e8 * 1.29e-7 = 30.96 m

This is the distance as measured by observer A.

3 0
3 years ago
Jupiter’s Great Red Spot rotates completely every six days. If the spot is circular (not quite true, but a reasonable approximat
artcher [175]

Answer:

v = 567.2 km/h

Explanation:

As we know that if Jupiter Rotate one complete rotation then it means that the it will turn by 360 degree angle

so here the distance covered by the point on its surface in one complete rotation is given by

distance = 2\pi r

distance = \pi D

now we will have the time to complete the rotation given as

t = 6 days

t = 6 (24 h) = 144 h

now the speed is given by

speed = \frac{distance}{time}

speed = \frac{\pi D}{t}

speed = \frac{\pi(26000 km)}{144}

v = 567.2 km/h

5 0
3 years ago
The refractive index of glass is 1.52
Ksenya-84 [330]

Explanation:

..upper side is glass ..

3 0
3 years ago
Can anyone help me with these questions? TIA!<br> (Don’t actually answer please! :) )
nataly862011 [7]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 7: </h2>

 \huge\text{Graphs:}

The graph of  

• The I-V for Ohmic Metal wire conductor at constant temperature  always shows a straight line between the Current(I) plotted at Y axis and Voltage(V) plotted at X axis. Picture 1

• The I-V graph for Diode shows that first the current is zero but as we increase the potential difference(voltage), it results in the increase in the current. Picture 2

<h2>_____________________________________ </h2><h2>Question 8: </h2>

\Large\textbf{Diode:}  

A diode is a device that allows current to flow in only one direction.

\Large\textbf{Forward and Reverse Biasing:}  

Forward Bias, When a diode is forward bias (a voltage in the "forward" direction) then the P-side of the diode is attached to the positive terminal and N-side is fixed to the negative side of the battery which is connected, current flows freely through the device. The forward bias decreases the thickness of potential barrier(The potential barrier barrier in which the charge requires additional force for crossing the region)

Reverse Bias, When a diode is Reverse bias(a voltage in the "backward direction) then the P-side of the diode is connected to the negative terminal and N-side is connected to the positive terminal of the battery which is connected. The reverse bias increases the thickness of the potential barrier resulting in the flow of no current.

 \Large\textbf{Answer to the Question "Resistance"}

The Forward bias decreases the resistance of the diode whereas the reversed bias increases the resistance of the diode. As in forward biasing the current is easily flowing through the circuit whereas reverse bias does not allow the current to flow through it.

<h2>_____________________________________ </h2><h2>Best Regards, </h2><h2>'Borz' </h2>

8 0
3 years ago
Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e
Nataliya [291]

Answer:

q=1.95*10^{-7}C

Explanation:

According to the free-body diagram of the system, we have:

\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)

So, we can solve for T from (1):

T=\frac{mg}{cos(15^\circ)}(3)

Replacing (3) in (2):

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e

The electric force (F_e) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)

Finally, we replace (5) in (4) and solving for q:

mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C

5 0
3 years ago
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