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3 years ago

The discoverer of radioactivity was

2 answers:

8 0

The dicoverer of radioactivity was in 1896 henri Becquerel was using naturally flurorescent minerals to study been discover in 1895 bu Wilhelm Roentgen.

Vikentia [17]3 years ago

5 0

Answer:

Henri Becquerel

Explanation:

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Jvhftfcticgvigybvhbjnjndjanvjisn

nirvana33 [79]

Answer:

dnndmssnxnxdnskkaamzmzma nzz.

4 0

3 years ago

Read 2 more answers

Any help would be great! Thank you x<br><br><br> Giving brainliest answer xoxo

garri49 [273]

Answer:

A vacuum would have been created. I hope this helps have a great day

3 0

3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 5.7 m from the center of the ride. Then the operator turns on the rid

tatyana61 [14]

Answer:

0.76

Explanation:

we are given:

radius (r) =5.7 m

speed (s) = 1 revolution in 5.5 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

coefficient of friction (Uk) = ?

we can get the minimum coefficient of friction from the equation below

centrifugal force = frictional force

m x r x ω^{2} = Uk x m x g

r x ω^{2} = Uk x g

Uk = $\frac{ r x ω^{2} }{g}$

where ω (angular velocity) = $\frac{2π}{time}$

= $\frac{2π }{5.5}$ = 1.14

Uk = $\frac{ 5.7 x 1.14^{2} }{9.8}$ = 0.76

6 0

3 years ago

Eric has a mass of 70 . He is standing on a scale in an elevator that is accelerating downward at 1.7 . What is the approximate

pychu [463]

Answer:

Explanation:

The value the scale shows is the reaction force to the normal force (they are equal by Newton's 3rd Law) that the scale exerts on Eric.

The forces on Eric are his weight (downward) and this normal force (upward), so we can write the net force over him as (also using Newton's 2nd Law):

$F=W-N=ma$

which means

$N=W-ma=mg-ma=m(g-a)$

and for our values this is:

$N=mg-ma=(70kg)(9.8m/s^2-1.7m/s^2)=567N$

5 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

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

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

4 years ago