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NeTakaya
3 years ago
11

Radioactivity was discovered by Albert Einstein. True or False

Physics
2 answers:
asambeis [7]3 years ago
6 0
Hello there!
The answer is <span>False.
</span><span>Henri Becquerel was the one to discover radioactivity.
</span>Hope I helped!
Vedmedyk [2.9K]3 years ago
3 0
False the man who discovered radioactivity was <span>Henri Becquerel</span>
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The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that
pav-90 [236]

Answer:

The permittivity of rubber is  \epsilon  = 8.703 *10^{-11}

Explanation:

From the question we are told that

     The  magnitude of the point charge is  q_1 =  70 \ nC  =  70 *10^{-9} \  C

      The diameter of the rubber shell is  d = 32 \ cm  =  0.32 \ m

       The Electric field inside the rubber shell is  E =  2500 \ N/ C

The radius of the rubber is  mathematically evaluated as

              r =  \frac{d}{2} =  \frac{0.32}{2}  =  0.16 \ m

Generally the electric field for a point  is in an insulator(rubber) is mathematically represented as

         E =  \frac{Q}{ \epsilon }  *  \frac{1}{4 *  \pi r^2}

Where \epsilon is the permittivity of rubber

    =>     E  *  \epsilon  *  4 * \pi *  r^2 =  Q

   =>      \epsilon  =  \frac{Q}{E *  4 *  \pi *  r^2}

substituting values

            \epsilon  =  \frac{70 *10^{-9}}{2500 *  4 *  3.142 *  (0.16)^2}

            \epsilon  = 8.703 *10^{-11}

7 0
3 years ago
A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p
levacccp [35]

Answer:

Q = 8,345 * v

Explanation:

So, we are looking for an expression of the amount of water that has been drained from the tub. The expression is in terms of v that represent the number of gallons of water drained since the plug was pulled. Since we are interested in the pounds of water that has been drained from the tub we need to take into account that for every gallon of water drained, 8.345 pounds have left the tub. Therefore, the expression for the weight of water Q that has been drained from the tub in terms of v is simply :

Q = 8,345 * v

Where v is the amount of gallons that has been drained from the tub.

Have a nice day. let me know if I can help with anything else

8 0
3 years ago
Pascal's Principle states that (a) If we apply pressure to a fluid in a sealed container, the pressure will be felt undiminished
nika2105 [10]

Answer:

a) If we apply pressure to a fluid in a sealed container, the pressure will be felt undiminished at every point in the fluid and on the walls of the container.

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area(A2) of the piston.

P=F/A

P1=P2

F1/ A1= F2/ A2

F2= F1* A2/ A1

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

In an incompressible liquid, the volume and amount of mass does not vary when pressure is applied.

5 0
2 years ago
A car is moving at 25.5 m/s when it accelerates at 1.94 m/s^2 for 2.3 s. What is the car's final speed? (Keep in mind direction
Stolb23 [73]

Answer:

29.96m/s

Explanation:

Given parameters:

Initial speed  = 25.5m/s

Acceleration  = 1.94m/s²

Time  = 2.3s

Unknown:

Final speed of the car  = ?

Solution:

To solve this problem, we are going to apply the right motion equation:

    v = u  + at

v is the final speed

u is the initial speed

a is the acceleration

t is the time taken

 Now insert the parameters and solve;

      v  = 25.5 + (1.94 x 2.3)  = 29.96m/s

3 0
2 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a
Vesna [10]

Answer:

Tangential speed=5.4 m/s

Radial acceleration=88.6m/s^2

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution=2\pi rad

2.59 rev=2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s

By using \pi=3.14

Angular velocity=\omega=16.27rad/s

Distance from axis=r=0.329 m

Tangential speed=r\omega=16.27\times 0.329=5.4m/s

Radial acceleration=\frac{v^2}{r}

Radial acceleration=\frac{(5.4)^2}{0.329}=88.6m/s^2

6 0
3 years ago
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