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Alekssandra [29.7K]
1 year ago
6

A responder can protect himself/herself from radiation by using shielding as a response action. What materials are best for prot

ecting against beta particles?
Physics
1 answer:
Irina-Kira [14]1 year ago
3 0

Answer:

Few millimeter thick aluminium, water, wood, acrylic glass or plastic.

Explanation:

The materials that are best for protection against beta particles are few millimeter thickness of aluminium, but for the high energy beta-particles radiations the low atomic mass materials such as plastic, wood, water and acrylic glass can be used.

These materials can also be used in personal protective equipment which includes all the clothing that can be worn to prevent any injury or illness due to the exposure to radiation.

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A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
Leona [35]

Answer:

0.6

Explanation:

The volume of a sphere = \frac{4}{3} \pi (\frac{D}{2})^3

Therefore \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

r of the disc = 1.15(\frac{ D}{2} )

Using conservation of angular momentum;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disc = m*\frac{   \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{  m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

5 0
1 year ago
100 POINTS! I will mark brainliest! Record your hypothesis as an “if, then” statement for the rate of dissolving the compounds:
love history [14]

Answer:

<u><em>Rate of dissolving compounds:</em></u>

If we increase the temperature of the solution, then the dissolving compound would dissolve more easily.

<u><em>Boiling Point of Compounds:</em></u>

If the inter-molecular forces of any compound is really strong, then the boiling point of the compound would be really high.

6 0
2 years ago
The weight of an object is the same on two different planets. The mass of planet A is only sixty percent that of planet B. Find
natka813 [3]

Answer:

0.775

Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

F=G\frac{Mm}{R^2}

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

F_A=G\frac{M_Am}{R_A^2}

For planet B,

F_B=G\frac{M_Bm}{R_B^2}

We also know that the weight of the object on the two planets is the same, so

F_A = F_B

So we can write

G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}

We also know that the mass of planet A is only sixty percent that of planet B, so

M_A = 0.60 M_B

Substituting,

G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}

Now we can elimanate G, MB and m from the equation, and we get

\frac{0.60}{R_A^2}=\frac{1}{R_B^2}

So the ratio between the radii of the two planets is

\frac{R_A}{R_B}=\sqrt{0.60}=0.775

6 0
1 year ago
Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. Th
Stels [109]

This question deals with the law of conservation of momentum, which basically says that the total momentum in a system must stay the same, provided there are no outside forces. Since you were given the mass and velocity of the two objects you can find the momentum (p=mv) of each and then add them together to find the total momentum of the system before they collide. This total momentum must be the same after they collide.  Since you have the mass and velocity of one of the objects after the collision you can find the its momentum after.  Subtract this from the the system total and you will have the momentum of the other object after the collision.  Now that you know the momentum of the other object you can find its velocity using p=mv and its mass from before.

Be careful with the velocities.  They are vectors, so direction matters.  Typically moving to the right is positive (+) and moving to the left is negative (-).  It is not clear from your question which direction the objects are moving before and after the collision.

6 0
1 year ago
Read 2 more answers
Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p
Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0
2 years ago
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