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3 years ago

What kind of radiation is emitted in the following nuclear reaction

Physics

1 answer:

miv72 [106K]3 years ago

4 0

Beta emission is occurring in the given nuclear reaction.

Answer: Option B

Explanation:

In this equation, the reactant is the Thorium atom, which is reduced to palladium. As the atomic number get decreased by one, so an electron will be emitted. This process of emission of electrons by radiation or decaying the reactant nuclei to form a new product nuclei is termed as beta emission.

So, the electrons are generally termed as beta particles while the positrons are termed as positive beta particles. So this is a kind of radioactive reactions where the reactant changes to new element by releasing an electron and thus there is a change in the atomic number of the product by one.

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Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value o

Elza [17]

Answer:

B or 2

Explanation:

5 0

2 years ago

A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

8 0

3 years ago

The carbon atom with 12 protons and 12 electrons has a charge of:

olga55 [171]

Answer:

Isotopes are the elements having the same atomic number i.e same number of protons which in turn is equal to same number of electrons. Number of protons and electrons are equal as the atom is electrically neutral. So, the only option is A.

I hope it helps you !

3 0

2 years ago

Determine the approximate force (N) used to pull a sled up a 400 m hill using 1900 J of work.

Sergeu [11.5K]

The work done to pull the sled up to the hill is given by
$W=Fd$
where
F is the intensity of the force
d is the distance where the force is applied.

In our problem, the work done is $W=1900 J$ and the distance through which the force is applied is $d=400 m$ , so we can calculate the average force by re-arranging the previous equation and by using these data:
$F= \frac{W}{d}= \frac{1900 J}{400 m} = 4.75 N \sim 5 N$

4 0

3 years ago

You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo

AlekseyPX

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

F is the applied force
Ff is the frictional force

ma = F - Ff

where;

a is acceleration of the cart
m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net) = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

Learn more about frictional force here: brainly.com/question/24386803

#SPJ1

4 0

2 years ago