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2 years ago

When an element ejects an alpha particle, the mass number of that element

Physics

1 answer:

Katyanochek1 [597]2 years ago

8 0

Answer:

A.reduce by 4.

Explanation:

In Chemistry, an alpha particle which is typically a helium nucleus has an atomic mass (number of nucleons) of four (4). Therefore, the ejection of an alpha particle by any chemical element simply means that, the mass number of that particular chemical element would be reduced by four (4).

Basically, this process is referred to as an alpha decay and can be defined as a radioactive disintegration of a chemical element that causes it to emit an alpha particle and as a result reducing its mass number by four (4).

Hence, when an element ejects an alpha particle, the mass number of that element reduces by 4.

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A person standing a certain distance from four identical loudspeakers is hearing a sound level intensity of 125 dB. What sound l

DENIUS [597]

Answer:

$\mathbf{\beta = 123.75 \ dB}$

Explanation:

From the question, using the expression:

$125 \ dB = 10 \ log (\dfrac{I}{I_o})$

where;

$I_o = 10^{-12} \ W/m^2$

$I = 10^{12.5} \times 10^{-12} \ W/m^2$

$I = 3.162 \ W/m^2$

This is a combined intensity of 4 speakers.

Thus, the intensity of 3 speakers = $\dfrac{3.162\times 3}{4}$

= 2.372 W/m²

Thus;

$\beta = 10 \ log ( \dfrac{2.372}{10^{-12}} ) \ W/m^2$

$\mathbf{\beta = 123.75 \ dB}$

7 0

2 years ago

12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it

vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

$v_{f} = v_{o}+a\cdot (t-t_{o})$ (1)

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$v_{f}$ - Final velocity, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

$t_{o}$ - Initial time, measured in seconds.

$t$ - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust ( $v_{o} = 0\,\frac{m}{s}$ , $a = 30\,\frac{m}{s^{2}}$ , $t_{o} = 0\,s$ , $0\,s\le t< 30\,s$ )

$v = 30\cdot t$ (2)

Free fall ( $v_{o} = 900\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s}$ , $t_{o} = 30\,s$ , $30\,s \le t \le 120\,s$ )

$v = 900-9.81\cdot (t-30)$ (3)

Now we created the graph speed-time, which can be seen below.

5 0

2 years ago

An athlete jumping vertically on a trampoline leaves the surface with a velocity of 8.5 m/s upward. what maximum height does she

Mumz [18]

Her center of mass will rise 3.7 meters. First, let's calculate how long it takes to reach the peak. Just divide by the local gravitational acceleration, so 8.5 m / 9.8 m/s^2 = 0.867346939 s And the distance a object under constant acceleration travels is d = 0.5 A T^2 Substituting known values, gives d = 0.5 9.8 m/s^2 (0.867346939 s)^2 d = 4.9 m/s^2 * 0.752290712 s^2 d = 3.68622449 m Rounded to 2 significant figures gives 3.7 meters. Note, that 3.7 meters is how much higher her center of mass will rise after leaving the trampoline. It does not specify how far above the trampoline the lowest part of her body will reach. For instance, she could be in an upright position upon leaving the trampoline with her feet about 1 meter below her center of mass. And during the accent, she could tuck, roll, or otherwise change her orientation so she's horizontal at her peak altitude and the lowest part of her body being a decimeter or so below her center of mass. So it would look like she jumped almost a meter higher than 3.7 meters.

8 0

2 years ago

When designing a roller coaster what are the biggest considerations?

kati45 [8]

If you are designing a roller coaster that goes upside down, you may consider of course seat belts or something that goes around you to keep yourself safe.

3 0

2 years ago

The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:

Furkat [3]

To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is s=r(B), that has its r=radius,B=angle in rad velocity v=ds/dt= w(r)
Do not forget about w = angular speed in rad/s and $w1 = 1 revolution/sec = 2Pi (rad/s)$
Now we can go to proportion
$v1=v2$
$w1*r1 = w2r2$ $w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)$
$w2 = w3 (which is the angular velocity of the rear wheel) 
$
SOLVING FOR A : $v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec$
$v3 = 14.66 ft/sec(1 mile/5280 ft)( 3600 sec/h)$ $= 9.99$ or something about 10 mph --- SOLVING FOR B.
I'm sure it helps!

7 0

2 years ago