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guapka [62]
2 years ago
6

An unstable nucleus which has a tendency to spontaneously change its form with the emission of high-energy particles or photons

is said to be _____________
Physics
2 answers:
Bezzdna [24]2 years ago
8 0

Answer: Radioactive nucleus.

Explanation: First, we can define the radiation as the emanation of particles or photons, for example, a flashlight is radiating photons when it is turned on.

So a neutron that emanates particles or photons when change form, is a nucleus that radiates, and those are called "radioactive nucleus"

Where radioactive implies that there is a presence of radiation in some way. In this case, the radiation is caused because the nucleus goes from an energetic state to other with lower energy, and that energy must be dissipated in some way, in this case, in form of radiation.

Pepsi [2]2 years ago
6 0

Answer:Radioactive

Explanation:

The radioactive nucleus is the one which does not has enough binding energy to hold the nucleus in a stable state and thus radiates either electron or proton to become a stable element.

A radioactive element is formed when after billions of years such as uranium and thorium. The stability of the nucleus depends upon the opposition of attractive and repulsive force among the nucleus.

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The seeds were sown (change the voice)​
MaRussiya [10]

Answer:

He sowed the seeds

Explanation:

While the question should have given the person who did the sowing for example the seeds were sown by him/ her/ the farmer/ or any name. Therefore, voice is given in passive and to change passive voice to active voice then the sentence will read as follows assuming that the seeds were sown by him

He sowed the seeds

7 0
1 year ago
A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit
makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

            2.87 = -1.442 + 1.8 u

               u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0
2 years ago
A spring stretches by 15cm when a mass of 300g hangs down from it,if the spring is then stretched an additional 10cm and release
bearhunter [10]

Answer:

0.1 m

Explanation:

It is given that,

Mass of the object, m = 350 g = 0.35 kg

Spring constant of the spring, k = 5.2 N/m

Amplitude of the oscillation, A = 10 cm = 0.1 m

Frequency of a spring mass system is given by :

Time period:

4 0
2 years ago
A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The t
azamat

Answer:

(a) x'(t)= 142

(b) 142

(c) y'(t)= -32t+5

(d) 96.8 mph

(e) 0.426 s

(f) 0.061 rad

Explanation:

Velocity is a time-derivative of position.

(a) x(t) = 142t

x'(t)= 142

(b) Since x'(t)= 142 is independent of t, it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

(c) y(t) = - 16t^2+5t+5

y'(t)= -32t+5

(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to x(t).

x(t)= 142t = 60.5

t=\dfrac{60.5}{142}= 0.426.

In this time, the vertical velocity, y'(t) is

y(t)= -32\times0.426+5 = -8.632

The speed of the ball at thus point is s=\sqrt{142^2+(-8.632)^2}=142 ft/s

To convert this to mph, we multiply the factor 3600/5280

s=142\times\dfrac{3600}{5280}=96.8 \text{ mph}

(e) The time has been determined from (d) above.

t= 0.426

(f) This angle is given by

\theta=\tan^{-1}\dfrac{y'(t)}{x'(t)}

\theta=\tan^{-1}\dfrac{-8.632}{142}=\tan^{-1}-0.0607=3.47 (Note here we are considering the acute angle so we ignore the negative sign)

In radians, this is

\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}

6 0
2 years ago
Calculate the mass of the air contained in a room that measures 2.50 m x 5.50 m x 3.00 m if the density of air is 1.29 g/dm3.53.
Law Incorporation [45]

Answer:

5.32\cdot 10^4 g

Explanation:

First of all, we need to find the volume of the room, which is given by

V=2.50 m \cdot 5.50 m \cdot 3.00 m =41.3 m^3

Now we  can find the mass of the air by using

m=dV

where

d=1.29 g/dm^3 is the density of the air

V=41.3 m^3 = 41,300 dm^3 is the volume of the room

Substituting,

m=(1.29)(41300)=5.32\cdot 10^4 g

6 0
2 years ago
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