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2 years ago

What is radioactivity?

Physics

2 answers:

alekssr [168]2 years ago

7 0

Answer:

Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation. A material containing unstable nuclei is considered radioactive. Three of the most common types of decay are alpha decay, beta decay, and gamma decay, all of which involve emitting one or more particles.

the giving off of rays of energy or particles by the breaking apart of atoms of certain elements (as uranium) 2 : the rays or particles that are given off when atoms break apart.

The three most common types of radiation are alpha particles, beta particles, and gamma rays.

HOPE IT HELPS

BRAINLIEST PLEASR

Olin [163]2 years ago

6 0

Answer:

Radioactivity is the spontaneous disintegration of heavier unstable nuclei to form lighter stable nuclei with release of energy particles such as alpha particles, beta particles and gamma rays

Explanation:

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Read the scenario and solve these two problems.

Burka [1]

Answers:

a) 5400000 J

b) 45.92 m

Explanation:

a) The kinetic energy $K$ of an object is given by:

$K=\frac{1}{2}mV^{2}$

Where:

$m=12000 kg$ is the mass of the train

$V=30 m/s$ is the speed of the train

Solving the equation:

$K=\frac{1}{2}(12000 kg)(30 m/s)^{2}$

$K=5400000 J$ This is the train's kinetic energy at its top speed

b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:

$E_{i}=E_{f}$

$K_{i}+P_{i}=K_{f}+P_{f}$

Where:

$K_{i}=5400000 J$ is the train's initial kinetic energy

$P_{i}=0 J$ is the train's initial potential energy

$K_{f}=0 J$ is the train's final kinetic energy

$P_{f}=mgh$ is the train's final potential energy, where $g=9.8 m/s^{2}$ is the acceleration due gravity and $h$ is the height.

Rewriting the equation with the given values:

$5400000 J=(12000 kg)(9.8 m/s^{2})h$

Finding $h$ :

$h=45.918 m \approx 45.92 m$

7 0

2 years ago

Read 2 more answers

Suppose you increase your walking speed from 5 m/s to 14 m/s in a period of 3 s. What is your acceleration

kirza4 [7]

5-14 m/s in 3 seconds
a=vf-vi/t
a=14-5/3
a=9/3
a=3 m/s^2

4 0

2 years ago

Sharp claws in wild animals is an example of which type of adaptation?

Sati [7]

Answer:

Structural adaptation

Explanation:

When any physical body part of an organism helps it in surviving, then this mechanism of survival is called a structural adaptation. For example short tail of penguins helps them in balancing on their heels and tails due to which they are able to reduce heat released from their feet. Thus, any physical parts of an organism that helps them in coping with the surrounding physical environment /predators are refereed to as a mode of structural adaptation.

3 0

3 years ago

The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c

Reika [66]

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ = 0.23805 N-m

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

7 0

2 years ago

In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of

Rashid [163]

Answer:

The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles

Explanation:

The parameters of the motion of the driver are;

The upgrade of the road, θ = 10°

The rate of constant acceleration of the bus driver = 5 ft./s²

The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s

The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s

The acceleration due to gravity, g ≈ 32.1740 ft./s²

Therefore, we have;

The acceleration due to gravity down the incline plane, gₓ = g·sinθ

∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²

The net acceleration of the bus, on the incline plane, $a_{Net}$ = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, $v_y$ = v × sin(θ)

∴ $v_y$ = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s

vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s

The velocity of the car, v₂, on the inclined plane is given as follows;

v₂ = v₁ - $a_{Net}$ × t

∴ t = (v₁ - v₂)/ $a_{Net}$ = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s

The distance covered, 's', is given as follows;

s = v₁·t - 1/2· $a_{Net}$ ·t²

∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.

The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles

5 0

2 years ago

What is radioactivity?​

What is radioactivity?