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Novay_Z [31]
3 years ago
5

Describe radioactive decay.

Physics
1 answer:
Snezhnost [94]3 years ago
7 0

The spontaneous transformation of an unstable atomic nucleus into a lighter one, in which radiation is released in the form of alpha particles, beta particles, gamma rays, and other particles. The rate of decay of radioactive substances such as carbon 14 or uranium is measured in terms of their half-life .

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Q1: We drop this 0.1 kg apple 1 m. What speed does it hit the ground
77julia77 [94]
Potential energy is measured using formula Ep=mgh
m=mass (kg)
g= acceleration due to gravity (which is 9.8 on earth)
h= height in metres above ground

For this question
m=0.1
g=9.8
h=1
So Ep=0.1(9.8)(1)
Ep=0.98 Joules

When it is dropped all of this potential energy is converted into kinetic energy which can be measured using formula
Ek=1/2m(v^2) (v=final velocity)

Since all potential energy in this q is converted to kinetic we know Ek=0.98Joules and our mass is the same (0.1kg)

So when we sub everything in we get
0.98=1/2(0.1)(v^2)
0.98=0.05(v^2)||divide both side by 0.05
19.6=v^2 ||square root both sides
v=4.4 m/s
5 0
3 years ago
The drawing shows a plot of the output emf of a generator as a function of time t. The coil of this device has a cross-sectional
adell [148]

This question is incomplete, the missing image uploaded along this answer below;

Answer:

a) frequency is 2.381 Hz

b) the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field is 0.5199 T

Explanation:

Given that;

cross sectional Area A = 0.018 m²

Number of turns N = 200

from the diagram. maximum time T = 0.42 sec

a) the frequency f of the generator in hertz

frequency = 1 / T

we substitute

frequency = 1 / 0.42 = 2.381 Hz

Therefore, frequency is 2.381 Hz

b)  the angular frequency in rad/s

angular speed ω = 2πf

we substitute

ω = 2π × 2.381

ω = 14.96 rad/s

Therefore, the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field.

to determine the magnitude of the magnetic field, we use the following expression;

e = NBAω

from the diagram, e = 28.0 V

so we substitute

28.0 V = 200 × B × 0.018 × 14.96

28 = 53.856B

B = 28 / 53.856

B = 0.5199 T

Therefore, the magnitude of the magnetic field is 0.5199 T

6 0
3 years ago
1. Plot the following graphs:
VLD [36.1K]

Answer:

(a) The distance-time graph for an object with uniform speed is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(b) The distance-time graph for an object with non-uniform speed is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

(c) The velocity-time graph for a car with uniform motion is giving by a horizontal line graph at the speed of constant motion with a zero gradient as shown in the attached diagram

(d) The velocity-time graph for a car moving with uniform acceleration is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(e) The velocity-time graph for a car moving with non-uniform acceleration is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

(f) According to Newton's first law of motion, an object at rest will remain at rest with no motion unless acted by a force, an therefore, will have no motion with time

Explanation:

8 0
3 years ago
If the frog lands with a velocity equal to its average velocity and comes to a full stop 0.25s later, what is the frog’s average
Nina [5.8K]

Answer:

Average accelation = -4V

Explanation:

a=\frac{V-V0}{t}

V=0 m/s (because the frog stopped)

V0 = V (average velocity)

t= 0,25 s

So;

a=\frac{V-V0}{t}=\frac{0-V}{0.25}=-4V

4 0
3 years ago
A schoolbus accelerates to 65 mph and enters the freeway. It travels for 2.3 hours at that speed while on the freeway. What's th
PtichkaEL [24]

Answer:

The distance travelled on the freeway is 149.5 miles.

Explanation:

The school bus travels on the freeway at constant speed. According to the statement, we need to calculate the distance travelled by the vehicle by means of the following formula:

x = v\cdot t (1)

Where:

x - Traveled distance, in miles.

v - Speed, in miles per hour.

t - Time, in hours.

If we know that v = 65\,\frac{mi}{h} and t = 2.3\,h, then the distance travelled by the school bus is:

x = v\cdot t

x =  \left(65\,\frac{mi}{h} \right)\cdot (2.3\,h)

x = 149.5\,mi

The distance travelled on the freeway is 149.5 miles.

5 0
2 years ago
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