Two descriptions about physical quantities are given below:

A radioactive substance decays according to the formula w = 20e¡kt grams where t is the time in hours. a find k given that after

blagie [28]

W=20 e(-kt)
A. Rearranging gives k= -(ln(w/20)/t
Substituting w= 10 and solving gives k=0.014
B. Using W=20e(-kt). After 0 hours, W=20. After 24 hours, W=14.29g. After 1 week (24x7=168h) W=1.9g
C. Rearranging gives t=-(ln(10/20)/k. Substituting w=1 and solving gives t=214 hours.
D. Differentiating gives dW/ dt = -20ke(-kt). Solving for t=100 gives dW/dt = 0.07g/h. Solving for t=1000 gives 0.0000002g/h
E. dW/dt = -20ke(-kt). But W=20e(-kt) so dW/dt = -kW

5 0

3 years ago

A particle whose speed is 50 m/sec moves along the line from A(2,1) to B (9,25)

WINSTONCH [101]

First, calculate for the distance between the given points A and B by using the equation,

D = sqrt ((x2 – x1)2 + (y2 – y1)2)

Substitute the known values:

D = sqrt((9 – 2)2 + (25 – 1)2)

D = 25 m

I assume the unknown here is the time it would require for the particle to move from point A to B. This can be answered by dividing the calculated distance by the speed given above.

t = (25 m)/ (50 m/s) = 0.5 s

Thus, it will take 0.5s for the particle to complete the route.

3 0

3 years ago

Newton’s law of gravitation says that gravity is a mutually attractive force. Explain the following observation: A small object

Airida [17]

Answer: See below

Explanation:

The Earth attracts the falling object with the same intensity of gravity as the object attracts the Earth, according to Newton's law of gravitation. The displacement of the two bodies, however, is inversely proportional to their respective masses.

Example: The Earth attracts a ball that falls 3 metres from the ground, even though the ball's mass is insignificant in comparison to the Earth's. Similarly, the ball draws the Earth with the same power, but the Earth's mass is enormously more than the ball's. As a result, the Earth collides with a billionth of a millimetre ball (or even less). Restart the Earth's descent on the ball you'll never see again.

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4 0

2 years ago

Read 2 more answers

In a wire with a 1.05 mm2 cross-sectional area, 7.93×1020 electrons flow past any point during 3.97 s. What is the current ????

34kurt

Answer:

The current in the wire is 31.96 A.

Explanation:

The current in the wire can be calculated as follows:

$I = \frac{q}{t}$

Where:

q: is the electric charge transferred through the surface

t: is the time

The charge, q, is:

$q = n*e$

Where:

n: is the number of electrons = 7.93x10²⁰

e: is the electron's charge = 1.6x10⁻¹⁹ C

$q = n*e = 7.93 \cdot 10^{20}*1.6 \cdot 10^{-19} C = 126.88 C$

Hence, the current in the wire is:

$I = \frac{126.88 C}{3.97 s} = 31.96 A$

Therefore, the current in the wire is 31.96 A.

I hope it helps you!

3 0

3 years ago

A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.796

Agata [3.3K]

Answer:

distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

Explanation:

Given that;

mass of block m = 0.200 kg

distance travelled d = 0.796 m

time t = 2.00 s

m₂ = 0.400 kg

If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?

Now, using the second equation of motion;

d = ut + ( $\frac{1}{2}$ × at²)

as the object started from rest, u=0

so, we substitute

0.796 = 0×2 + ( $\frac{1}{2}$ × a(2)²)

0.796 = 0 + ( $\frac{1}{2}$ × 4a)

0.796 = 2a

a = 0.796 / 2

a = 0.398 m/s²

using first equation of motion

$V_{f}$ = u + at

we substitute

$V_{f}$ = 0 + 0.398 × 2

$V_{f}$ = 0.796 m/s

now, average velocity is given as;

$V_{avg}$ = ( 0.796 m/s + 0 ) / 2

now, distance as the block moves in 2s will be;

D = [( 0.796 m/s + 0 ) / 2 ] × 2

D = 0.796 m

Therefore, distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

5 0

3 years ago