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Scorpion4ik [409]
2 years ago
10

A spherical snowball melts at a rate proportional to its surface area. (a) write a differential equation for its volume, v. (use

k for the constant of proportionality.) dv dt = correct: your answer is correct. (b) if the initial volume is s, solve the differential equation.
Physics
1 answer:
tresset_1 [31]2 years ago
6 0

Answer:

Explanation:

The rate of change in volume is proportional to the surface area:

dV/dt = kA

Integrating:

V = kAt + C

At t=0, V = s, so:

s = kA(0) + C

C = s

Therefore:

V = kAt + s

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A 2,000 g quantity of C-14 is left to undergo radioactive decay. The half-life of C-14 is approximately 5,700 years. After going
Katyanochek1 [597]
Start with 2,000 grams.
After 1 half-life, 1,000 grams are left.
After another half-life, 500 are left.
After another half-life, 250 are left.
After another half-life, 125 are left.

That was FOUR half-lifes.

X = 4 .
8 0
3 years ago
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Real life examples of atomic physics
Flauer [41]

Answer:

magnets

static electricity

heat

7 0
2 years ago
An orchestra is practicing a piece that is to be played at an allegro tempo. The conductor sets a metronome at 160 beats per min
omeli [17]

Answer:

375 ms

Explanation:

the frequency of metronome , f = 160 beats per minute

f = 160 /60 beats per sec

f = 2.67 beats /s

the period of a single beat , T = 1/f

T = 1/2.67 s

T = 0.375 s = 375 ms

the period of a single beat is 375 ms

3 0
3 years ago
A 180 lb crate is on the ground, and a strong rope is attached. You need to move it across the basement floor, which has a coeff
jekas [21]

Answer:

F = 505.13 N

Yes it is better to pull the rope rather than push it

Explanation:

Let the force is applied at an angle of 60 degree

so we will have net vertical force on the crate is given as

F_n + Fsin60 = mg

here we know

m = 180 lb

m = 81.65 kg

F_n = 81.65(9.81) - Fsin60

F_n = 801 - 0.866 F

now friction force on the crate is given as

F_x = \mu F_n

Fcos60 = 0.7(801 - 0.866 F)

0.5F + 0.61F = 560.7

F = 505.13 N

Yes it is better to pull the rope rather than push it

6 0
3 years ago
a skier is gliding along 3.0 m/s on horizontal, frictionless snow. he suddenly starts down a 10 degrees incline. his speed at th
Elza [17]
... find length
(way 1) determine acceleration using force
only force act on skier is mg vertically. spilt vector we get force along the incline = mgsin(10) and f= ma so
ma = mgsin(10) or a = gsin(10)
a (along the incline)= gsin(10) = 10sin(10) = 1.74
v^2 = u^2 + 2as
15^2 = 3^2 + 2(1.74)s
s = 62.06 m

(way 2) using conservation of energy
energy (KE+PE) on top = energy at bottom
0.5m3^2 + mgh = 0.5m15^2 +0
h (height of incline) = (112.5 - 4.5)/10 = 19.8 m
length of incline = h/sin(10) = 62.2 m ; trigonometry

... find time
s = (u+v)t/2
t = 2s/(u+v) = 2(62.2)/(3+15) = 6.91 s
4 0
3 years ago
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