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3 years ago

6

How long does it take for the velocity of the rain drop to reach 99% of its terminal velocity? (assume the conditions from part

(b). round your answer to two decimal places.)?

1 answer:

vodomira [7]3 years ago

3 0

If you think about it its part a and b

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How much heat (in calories) are needed to raise the temperature of 60 g of water

laila [671]

Answer:

Well, each ml of water requires one calorie to go up 1 degree Celsius, so this liter of water takes 1000 calories to go up 1 degree Celsius.

Explanation:

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2 years ago

(1) In nondestructive testing, a discontinuity may be defined as an interruption in the normal physical structure or configurati

QveST [7]

Answer: BOTH ARE TRUE

Explanation: Nondestructive testing or Evaluation is a term used in the field of science and technology to describe the evaluations, analysis or testing carried out on components of materials without destroying any part or components of the test materials. It is very useful in scientific research or industrial engineering environments. When any disruption of physical structure or configuration of a component will lead to discontinuing of the test, and it may not affect the usefulness of the affected parts.

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3 years ago

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

svetoff [14.1K]

Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Hence the charge at any time, t is $q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

6 0

3 years ago

When you walk at an average speed (constant speed, no acceleration) of 20.7 m/s in 75.8 sec you will cover a distance of______?

nlexa [21]

1,569.06m
(20.7•75.8)

3 0

3 years ago

A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at

pav-90 [236]

Answer:

Explanation:

Given

mass of drop $m=2 kg$

height of fall $h=1 m$

ball leaves the foot with a speed of 18 m/s at an angle of $55^{\circ}$

Velocity of ball just before the collision with the floor

$u^=2gh$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.8\times 1}=4.42 m/s$

Impulse delivered in Y direction

$J_y=m(v\sin (55)-(-u))$

$J_y=2(18\sin (55)+4.42)$

$J_y=38.32 kg-m/s$

Impulse in x direction

$J_x=m\times v\cos (55)$

$J_x=2\times 4.42\cos (55)=20.646$

$J_{net}=\sqrt{J_x^2+J_y^2}$

$J_{net}=\sqrt{(38.32)^2+(20.64)^2}$

$J_{net}=43.52 kg-m/s$

at an angle of $\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}$

$\phi =tan^{-1}(1.856)$

$\phi =61.7^{\circ}$

7 0

3 years ago