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3 years ago

5

sound transfers energy through solid liquid and gas what does ocean waves transfer energy through is it water or water and air a

s evaporation

1 answer:

trapecia [35]3 years ago

7 0

Hears the answer is endocytosis. <span />

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There are two blocks: one large one initially at rest, and a smaller one, initially moving to the right withsome speed. The smal

KATRIN_1 [288]

Answer:

Explanation:

Let the initial velocity of small block be v .

by applying conservation of momentum we can find velocity of common mass

25 v = 75 V , V is velocity of common mass after collision.

V = v / 3

For reaching the height we shall apply conservation of mechanical energy

1/2 m v² = mgh

1/2 x 75 x V² = 75 x g x 10

V² = 2g x 10

v² / 9 = 2 x 9.8 x 10

v² = 9 x 2 x 9.8 x 10

v = 42 m /s

small block must have velocity of 42 m /s .

Impulse by small block on large block

= change in momentum of large block

= 75 x V

= 75 x 42 / 3

= 1050 Ns.

6 0

3 years ago

a car travels at 15 m/s for 10 s. It then speeds up with a constant acceleration of 2.0 m/s? for 15 s. At the end of this time,

Firlakuza [10]

V = u + at where u is initial velocity (15 m/s), a is acceleration (2m/s^2) and t is time (15 seconds)

V = 15 + 2 X 15

V = 45 m/s

6 0

3 years ago

Copernicus and other astronomers before him thought that celestial bodies followed a _____ orbital path.

murzikaleks [220]

The correct answer is circular. Copernicus and other astronomers before him thought that celestial bodies followed a circular orbital path. Copernicus was a Polish astronomer that concluded that the sun is at rest near the center of the universe and the earth is revolving around it annually. This theory is called heliocentric.

6 0

2 years ago

Read 2 more answers

What is the power of a refrigerator with voltage 110 V and<br> current 0.8 A?

tia_tia [17]

Answer: 88

Explanation:

8 0

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