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LuckyWell [14K]
3 years ago
8

What is another unit for momentum besides kg-m/s? a. N b. N-s c. N-s2 d. N/s

Physics
1 answer:
lara [203]3 years ago
8 0
'Newton-second' is dimensionally equivalent to 'kilogram-meter/second'.
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You throw a rock straight up into the air with a speed of 14.2 m/s. how long does it take the rock to reach its highest point?
slega [8]

The acceleration of gravity on or near the Earth's surface is 9.8 m/s² downward.
Is that right ?           I don't hear any objection, so I'll assume that it is.

That means that during every second that gravity is the only force on an object,
the object either gains 9.8m/s of downward speed, or it loses 9.8m/s of upward
speed.   (The same thing.)

If the rock starts out going up at 14.2 m/s, and loses 9.8 m/s of upward speed
every second, it runs out of upward gas in (14.2/9.8) = <em>1.449 seconds</em> (rounded)

At that point, since it has no more upward speed, it can't go any higher.  Right ?

(crickets . . .)

4 0
3 years ago
A system is best defined as?
erma4kov [3.2K]

Answer:

a group of organs working together to perform a function.

Explanation:

6 0
3 years ago
Read 2 more answers
The third floor of a house is 8m above street level. How much work is needed to move a 136kg refrigerator to the third floor?
jonny [76]

m = Mass of the refrigerator to be moved to third floor = 136 kg

g = Acceleration due to gravity by earth on the refrigerator being moved = 9.8 m/s²

h = Height to which the refrigerator is moved  = 8 m

W = Work done in lifting the object

Work done in lifting the object is same as the gravitational potential energy gained by the refrigerator. hence

Work done = Gravitation potential energy of refrigerator

W = m g h

inserting the values

W = (136) (9.8) (8)

W = 10662.4 J



8 0
3 years ago
A series circuit consists of a 100-ω resistor, a 10.0-μf capacitor, and a 0.350-h inductor. the circuit is connected to a 120-v
Tpy6a [65]
Current will be I=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{\sqrt{ R^{2}+(X_{C}-X_{L})^{2}}}\\where~X_{c}=\dfrac{1}{j.\omega .C}~and~X_{L}=j.\omega.L~where~\omega=2.\pi f~and~f=60Hz
now just pluf in the values and Voila..
7 0
3 years ago
A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
3 years ago
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