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8090 [49]
3 years ago
6

If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :

Physics
2 answers:
rewona [7]3 years ago
7 0

Answer:

<u>(Frequency)⁻³(velocity)⁴Density</u>

Explanation:

See attachments.

gizmo_the_mogwai [7]3 years ago
5 0

Let's see

Momentum be P

\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c

\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c

\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c

\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}

\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c

On comaparing

  • c=1

So

  • b-3c=1
  • b-3=1
  • b=1+3
  • b=4

and

  • -a-b=-1
  • -a-4=-1
  • -a=-1+4=3
  • a=-3

So the unit is

  • DV⁴/F³
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Two balloons (m = 0.021 kg) are separated by a distance of d = 16 m. They are released from rest and observed to have an instant
evablogger [386]

(a) 2.56\cdot 10^{-5} C

According to Newton's second law, the force experienced by each balloon is given by:

F = ma

where

m = 0.021 kg is the mass

a = 1.1 m/s^2 is the acceleration

Substituting, we found:

F=(0.021)(1.1)=0.0231 N

The electrostatic force between the two balloons can be also written as

F=k\frac{Q^2}{r^2}

where

k is the Coulomb's constant

Q is the charge on each balloon

r = 16 m is their separation

Since we know the value of F, we can find Q, the magnitude of the charge on each balloon:

Q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(0.0231)(16)^2}{9\cdot 10^9}}=2.56\cdot 10^{-5} C

(b) 1.6\cdot 10^{14} electrons

The magnitude of the charge of one electron is

e=1.6\cdot 10^{-19}C

While the magnitude of the charge on one balloon is

Q=2.56\cdot 10^{-5} C

This charge can be written as

Q=Ne

where N is the number of electrons that are responsible for this charge. Solving for N, we find:

N=\frac{Q}{e}=\frac{2.56\cdot 10^{-5}}{1.6\cdot 10^{-19}}=1.6\cdot 10^{14}

5 0
3 years ago
Marcy pulls a backpack on wheels down the 100 m hall. The 60N force is applied at an angle of 30° above the horizontal. How much
shutvik [7]

Work= Fcos∆×S

W=60N×cos 30⁰×100

W=60×0.866×100=5196.1J

6 0
3 years ago
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jeka94

Answer:

Ive left an image here for use, I hope its helpful

Explanation:

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5 0
3 years ago
Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s
muminat

Answer:

Acceleration = 10.06 m/s²

Explanation:

1 mile = 1.6093km

1609.3m = 1 mile

1 m = \frac{1}{1609} mile

50.0 miles/hour = \frac{50 * 1609.3}{60 * 60} m/s

                          = 22.35m/s

from equation

S = Ut + 1/2 at²

v = U + at

22.35 = 0 + a * 2.22

a = 22.35 ÷ 2.22

= 10.06 m/s²

4 0
3 years ago
Which of the following weather conditions would result in the greatest rate of evaporation from the Earth's surface? A. hot and
lisov135 [29]
A: A hit and dry weather

Hope this helps :)
6 0
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