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3 years ago

A car drives 30 m north in 1.5 seconds and drives 60 east in 2 seconds. What is the average velocity

Physics

1 answer:

Bad White [126]3 years ago

4 0

Answer:

19.2 m/s

Explanation:

From the question,

Velocity = displacement/time = d/t

V = d/t.................................. Equation 1

Average velocity = Total displacement/ total time

Total dispalcement = √(30²+60²) = √4500 = 67.08 m

Total time = 1.5+2 = 3.5 s.

Substitute these value into equation 1

V = 67.08/3.5

V = 19.2 m/s.

Hence the average velocity is 19.2 m/s

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Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m

maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension in the string

$\mu$ is the mass per unit length

For the string in the problem,

L = 30.0 m

$\mu=9.00\cdot 10^{-3} kg/m$

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

$f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz$

The next frequencies (harmonics) are given by

$f_n = nf$

with n being an integer number and f being the fundamental frequency.

So we get:

$f_2 = 2 (0.786 Hz)=1.572 Hz$

$f_3 = 3 (0.786 Hz)=2.358 Hz$

$f_4 = 4 (0.786 Hz)=3.144 Hz$

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4 years ago

Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a

weeeeeb [17]

Answer:

force becomes one - ninth

Explanation:

According to Coulomb's law in electrostatics, two charges can exert a force of attraction or repulsion on each other which is directly proportional to the product of two charges and inversely proportional to the square of distance between them.

Here both the charges remains same but the distance is variable.

So, we can say that

$F \alpha \frac{1}{d^{2}}$ .... (1)

Where d be the distance between the tow charges

As the distance between two charges increases by factor of three, let the new force be F'.

$F' \alpha \frac{1}{9d^{2}}$ .... (2)

Divide equation (2) by equation (1), we get

$\frac{F'}{F}=\frac{d^{2}}{9d^{2}}$

$F'=\frac{F}{9}$

Thus, the force becomes one - ninth times the initial force.

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Answer:

0.45

Explanation:

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inysia [295]

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