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joja [24]
3 years ago
10

EASY BRAINLIEST PLEASE HELP!!

Physics
2 answers:
Kryger [21]3 years ago
7 0

Answer:

solution given:

frequency[f]=215.0Hz

velocity[V]=x

wave length=2.4m

we have

wave length=\frac{V}{f}

2.4m=\frac{x}{215}

2.4×215=\frac{x}{}

x=516 m/s

velocity=speed =516m/s

Musya8 [376]3 years ago
3 0

Answer:

look at the picture i have sent

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A uniform rod of length l and mass m is pivoted around a line perpindicular to the rod at location l/3
beks73 [17]

Answer:

I for a rod about its center is I = M L^2 / 12

A = L (1/2 - 1/3) = L / 6     distance from middle to location L/3

A^2 = L^2 / 36

I = M L^2 (1/12 + 1/36) = M L^2 (4 / 36) = M L^2 / 9

I about given location by parallel axis theorem

3 0
2 years ago
Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7
sleet_krkn [62]

Answer : \underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Explanation :

Given that,

Charge of particle 1 =  -1.50\times10^{-7} c

Distance x = 6 cm

Charge of particle 2 = 1.50\times10^{-7} c

Distance x = 27 cm

Total distance = \dfrac{6+27}{2}

r = 16.5\ cm

Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm

Formula of electric field

E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}

Now, the the electric field due to  particle 1

\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i}  \dfrac{N}{C}

\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i}  \dfrac{N}{C}

\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Similarly, the electric field due to particle 2

\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Resultant Electric field

\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}

\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Hence, this is the required answer.






3 0
3 years ago
A spherical snowball is melting in such a way that its radius is decreasing at a rate of 0.1 cm/min. At what rate is the volume
Ne4ueva [31]

Answer: 80.384 cubic cm /min

Explanation:

Let V denote the volume and r denotes the radius of the spherical snowball .

Given : \dfrac{dr}{dt}=-0.1\text{cm/min}

We know that the volume of a sphere is given by :-

V=\dfrac{4}{3}\pi r^3

Differentiating on the both sides w.r.t. t (time) ,w e get

\dfrac{dV}{dt}=\dfrac{4}{3}\pi(3r^2)\dfrac{dr}{dt}\\\\\Rightarrow\ \dfrac{dV}{dt}=4\pi r^2 (-0.1)=-0.4\pi r^2

When r= 8 cm

\dfrac{dV}{dt}=-0.4(3.14)(8)^2=-80.384

Hence, the volume of the snowball decreasing at the rate of 80.384 cubic cm /min.

6 0
3 years ago
Which of the following statements is true of space exploration?
elena-s [515]
You don't have a following space exploration
6 0
3 years ago
Coulomb's law is expressed mathematically as
nataly862011 [7]
Force between two charges  = 

     ( 1/4πε₀ ) · (Charge #1) · (Charge #2) / (Distance between them)²

  in the direction away from each other.

In other words, if the force is positive, the charges are repelling.
If the force is negative, the charges are attracting.
4 0
3 years ago
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