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3 years ago

It is known that a shark can travel at a speed of 15 m/s.how far can a shark go in 10 seconds?

Physics

2 answers:

MrRissso [65]3 years ago

8 0

If a shark can travel 15 miles per second, then it can go 150 miles in 10 seconds.

Alla [95]3 years ago

6 0

Start with 15 m/s. There are 2 units and you need to cancel the seconds to get meters.

10 seconds has seconds as a unit. This can be used to cancel the sec in the denominator of 15 m/s. This just a little tip I use when I'm unsure about what to do.

15 m/s x 10 s = 150 meters

Hope I helped :)

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A laser beam is incident at an angle of 33.0° to the vertical onto a solution of cornsyrup in water.(a) If the beam is refracted

IrinaK [193]

Answer:

1.29649

488.08706 nm

$6.14644\times 10^{14}\ Hz$

231715700.28346 m/s

Explanation:

n denotes refractive index

1 denotes air

2 denotes solution

$\lambda_0$ = 632.8 nm

From Snell's law we have the relation

$n_1sin\theta_1=n_2sin\theta_2\\\Rightarrow n_2=\dfrac{n_1sin\theta_1}{sin\theta_2}\\\Rightarrow n_2=\dfrac{1\times sin33}{sin24.84}\\\Rightarrow n_2=1.29649$

Refractive index of the solution is 1.29649

Wavelength is given by

$\lambda=\dfrac{\lambda_0}{n_2}\\\Rightarrow \lambda=\dfrac{632.8}{1.29649}\\\Rightarrow \lambda=488.08706\ nm$

The wavelength of the solution is 488.08706 nm

Frequency is given by

$f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{488.08706\times 10^{-9}}\\\Rightarrow f=6.14644\times 10^{14}\ Hz$

The frequency is $6.14644\times 10^{14}\ Hz$

$v=\dfrac{c}{n_2}\\\Rightarrow v=\dfrac{3\times 10^8}{1.29469}\\\Rightarrow v=231715700.28346\ m/s$

The speed in the solution is 231715700.28346 m/s

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Calculate the electric field at point A, located at coordinates (0 m, 12.0 m ). Give the x and y components of the electric fiel

Yanka [14]

Answer:

The correct answer is " $(0,300\times 10^{-3} \ N/C)$ ".

Explanation:

The given problem seems to be incomplete. Please find the attachment of the complete query.

According to the question,

At point A, we have

⇒ $E_x = \frac{k q_1}{d_1^2} Cos \theta_1 - \frac{k q_2}{d_2^2} Cos \theta_2$

or,

⇒ $E_x = 9\times 10^9\times [\frac{6\times 10^{-9}}{15^2}\times \frac{9}{15}-\frac{8\times 10^{-9}}{20^2}\times \frac{16}{20} ]$

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and,

⇒ $E_y = \frac{kq_1}{d_1^2}Sin \theta_1 +\frac{kq_2}{d_2^2}Sin \theta_2$

or,

⇒ $E_y = 9\times 10^9\times [\frac{6\times 10^{-9}\times 12}{15^2\times 15}+ \frac{8\times 10^{-9}\times 12}{20^2\times 20} ]$

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