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3 years ago

5

Lightning flashed, and after 7 seconds there was a roll of thunder. To the nearest meter, determine how far lightning flashed fr

om us,
if the speed of sound is 337 m / s.

1 answer:

Nadya [2.5K]3 years ago

3 0

Let’s come back to the equation of SPEED OF AN ECHO
SPEED =2XDISTANCE/ TIME
So 337*7/2=1179.5 nearest meter =1800 m away
Hope you will get it right

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I need a diagram for how a scrap heap magnet works

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Which of the following describes electric current in the ocean?

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Electric fields cause electric currents in the ocean. When electromagnetic fields are induced in the conducting oceans and Earth, Ionospheric and Magnetospheric electric current systems flow above the surface of the ocean and by the action of moving sea water.

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4 years ago

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In her hand, a softball pitcher swings a ball of mass 0.245 kg around a vertical circular path of radius 59.8 cm before releasin

GuDViN [60]

Answer:

The velocity is $v_b = 20.17 \ m/s$

Explanation:

From the question we are told that

The mass of the ball is $m = 0.245 \ kg$

The radius is $r = 59.8 \ cm = 0.598 \ m$

The force is $F = 30.9 \ N$

The speed of the ball is $v = 16.0 \ m/s.$

Generally the kinetic energy at the top of the circle is mathematically represented as

$K_t = \frac{1}{2} * m * v^2$

=> $K_t = \frac{1}{2} * 0.245 * 16.0 ^2$

=> $K_t = 31.36 \ J$

Generally the work done by the force applied on the ball from the top to the bottom is mathematically represented as

$W = F * d$

Here d is the length of a semi - circular arc which is mathematically represented as

$d = \pi * r$

So

$W = 30.9 * 0.598$

$W = 18.48 \ J$

Generally the kinetic energy at the bottom is mathematically represented as

$K_b = \frac{1}{2} * m * v_b^2$

=> $K_b = \frac{1}{2} * 0.245 * v_b^2$

=> $K_b = 0.1225 * v_b^2$

From the law of energy conservation

$K_t + W =K_b$

=> $31.36+ 18.48 = 0.1225 * v_b^2$

=> $v_b = 20.17 \ m/s$

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3 years ago

For a galvanic cell to generate an electric current flowing from anode to cathode, what must be true

photoshop1234 [79]

Complete question:

For a galvanic cell to generate an electric current flowing from anode to cathode, what must be true?

(a) Electrons flow from the anode to the cathode

(b) Electrons flow from the more negatively charged electrode to the more positively charged electrode

(c) Electrons flow from higher potential energy to lower potential energy

(d) All of the above are true.

Answer:

(d) All of the above are true.

Explanation:

A galvanic or Voltaic cell is a primary type of electrochemical cell that is used to generate electrical energy from the chemical reactions that take place in it.

It consists of a positive electrode (cathode) and a negative electrode (anode) for the movement of charges.

(a) Electrons flow from the anode to the cathode. TRUE

Anode is the negative electrode and for electron current, electrons flow from negative electrode to positive electrode.

(b) Electrons flow from the more negatively charged electrode to the more positively charged electrode. TRUE

Based on electron current flow.

(c) Electrons flow from higher potential energy to lower potential energy. TRUE

The driving force of the electron flow is the potential difference. Electrons must flow from higher potential to lower potential.

All the options are correct, so we select option "D"

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3 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

4 years ago