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1 year ago

6

You are learning about energy transformations in science class. Mel and Sam's built this set-up to see if light energy could be

transformed into _________ energy.

1 answer:

Allisa [31]1 year ago

5 0

I think it might be heat energy. light transforms into heat energy

You might be interested in

A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x=0 marking the city limits. His

adoni [48]

Answer:

1) $v_2=23\ m.s^{-1}$ & $x_2=43\ m$ east of sign post

2) $x'=55\ m$ east of sign post

3) $x_n=205\ m$ east of the signpost.

4) $v_z=35\ m.s^{-1}$

Explanation:

Given:

position of motorcyclist on entering the city at the signpost, $x_0=0\ m$

time of observation after being at x=5m east of the signpost, $t_m=0\ s$

constant acceleration of the on entering the city, $a=4\ m.s^{-2}$

distance of the motorcyclist moments later after entering, $s_m=5\ m$

velocity of the motorcyclist moments later after entering, $u_m=15\ m.s^{-1}$

<u>Now the initial velocity on at the sign board:</u>

$u_m^2=u^2+2.a.x_m$

where:

$u=$ initial velocity of entering the city at the signpost

Putting respective values:

$15^2=u^2+2\times 4\times 5$

$u=13.6015\ m.s^{-1}$

1)

Position at time $t_2=2\ s$ sec.:

Using equation of motion,

$x_2=u_m.t_2+\frac{1}{2} a.(t_2)^2+5$ because it has already covered 5m before that point

$x_2=15\times 2+0.5\times 4\times 2^2+5$

$x_2=43\ m$ east of sign post

Velocity at time $t_2=2\ s$ sec.:

$v_2=u_m+a.t_2$

$v_2=15+4\times 2$

$v_2=23\ m.s^{-1}$

2)

Position when the velocity is $v'=25\ m.s^{-1}$ :

using equation of motion,

$v'^2=u_m^2+2.a.x'+5$

$25^2=15^2+2\times 4\times x'+5$

$x'=55\ m$ east of sign post

3)

Given that:

acceleration be, $a_n=2\ m.s^{-2}$

time, $t_n=5\ s$

Position after the new acceleration and the new given time:

using equation of motion,

$x_n=u_m.t_n+\frac{1}{2} a_n.(t_n)^2+5$

$x_n=15\times 5+0.5\times 2\times 5^2+5$

$x_n=205\ m$ east of the signpost.

4)

now time of observation, $t_z=5\ s$

$v_z=u_m+a.t_z$

$v_z=15+4\times 5$

$v_z=35\ m.s^{-1}$

8 0

2 years ago

What charge accumulates on the plates of a 2.0-μF air-filled capacitor when it is charged until the potential difference across

enot [183]

Answer:

0.0002 C.

Explanation:

Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)

Mathematically, charge can be expressed as

Q = CV ................................. Equation 1

Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.

Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.

Substitute into equation 1

Q = 2×10⁻⁶× 100

Q = 2×10⁻⁴ C

Q = 0.0002 C.

The amount of charge accumulated = 0.0002 C

7 0

1 year ago

Read 2 more answers

When 30 V is applied across a resistor it generates 600 W of heat: what is the magnitude of its resistance?

grandymaker [24]

Answer:

<h2>1.5 ohms</h2>

Explanation:

Power is expressed as P = V²/R

R = resistance

V = supplied voltage

Given P = 600W and V = 30V

R = V²/P

R = 30²/600

R = 900/600

R = 1.5ohms

magnitude of its resistance is 1.5ohms

3 0

2 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

2 years ago

Mrs. Gonzalez is about to give birth and Mr. Gonzalez is rushing her to the hospital at a speed of 30.0 m/s. Witnessing the spee

valina [46]

Answer: The frequency = 1714.3Hz

Explanation: The solution can be achieved by using doppler effect formula.

Since the source is moving toward the observer, the velocity of the observer will be positive.

Please find the attached file for the solution

3 0

1 year ago