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2 years ago

6

I've been getting this question wrong and i still don't know it.

2 answers:

Diano4ka-milaya [45]2 years ago

8 0

Answer:

the answer is B - He can use the laser to point to specific information on the slides.

Explanation:

laser pointers are used and its on edge

RideAnS [48]2 years ago

7 0

Answer: Option (A) is the correct answer.

Explanation:

Laser helps in zooming an image on which it is directed or falls so that it becomes easier for a large population to understand what the teacher or guide is talking about.

For example, when Greg is giving slide show presentation to a large audience then in order to make his slides visualized by everyone in the room he uses laser.

Therefore, we can conclude that a laser best help him with the presentation as he can use the laser to enlarge images on the slides.

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In the following equation , which compounds are the reactants? NaCL+AgNo>NaNo+AgCl

NeX [460]

The first two are always the reactants the products come after so they are last

3 0

2 years ago

You have probably noticed that carrying a person in a pool of water is much easier than carrying a person through air. To unders

Kruka [31]

Answer:

The answer is "0.91238 and 744.8"

Explanation:

In this scenario it is easier to take a person to the water-pool than to transport the people in the air, as the person's strength is increased by water upwards:

$f_b \to m \to mg =person \\\\F_B \ in\ air = v\ & air\ g \\\\$

$=0.076 \times 1.225 \times 9.8 \\\\ =0.91238 \ N\\\\$

$F_B \ in \ water = v \& water \ g \\\\$

$=0.076 \times 1000 \times 9.8\\\\= 744.8 \ N\\$

6 0

2 years ago

A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and

Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

$q_{out}=q_{2}-q_{1}$

$q_{out}=-6.43-(-4.00)$

$q_{out}=-2.43\ \mu C$

The charge on the inner surface is q.

$q+(-2.43)=-6.43$

$q=-6.43+2.43= 4.00\ \mu C$

We need to calculate the electric field outside the shell

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}$

$E=33927618.73\ N/C$

$E=3.39\times10^{7}\ N/C$

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

5 0

2 years ago

A sort of "projectile launcher" is shown below. A large current moves in a closed loop composed of fixed rails, a power supply,

Zarrin [17]

Answer:

wallah i don't understand anything with my stoopid brain

Explanation:

3 0

2 years ago

A certain AM radio wave has a frequency of 1.12 x 100 Hz. Given that radio waves travel at

riadik2000 [5.3K]

Answer: 267 m

Explanation:

2.99x10^8 m/s

———————-

1.12 x 10^6 Hz

3 0

2 years ago