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denis23 [38]
3 years ago
6

44.2 cm + 0.123 cm = cm

Physics
2 answers:
aalyn [17]3 years ago
7 0
The answer is most likely 44.323cm
trapecia [35]3 years ago
3 0
The answer is 44.323 cm
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Trace the path of a ray emitted from the tip of the object toward the focal point of the mirror and then the reflected ray that
Crazy boy [7]

Answer: find the attached files for the answer

Explanation:

The reflected ray appears to have originated from the focal point. We should actually draw a vector from the focal point through the point where the incident ray hits the mirror but we shorten the vector so that its starting point is on the mirror, without changing its angle.

Please find the attached files for the solution

5 0
3 years ago
A 10- kilogram block is pushed across a horizontal surface with a horizontal force of 20 N against a friction force of 10 N. The
Temka [501]

Answer:

1m/s^2

Explanation:

Mass of block=10 kg

Applied horizontal force =F=20 N

Friction force=f=10 N

We have to find the acceleration of block.

Net force=Applied horizontal force-friction force

ma=F-f

Where F= Horizontal force

f=Friction force

m=Mass of object

a=Acceleration of object

10a=20-10=10

a=\frac{10}{10}=1 m/s^2

Hence, the acceleration of the block=1m/s^2

4 0
3 years ago
An insulating sphere is 8.00 cm in diameter and carries a 6.50 µC charge uniformly distributed throughout its interior volume.
Kobotan [32]

Explanation:

(a)   Formula to calculate the density is as follows.

            \rho = \frac{Q}{\frac{4}{3}\pi a^{3}}

                       = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}

                     = 2.42 \times 10^{-2} C/m^{3}

Now, calculate the charge as follows.

            q_{in} = \rho(\frac{4}{3} \pi r^{3})

                      = 2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}

                      = 10.106 \times 10^{-8} C

or,                   = 101.06 nC

(b)  For r = 6.50 cm, the value of charge will be calculated as follows.

                q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}

                          = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}

                          = 7.454 \mu C

7 0
3 years ago
A train station bell gives off a fundamental tone of 500 Hz as the train approaches the station at a speed of 20 m/s. If the spe
Brut [27]

Answer: 529.9 Hz

Explanation:

Here we need to use the Doppler equation, so we have:

f' = f*(v + v0)/(v - vs)

Here, f is the frequency = 500Hz

v is the velocity of the wave, = 334m/s

v0 is the velocity of the observer = 20m/s

vs is the velocity of the source = 0m/s

Then we have:

f' = 500Hz*(334m/s + 20m/s)/(334m/s) = 529.9 Hz

8 0
3 years ago
When two magnets are brought close together their magnetic
Gnoma [55]

Answer:

You know the saying "Opposites attract" well that is how you can remember that South and North Magnetic Poles connect.  

Explanation:

Hope this helps ya

6 0
3 years ago
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