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Nadusha1986 [10]
2 years ago
14

3. Check that the dimensions of both side of the equations below are agreeing to each other:

Physics
1 answer:
Reptile [31]2 years ago
4 0

Answer:

b is the answer of that question answer sapai

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3. Why is the term cold blooded a misconception? Explain​
jok3333 [9.3K]

Answer:

The term “cold-blooded” implies that these animals are in a never-ending struggle to stay warm. That really isn't correct. A cold-blooded animal can warm up their blood by being in the sun for hours.

7 0
3 years ago
How can I covert these 2,34 meters in decimeters<br>​
zysi [14]

Answer:

If you meant 2.34, 2.34 meters = 23.4 decimeters.

Formula: multiply the value in meters by the conversion factor '10'.

So, 2.34 meters = 2.34 × 10 = 23.4 decimeters.

Hope that helps. x

4 0
2 years ago
Read 2 more answers
A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50
Lelechka [254]

Answer:1.902 m

Explanation:

Given

height of apartment=1.5 m

It takes 0.21 sec to reach the bottom from apartment

So

s=u_1t+\frac{gt^2}{2}

1.5=u_1\times 0.21+\frac{9.81\times 0.21^2}{2}

u_1=6.11 m/s

i.e. if ball is dropped from top its velocity at window is 6.11 m/s

So height of upper floor above window

v^2-u^2=2as

where s= height of upper floor above window

here u=0

6.11^2=2\times 9.81\times s

s=\frac{6.11^2}{2\times 9.81}

s=1.902 m

5 0
3 years ago
If a certain shade of blue has a frequency of 7.33 x 10^14Hz what is the energy of exactly one photon of this light. Is this as
geniusboy [140]
In calculating the energy of a photon of light, we need the relationship for energy and the frequency which is expressed as:
 
E=hv

where h is the Planck's constant (6.626 x 10-34 J s)and v is the frequency.
E = 6.626 x 10-34 J s (<span>7.33 x 10^14 /s) = 4.857 x 10^-19 J</span>
6 0
3 years ago
A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 33.0 nC , respecti
matrenka [14]

Explanation:

Formula depicting relation between total flux and total charge Q is as follows.

              \phi  = \frac{Q}{\epsilon_{o}}    (Gauss's Law)

Putting the given values into the above formula as follows.

            Q = \phi \times \epsilon_{o}

                = -953 Nm^{2}/C \times 8.854 \times 10^{-12}

                = -8.4 \times 10^{-9} C

                = -8.4 nC

Therefore, when the unknown charge is q  then,

         -14.0 nC + 33.0 nC + q = -8.4 nC

               q = -27.4 nC

Thus, we can conclude that charge on the third object is -27.4 nC.

7 0
3 years ago
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