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Vika [28.1K]
3 years ago
5

In which situation is a person doing scientific work on an object?

Physics
1 answer:
viktelen [127]3 years ago
7 0
When they’re in a science lab or in school
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A university is planning to send an optical telescope into orbit. stuents attending the university will be able to view which of
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Help please! This question is driving me crazy
Varvara68 [4.7K]

Answer:

-10.8°, or 10.8° below the +x axis

Explanation:

The x component of the resultant vector is:

x = 3.14 cos(30.0°) + 2.71 cos(-60.0°)

x = 4.07

The y component of the resultant vector is:

y = 3.14 sin(30.0°) + 2.71 sin(-60.0°)

y = -0.777

Therefore, the angle between the resultant vector and the +x axis is:

θ = atan(y / x)

θ = atan(-0.777 / 4.07)

θ = -10.8°

The angle is -10.8°, or 10.8° below the +x axis.

3 0
3 years ago
Which evidence taken from a crime scene would be the best evidence for a District Attorney to gain a conviction with DNA evidenc
mezya [45]

Answer:

Blood

Explanation:

All the mentioned option will yield the same DNA for the suspect but blood, besides the DNA also contain further information that can be matched to the victim like blood group, genotype, some antibodies present and even traces of substances and disease.

6 0
3 years ago
Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini
amid [387]

Answer:

Recall that the electric field outside  a uniformly charged solid sphere  is exactly the same as if the charge were all at a point in the centre of the  sphere:

E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2}  } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r

=\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2}  }{2R^{3} } ]

∴NOTE: Graph is attached

8 0
3 years ago
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