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NISA [10]
3 years ago
14

What does a physicist study?

Physics
1 answer:
MissTica3 years ago
5 0

Answer:

A. Matter and energy

Explanation:

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What is the difference between celestial and terrestrial in planetary terms? Someone please answer
andriy [413]

"Celestial"  =  anything to do with the sky

("Cielo" ..... Spanish for "sky"
 "Ceiling" ... that thing up over your head
 "Caelum" .. Latin for "heaven")


"Terrestrial"  =  anything to do with the Earth

("Terra" ... Latin for "Earth")
5 0
3 years ago
An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo
Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=<u>16321.0769 N/C</u>

 Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷  0.000000152881

=<u>35321.58999 N/C</u>

8 0
2 years ago
A body is travelling with a velocity 30 m/s².what will be its velocity after 4s?​
Ipatiy [6.2K]

Answer:

70m/s²

Explanation:

we will use the first equation of Dalton to find it

3 0
3 years ago
A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.2 meters/second. At what horizontal distance fr
NARA [144]
To find the horizontal distance multiple the horizontal velocity by the time. Since there is no given time it must be calculated using kinematic equation.

Y=Yo+Voyt+1/2at^2
0=.55+0+1/2(-9.8)t^2
-.55=-4.9t^2
sqrt(.55/4.9)=t
t=0.335 seconds

Horizontal distance
=0.335s*1.2m/s
=0.402 meters
8 0
3 years ago
PLEASE HELLLLLLLLLLLLPPPPPPPPPP!!!!!! I will give brainliest
alexandr1967 [171]

Answer:

its C. The north pole of one magnet attracts the south pole of another

Explanation:

I JUST TOOK THE TEST

4 0
3 years ago
Read 2 more answers
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