Answer:
Weight of body = 490 Newton
Explanation:
Mass of the body is the total amount of matter contained in the body but Weight of the body is the force exerted by gravity on the body. Weight of the body depends upon acceleration due to gravity of the planet.
Thus,
Weight = mass × acceleration due to gravity
W = mg
W = weight of the body
m = mass of the body
g = acceleration due to gravity
W = 50 × 9.8
W = 490 
W = 490 Newton
Hence, weight of the body is 490 Newton.
Answer:
a) B = 10⁻¹ r
, b) B = 4 10⁻⁹ / r
, c) B=0
Explanation:
For this exercise let's use Ampere's law
∫ B. ds = μ₀ I
Where I is the current locked in the path. Let's take a closed path as a circle
ds = 2π dr
B 2π r = μ₀ I
B = μ₀ I / 2μ₀ r
Let's analyze several cases
a) r <Rw
Since the radius of the circumference is less than that of the wire, the current is less, let's use the concept of current density
j = I / A
For this case
j = I /π Rw² = I’/π r²
I’= I r² / Rw²
The magnetic field is
B = (μ₀/ 2π) r²/Rw² 1 / r
B = (μ₀ / 2π) r / Rw²
calculate
B = 4π 10⁻⁷ /2π r / 0.002²
B = 10⁻¹ r
b) in field between Rw <r <Rs
In this case the current enclosed in the total current
I = 0.02 A
B = μ₀/ 2π I / r
B = 4π 10⁻⁷ / 2π 0.02 / r
B = 4 10⁻⁹ / r
c) the field outside the coaxial Rs <r
In this case the waxed current is zero, so
B = 0
Answer:
a. Disk, 28 thousand light-years from
Explanation:
Since, a galactic disc is a component of disc galaxies, for instance spiral galaxies and lenticular galaxies. It consists of a stellar component and a gaseous component.
Also, the Sun lies within the galactic disk or in other words it is thought to be located in the galactic disk,
The Sun is located about 26,000 light-years away from the centre of the galaxy.
∵ From the given options 28,000 is nearest to 26,000
Hence, the sun is about 28 thousand light-years from the centre of the galaxy.
i.e. OPTION 'a' would be correct.
Answer:
37.125 m
Explanation:
Using the equation of motion
s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration
<u>Distance during acceleration</u>
Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.
Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be
a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}
Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion
s=0*4.5+ 0.5*1*4.5^{2}=0+10.125
=10.125 m
<u>Distance at a constant speed</u>
At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time
Distance=4.5 m/s*6 s=27 m
<u>Total distance</u>
Total=27+10.125=37.125 m
The answer would be B.
<span>
Standard deviation basically measures how spread out the values are. Without solving, you can easily tell which one among your choices have a smaller deviation. The closer the values are to each other the smaller the standard deviation. The values of choice B are the closest together, so you can assume that they have the smallest standard deviation. </span>
