Explanation:
the spped of that we can't measure with own that measure a machine is called
Answer:
the magnitude of the total electric field is 25 V/m
Explanation:
Given data
point above = 0.5 m
charges = 18 V/m
to find out
the magnitude of the total electric field
solution
the magnitude of the total electric field is 25 V/m because number is less than the twice magnitude of field by every charge and the horizontal component in electric field by every charge cancel out and vertical component in field added together
For this case, the first thing you should do is write the kinematic motion equation of the block.
We have then:
vf = vo + a * t
Where,
vf: Final speed.
vo: Initial speed.
a: acceleration.
t: time.
Substituting the values:
(16) = (0) + a * (16)
Clearing the acceleration:
a = 16/16 = 1m / s ^ 2
Note: the other data for this case are not used in this problem.
answer:
The acceleration of the box is 1m / s ^ 2
Given that,
Current = 4 A
Sides of triangle = 50.0 cm, 120 cm and 130 cm
Magnetic field = 75.0 mT
Distance = 130 cm
We need to calculate the angle α
Using cosine law




We need to calculate the angle β
Using cosine law




We need to calculate the force on 130 cm side
Using formula of force



We need to calculate the force on 120 cm side
Using formula of force


The direction of force is out of page.
We need to calculate the force on 50 cm side
Using formula of force


The direction of force is into page.
Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.
Answer:
B. About 12 degrees
Explanation:
The orbital period is calculated using the following expression:
T = 2π*(
)
Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.
Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.
Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:
= 
Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be
years, or 31.62 years.
We find the amount of degrees it moves in 1 year:

or about 12 degrees.