Answer:
The electric current in the wire is 0.8 A
Explanation:
We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:
B= Magnetic field due to a straight and long wire that carries current
u= Free space permeability
I= Electrical current passing through the wire
a = Perpendicular distance from the wire to the point where the magnetic field is located
Magnetic Field Calculation
We cleared (I) of the formula (1):
Formula(2)
a =8cm=0.08m
We replace the known information in the formula (2)
I=0.8 A
Answer: The electric current in the wire is 0.8 A
Answer:
<em>11.06m/s²</em>
Explanation:
According to Newtons second law of motion
Given
Mass m = 17kg
Fm = 208N
theta = 36 degrees
g = 9.8m/s²
a is the acceleration
Substitute
208 - 0.148(17)(9.8)cos 36 = 17a
208 - 24.6568cos36 = 17a
208 - 19.9478 = 17a
188.05 = 17a
a = 188.05/17
a = 11.06m/s²
<em>Hence the the magnitude of the resulting acceleration is 11.06m/s²</em>
Answer:
option A
Explanation:
the correct answer is option A
The given blanks in the question will be filled with Absolute and relative receptively.
time is absolute means the numerical representation of time example 12:00 pm is absolute time.
Measurement of distance is relative. A relative distance is one which is represented with the reference of some other place, it is not an exact value.
The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.
Final velocity(v3) of the red barge is calculated by following formula
m1×v1+ m2×v2= (m1+m2)v3
Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s
150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3
37500+ 320000= 1150000×v3
357500= 1150000×v3
v3= 0.311 m/s
<h3>What is elastic collision velocity? </h3>
- The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.
For more information on elastic collision velocity kindly visit to
brainly.com/question/29051562
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