The image shows an electromagnet made with a 1.5-volt battery how would the magnetic field be affected if you replaced its 1.5-v
olt battery (on the left) with a 9-volt battery (shown on the right) while using the same nail and coil?
A. The magnetic field strength would increase
B. The magnetic field would not be affected by this change
C. The magnetic field strength would decrease
D. The magnetic field would change direction more frequently
A. The magnetic field strength would increase
B. The magnetic field would not be affected by this change
C. The magnetic field strength would decrease
D. The magnetic field would change direction more frequently
2 answers:
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Answer:
the work done to lift the bucket = 3491 Joules
Explanation:
Given:
Mass of bucket = 10kg
distance the bucket is lifted = height = 11m
Weight of rope= 0.9kg/m
g= 9.8m/s²
initial mass of water = 33kg
x = height in meters above the ground
Let W = work
Using riemann sum:
the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)
=
Work done in lifting the bucket (W) = force × distance
Force (F) = mass × acceleration due to gravity
Force = 9.8 * 10 = 98N
W done by bucket = 98×11 = 1078 Joules
Work done to lift the rope:
At Height of x meters (0≤x≤11)
Mass of rope = weight of rope × change in distance
= 0.8kg/m × (11-x)m
W done = integral of (mass×g ×distance) with upper 11 and lower limit 0
W done =
Note : upper limit is 11 not 1, problem with math editor
W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0
W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]
=7.84(60.5 -0) = 474.32 Joules
Work done in lifting the water
At Height of x meters (0≤x≤11)
Rate of water leakage = 36kg ÷ 11m = kg/m
Mass of rope = weight of rope × change in distance
= kg/m × (11-x)m = 3.27kg/m × (11-x)m
W done = integral of (mass×g ×distance) with upper 11 and lower limit 0
W done =
Note : upper limit is 11 not 1, problem with math editor
W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0
W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]
= 32.046(60.5 -0) = 1938.783 Joules
the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)
the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103
the work done to lift the bucket = 3491 Joules
Answer:
t = 2 v₀ / g
Explanation:
For this projectile launch exercise we use the displacement equations
x = vox t
y = y₀ + t - ½ g t²
As it is launched horizontally the vertical velocity is zero and the point of origin of the coordinate system is here, so y₀ is zero.
x = v₀ t
y = ½ g t²
They ask us for the time for which
x = y
vo t = ½ g t²
t = 2 v₀ / g