Answer:
a) the magnitude of r is 184.62
b) the direction is 37.74° south of the negative x-axis
Explanation:
Given the data in the question;
as illustrated in the image blow;
To find the the magnitude of r, we will use the Pythagoras theorem
r² = y² + x²
r = √( y² + x²)
we substitute
r = √((-113)² + (-146)²)
r = √(12769 + 21316 )
r = √(34085 )
r = 184.62
Therefore, the magnitude of r is 184.62
To find its direction, we need to find ∅
from SOH CAH TOA
tan = opposite / adjacent
tan∅ = -113 / -146
tan∅ = 0.77397
∅ = tan⁻¹( 0.77397 )
∅ = 37.74°
Therefore, the direction is 37.74° south of the negative x-axis
The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:
where
m=225 kg is the mass of the boat
is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:
and the negative sign means the direction of the impulse is against the direction of motion of the boat.
Answer:
depth of well is 163.30 m
Explanation:
Given data
speed of sound = 343 m/s
timer = 6.25 s
to find out
depth of well
solution
let us consider depth d
so equation will be
depth = 1/2 ×g ×t² ..............1
and
depth = velocity of sound × time .................2
here we have given time 6.25 that is sum of 2 time
when stone reach at bottom that time
another is sound reach us after stone strike on bottom
so time 1 + time 2 = 6.25 s
so from equation 1 and 2 we get
1/2 ×g ×t² = velocity of sound × time
1/2 ×9.8 × t1² = 343 × (6.25 - t1 )
t1 = 5.77376 sec
so height = 1/2 ×g ×t²
height = 1/2 ×9.8 × (5.773)²
height = 163.30 m
Answer:
k = 2.279
Explanation:
Given:
Magnitude of charge on each plate, Q = 172 μC
Now,
the capacitance, C of a capacitor is given as:
C = Q/V
where,
V is the potential difference
Thus, the capacitance due to the charge of 172 μC will be
C = 
Now, when the when the additional charge is accumulated
the capacitance (C') will be
C' = 
or
C' = 
now the dielectric constant (k) is given as:
substituting the values, we get
or
k = 2.279
If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.
20 J/s = 10% of Supply
20 J/s = (0.1) x (Supply)
Divide each side by 0.1:
Supply = (20 J/s) / (0.1)
<em>Supply = 200 J/s </em>(200 watts)
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Here's something to think about: What could you do to make the lamp more efficient ? Answer: Use it for a heater !
If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about. Suddenly ... bada-boom ... the lamp is 90% efficient !
