The frequency will not change
The hawk’s centripetal acceleration is 2.23 m/s²
The magnitude of the acceleration under new conditions is 2.316 m/s²
radius of the horizontal arc = 10.3 m
the initial constant speed = 4.8 m/s
we know that the centripetal acceleration is given by
=
= 23.04/10.3
= 2.23 m/s²
It continues to fly but now with some tangential acceleration
= 0.63 m/s²
therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration
so
=
=
= 2.316 m/s²
So the magnitude of net acceleration will become 2.316 m/s².
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the weight of the balloon is .030 * 10 = 0.3 N
the weight of the gas of volume v is 0.54*10 N
The lifting force of a volume of v m³ of displaced air is 1.29v N
so, we need
1.29*10*v = 0.3 + 0.54*10*v
or
1.29v = 0.03+0.54v
Answer:
c. 0.02 C and 4 J
Explanation:
Applying,
Q = CV................ Equation 1
Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.
From the question,
Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V
Substitute these values into equation 1
Q = (50×10⁻⁶)(400)
Q = 0.02 C.
Also Applying
E = CV²/2............. Equation 2
Where E = Energy stored.
Therefore,
E = (50×10⁻⁶ )(400²)/2
E = 4 J
Hence the right option is c. 0.02 C and 4 J