Answer:
104 N
Explanation:
m = 1300 kg
a = 0.08m/s^2
F = 1300*0.08
F = 104 N
Newtons is the unit of force.
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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Answer:
8.97 Watt
Explanation:
Resistance, R = 20 ohm
Inductance, L = 10 mH
V(t) = 20 Cos (1000 t + 45°)
Compare with the standard equation
V(t) = Vo Cos(ωt + Ф)
Ф = 45°
ω = 1000 rad/s
Vo = 20 V
Inductive reactance, XL = ωL = 1000 x 0.01 = 10 ohm
impedance is Z.


Z = 22.36 ohm



Apparent power is given by
P = Vrms x Irms
P = 14.144 x 0.634
P = 8.97 Watt
Answer:
1.36 x 10^-3 cm
Explanation:
Area = 50 ft^2 = 46451.5 cm^2
mass = 6 oz = 170.097 g
density = 2.70 g/cm^3
Let t be the thickness of foil in cm.
mass = volume x density
mass = area x thickness x density
170.097 = 46451.5 x t x 2.70
t = 1.36 x 10^-3 cm
Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.
A :-) for this question , we should apply
F = ma
( i ) Given - m = 2 kg
a = 15 m/s^2
Solution :
F = ma
F = 2 x 15
F = 30 N
( ii ) Given - m = 2 kg
a = 10 m/s^2
Solution :
F = ma
F = 2 x 10
F = 20 N
.:. The net force of object ( i ) has greater force compared to object ( ii ) by
( 30 - 20 ) 10 N