A car starts from the origin and is driven 1.88 km south, then 9.05 km in a direction 47° north of east. Relative to the origin,
1 answer:
Answer:
(a) θ = 55.85 degree
(b) 7.89 km
Explanation:
Using vector notations
A = 1.88 km south = 1.88 (- j) km = - 1.88 j km
B = 9.05 km 47 degree north of east
B = 9.05 ( Cos 47 i + Sin 47 j) km
B = (6.17 i + 6.62 j) km
Net displacement is
D = A + B
D = - 1.88 i + 6.17 i + 6.62 j = 4.29 i + 6.62 j
(a) Angle made with positive X axis
tanθ = 6.62 / 4.29 = 1.474
θ = 55.85 degree
(b) distance =
distance = 7.89 km
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Answer:
6.19 x m/
Explanation:
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∑ = N - mg = m.
= 0
From the exercise, we deduce there is no motion in y-axis, so:
N = mg
Then for x-axis we have:
∑ = H -
= m.
= 0
Now, from the exercise we deduce that we are looking for the greatest static friction which means to have the maximun static friction to start moving, so at this point the acceleration is zero, so we can find horizontal force (H), which then will act in the airplane to move it. Therefore we have:
H = =
=
N =
mg
H = (0.76)(84Kg)(9.8m/)
H = 625.63 N
Now we apply this force to the weight of the plane to find the greatest acceleration the mann can give to start moving the plane.
a = =
a =
a = 6.19 x m/
Answer:
mass conservation is valid for all closed system where the mass will remain in the system always
Explanation:
Conservation of mass is applicable everywhere in classical physics
Here we can also apply mass conservation as we know that the initially the beaker and its water content is having total mass of 109.44 g
now when we heated it to higher temperature then its total mass will be lesser than the initial mass this is because some of the water may evaporate from the system.
Here if we repeat the same experiment with closed boundary then we can see that total mass will be conserved
So here mass conservation is valid for all closed system where the mass will remain in the system always