Answer:
Speed of the wave in the string will be 3.2 m/sec
Explanation:
We have given frequency in the string fixed at both ends is 80 Hz
Distance between adjacent antipodes is 20 cm
We know that distance between two adjacent anti nodes is equal to half of the wavelength
So
We have to find the speed of the wave in the string
Speed is equal to
So speed of the wave in the string will be 3.2 m/sec
Answer:
(a) reaction at each front wheel is 5272N (upward)
(b) force between boulder and pallet is 4124N (compression)
Explanation:
Acceleration of the truck = 1 m/
(to the left)
when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,
= 0.5 m/
(upward) ,
= 0.5 m/
(upward)
Let T be tension in the cable
pallet and boulder: ∑fy = ∑(fy)eff = 2T- ( +
)g = (
+
)
= 2T- (400 + 50)*(9.81 m/) = (400 + 50)*(0.5 m/
)
T = 2320N
Truck: = ∑(
)eff: =
(3.4m) +
(2.0m) - T (0.6m)=
(1.0m)
Nf = (2.0m)(2000 kg)(9.81 m/ )/3.4m - (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/
) = 11541.2N - 409.4N - 588.2N = 10544N
∑fy (upward) = ∑(fy)eff: +
-
g = 0
10544 + - (2000kg)(9.81 m/
) = 0
= 9076N
∑fx (to the left) = ∑(fx)eff: - T =
= 2320N + (2000kg)(9.81 m/
) = 4320N
(a) reaction at each front wheel:
1/2 (upward): 1/2 (10544N) = 5272N (upward)
(b) force between boulder and pallet:
∑fy (upward) = ∑(fy)eff: +
g -
= (400kg)(9.81 m/
) + (400kg)(0.5 m/
) = 4124N (compression)