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Free_Kalibri [48]
4 years ago
13

A sine wave with maximum amplitude after 1/4 cycle

Physics
1 answer:
frosja888 [35]4 years ago
4 0
Yeah what do you want me to answer for this question?
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Philip was diagnosed with __________ vestibular disorder, which is a dysfunction of the nervous system that processes balance. A
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D. Central

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3 years ago
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A 2000-kg truck is being used to lift a 400-kg boulder B that is on a 50-kg pallet A. Knowing the acceleration of the rear-wheel
anzhelika [568]

Answer:

(a) reaction at each front wheel is 5272N (upward)

(b) force between boulder and pallet is 4124N (compression)

Explanation:

Acceleration of the truck a_{t = 1 m/s^{2}  (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,

a_{A} =  0.5 m/s^{2} (upward) , a_{B} =  0.5 m/s^{2} (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- (m_{A} + m_{B})g =  (m_{A} + m_{B})a_{B}

                                  = 2T- (400 + 50)*(9.81 m/s^{2}) = (400 + 50)*(0.5 m/s^{2})

                        T = 2320N

Truck:  M_{R} = ∑(M_{R})eff: = -N_{f} (3.4m) + m_{T} (2.0m) - T (0.6m)= m_{T} a_{T} (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/s^{2} )/3.4m -  (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/s^{2})  = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: N_{f} + N_{R} - m_{T}g = 0

                                       10544 + N_{R}  - (2000kg)(9.81 m/s^{2} ) = 0

                N_{R} = 9076N

   ∑fx (to the left) = ∑(fx)eff:  F_{R} - T = m_{T} a_{T}

                                      F_{R} = 2320N + (2000kg)(9.81 m/s^{2} ) = 4320N

(a) reaction at each front wheel:

1/2 N_{f} (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: N_{B} + M_{B}g - m_{B}a_{B}

            N_{B} = (400kg)(9.81 m/s^{2}) + (400kg)(0.5 m/s^{2}) = 4124N (compression)

3 0
4 years ago
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At 0.0 degrees Celsius, a wire cable is 410.0000 meters in length. When the temperature increased to 30.0 degrees Celsius, the w
zzz [600]

Answer: The coefficient of expansion for the wire 0.000028 (^oC)^{-1}.

Explanation:

Original length of the wire = L= 410.0000 m

Initial temperature =T_1=0.0^oC

Final Temperature = T_2=30.0^oC

Increase in Length of the wire = \Delta L=0.3444 m

\frac{\Delta L}{L}=\alpha _{L}\times \Delta T=\alpha _{L}\times (T_2-T_1)

\alpha _L=\frac{\Delta L}{L\times (T_2-T_1)}=\frac{0.3444 m}{410.0000 m\times (30.0^oC-0.0^oC)}

\alpha _L=0.000028 (^oC)^{-1}

The coefficient of expansion for the wire 0.000028 (^oC)^{-1}.

3 0
3 years ago
An electric field of intensity 3.7 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.3
jekas [21]

Answer:

a)906.5 Nm^2/C

b) 0

c) 742.56132 N•m^2/C

Explanation:

a) The plane is parallel to the yz-plane.

We know that

flux ∅= EAcosθ

3.7×1000×0.350×0.700=906.5 N•m^2/C

(b) The plane is parallel to the xy-plane.

here theta = 90 degree

therefore,

0  N•m^2/C

(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.

therefore, applying the flux formula we get

3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C

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3 years ago
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What cloud brings thunderstorms and lightning?
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<span>Cumulonimbus clouds is what you're looking for.</span>
6 0
4 years ago
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