Determine the tension developed in cables OD and OB and the strut OC, required to support the 500-lb crate. The spring OA has an
unstretched length of 0.2 ft and a stiffness of kOA
1 answer:
Answer:Fob= 86.2 lb, Foc= 175 lb, Fod= 506 lb
Explanation:firstly we need to get the spring stretch by subtracting the unstretched length from 1. The result will be (1-0.2=0.8ft). We then get the spring force by multiplying kOA by 0.8ft. therefore the spring force will be {350x0.8}= 280 lb. After that we can resolve each force as shown in the attachment.
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Answer:
E = 1.04*10⁻¹ N/C
Explanation:
Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:
As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.
Replacing this values in the equation above, we can solve for a, as follows:
According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:
F = mp*a = q*E
For a proton, we have the following values:
mp = 1.67*10⁻²⁷ kg
q = e = 1.6*10⁻¹⁹ C
So, we can solve for E (in magnitude) , as follows:
⇒ E = 1.04*10⁻¹ N/C