The correct answer is A; True.
Further Explanation:
This is a correct phrase that is important to learn when owning any type of vehicle. When a car battery is dead, it can usually be jump started by using another cars battery or a portable battery charger. It is extremely important to put the positive battery cable on the positive battery post. Then the negative cable will be placed on the negative car battery post and the negative ground wire can be anywhere on the car except on the battery.
The car needs to be connected properly for a few minutes before trying to start the car. This helps the car battery to get enough "juice" to start. If the battery cables are placed wrong this can cause sparks to come out of the cables/battery and cause bodily harm.
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Where is Eq.(28) ?? You should show it to find the result
Answer:
Explanation gives the answer
Explanation:
% Using MATLAB,
% Matlab file : fieldtovar.m
function varargout = fieldtovar(S)
% function that accepts single structure as input, assigning each
% of the field values to user-defined variables
fields = fieldnames(S); % get the field names of the input structure
% check if number of user-defined variables and number of fields in
% structure are equal
if nargout == length(fields)
% if equal assign each value of structure to user-defined varable
for i=1:nargout
varargout{i} = getfield(S,fields{i});
end
else
% if not equal display an error message
error('The number of output variables does not equal the number of fields');
end
end
%This brings an end to the program
The maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing is 28.98 MPa.
<h3>What is power?</h3>
Power is the energy transferred per unit time.
Torque is find out by
P = 2πNT/60
10000 = 2π x 2000 x T / 60
T =47.74 N.m
The gear ratio Ne / Ns =4/1
Ns =2000/4 = 500
Ts =Ps x 60/(2π x 500)
Ts =190.96 N.m
Maximum shear stress τ = 16/π x (T / (d₀⁴ - d₁⁴))
τ max =T/J x D/2
where d₁ = 30mm = 0.03 m
d₀ = 30 +(2x 4) = 38mm =0.038 m
Substitute the values into the equation, we get
τ max = 16 x 190.96 x 0.038 /π x (0.038⁴ - 0.03⁴)
τ max = 28.98 MPa.
Thus, the maximum shear stress in the tube is 28.98 MPa.
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