A person exerts a horizontal force of F=45N on the end of an 86cm wide door. The magnitude of the torque due to F about the pivot point is determined by |τ|=|rxF|=rFsinθ . Determine the magnitude of the torque, |τ| , on the door about its hinges due to F . |τ|=0Nm |τ|=38.7Nm |τ|=3870Nm
Answers
Answer: The magnitude of torque is 38.7Nm
Explanation: Please see the attachment below
The magnitude of the torque on the door about its h1nges due to the applied force is 38.7 Nm.
How to calculate the magnitude of the torque?
The magnitude of the torque on the door about its h1nges due to the applied force is calculated by applying the following formula as shown below;
τ = rF
where;
- r is the perpendicular distance of the applied force
- F is the applied force
The given parameters include;
perpendicular distance, r = 86 cm = 0.86 m
the applied force , F = 45 N
The magnitude of the torque on the door about its h1nges due to the applied force is calculated as;
τ = rF
τ = 0.86 m x 45 N
τ = 38.7 Nm
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Answers
Answer:
= 2630.6 N.m
Explanation:
(FR)x = ΣFx = -F4 = -407 N
(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N
(MR)B =ΣM + Σ(±Fd)
= MA + F1(d1 +d2) + F2d2 - F4d3
= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)
= 2630.64 N.m (counterclockwise)
Final answer:
The Cartesian components of the resultant force and the couple moment are calculated by summing up all the forces and moments acting on the object. The resultant force is 1724 N and the couple moment is 29.764 N*m.
Explanation:
The resultant force and couple moment in the Cartesian coordinate system can be obtained by summing up all the forces and moments acting on the object. In this case, we have the forces F1, F2, F3, F4 and the couple moment MA acting on the object. The resultant force (FR) can be calculated as the sum of all the forces, i.e., FR = F1 + F2 + F3 + F4. Using the values given, FR = 510 N + 306 N + 501 N + 407 N = 1724 N. The resultant moment (MR) can be calculated as the sum of all the moments, i.e., MR = d1*F1 + d2*F2 + d3*F3 + d4*F4 - MA. Using the values given, MR = 0.880 m * 510 N + 1.11 m * 306 N + 0.560 m * 501 N + 2.08 m * 407 N - 1504 N*m = 29.764 N*m. Therefore, the Cartesian components of the resultant force and the couple moment are 1724 N and 29.764 N*m respectively.
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Answers
To solve the problem it is necessary to apply the concepts related to thermal expansion of solids. Thermodynamically the expansion is given by
Where,
Original Length of the bar
= Change in temperature
= Coefficient of thermal expansion
On the other hand our values are given as,
Replacing we have,
The width of the expansion of the cracks between the slabs is 0.5832cm
The width of the expansion cracks between the slabs to prevent buckling should be 0.5832cm.
How to calculate width?
According to this question, the following information are given:
- Lo = Original length of the bar
- ∆T = Change in temperature
- α = Coefficient of thermal energy
The values are given as follows:
- Lo = 18m
- T1 = 25°C, T2 = 52°C
- α = 12 × 10-⁶/°C
∆L = Loα (T2 - T1)
∆L = 18 × 12 × 10-⁶ (27)
∆L = 3.24 × 10-⁴ × 18
∆L = 5.832 × 10-³m
Therefore, the width of the expansion of the cracks between the slabs is 0.5832cm.
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Answers
Answer:
2.70
Explanation:
pH = -log[H+]
pH = -log[2.0x10^-3]
pH = 2.70
Answers
Answer:
a=-2.77 m/s^2
Explanation:
Assuming constant acceleration,
v=at + v_0
where v_0 is the initial velocity.
At rest, v=0, so
0=at+v_0
So solving the equation for a:
a=(-v_0)/t
Inserting the numbers yields
a=-2.77 m/s^2
Answers
the answer is false. Hope this helps
Answers
Answer:
0.158 A.
Explanation:
Mass of gold deposited = 3 x 10^-3 kg
= 3 g
Molar mass = 196 g/mol
Number of moles = 3/196
= 0.0153 mol.
Faraday's constant,
1 coloumb = 96500 C/mol
Quantity of charge, Q = 96500 * 0.0153
= 1477.04 C.
Remember,
Q = I * t
t = 2.59 hr
= 2.59 * 3600 s
= 9324 s
Current, I = 1477.04/9324
= 0.158 A.
Answer:
0.158A
Explanation:
Using Faraday's first law of electrolysis which states that the mass(m) of a substance deposited or liberated at any electrode is directly proportional to the quantity of charge or electricity (Q) passed. i.e
m ∝ Q
m = Z Q
Where;
Z is the proportionality constant called electrochemical equivalent.
Faraday also observed that when 1 Faraday of electricity is equivalent to 96500C of charge.
Also,
Quantity of charge (Q), which is the product of current (I) passing through and the time taken (t) for the electrolysis, is given by;
Q = I x t; ----------------------(i)
With all of these in place, now let's go answer the question.
Since the gold ions carry one elementary unit of positive charge, now let's write the cathode-half reaction for gold (Au) as follows;
Au⁺ + e⁻ = Au ---------------------(ii)
From equation (ii) it can be deduced that when;
1 Faraday (96500C) of electricity is passed, 1 mole of Au forms ( = 197 grams of Au) [molar mass of Au = 197g]
Then, 3.00 x 10⁻³ kg (= 3 g of Au) will be formed by 3g x 96500C / 197g = 1469.5C
Therefore, the quantity of charge (Q) deposited is 1469.5C
Substitute this value (Q = 1469.5C) and time t = 2.59h (= 2.59 x 3600 s) into equation (i);
Q = I x t
1469.5 = I x 2.59 x 3600
1469.5 = I x 9324
Solve for I;
I = 1469.5 / 9324
I = 0.158A
Therefore, the current in the cell during that period is 0.158A
Note:
1 mole of gold atoms = 176g
i.e the molar mass of gold (Au) is 176g