Physics
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Real physics questions asked by students, solved step by step Every question below was submitted by a real student and answered step by step.
The answer is 6.4 hours. Average speed is total distance over total time, so setting (200 + 37t)/(4 + t) = 42 and clearing the fraction gives 5t = 32.
Equal-time segments require weighting by time: distance = 2(60) + 3(40) = 240 km over 5 h, giving 48 km/h, not the naive arithmetic mean of 50 km/h.
A car drives 10 km from A and immediately returns, taking 20 minutes total. Total path is 20 km and 20 min = 1/3 h, so the average speed is 20 / (1/3) = 60 km/h.
A jogger runs 3.0 km in 10 min, rests 5 min, then runs 1500 m in 5 min. Total 4500 m over 20 min gives an average speed of 225 m/min, since rest time still counts.
A 120 km trip averages 60 km/h, so it lasts 2 h. The first 30 km at 40 km/h take 0.75 h and 70 km at 70 km/h take 1 h, leaving 20 km in 0.25 h = 80 km/h.
The average speed is 54.5 km/h, not 55. Equal distances at different speeds call for the harmonic mean 2*v1*v2/(v1 + v2) = 2(50)(60)/110 = 600/11 km/h.
Split the road into fractions of s: time = (s/4)/20 + (3s/4)/40 = s/32, so the average speed is exactly 32 km/h, well below the naive mean of 35 km/h.
Maximum shear is 3800 kN and maximum bending moment is 13000 kN·m, both at the fixed end. Sum the vertical loads for the reaction, then take moments about the support.
A car passes A, B, C spaced 12 km apart, taking 20 min for AB and 30 min for BC. Then v(AB) = 36 km/h, v(BC) = 24 km/h and v(AC) = 28.8 km/h, so AB is fastest.
45 mi/h equals 20.1168 m/s, about 20.12 m/s. Chain two conversion factors — 1 mile = 1609.344 m and 1 hour = 3600 s — so the unwanted units cancel on paper.
At 1040 km/h a plane moves 1040000/3600 = 288.9 m/s. Over a 500 ms blink that is 288.9 x 0.5 = 144.4 m, so the closest listed value is 144 m.
Divide each weight fraction by its density (0.55/1.7 and 0.45/1.1), then normalise. The less dense resin occupies more volume than its 45 percent weight share suggests.
G12 is 14.40 GPa. Shear loads the fibre and the matrix in series, so their compliances add: 1/G12 = 0.75/27 + 0.25/6 = 0.069444 per GPa, and inverting gives 14.40.
The Krenchel factor is 0.955. Take the cosine of 8.7 degrees (0.98849) and raise it to the fourth power: even a small 8.7 degree misalignment costs about 4.5 percent of stiffness.
The efficiency factor is 0.305. Half the fibres sit at 31 degrees to the load and half at 59 degrees, so sum a_n cos^4(theta_n) over both directions.
The laminate stiffness is 48.8 GPa. Plies aligned with the load count fully at 106 GPa, each off-axis ply counts as only 10 percent, and the total is averaged over five.
Two walkers leave A at 1.5 m/s and 2.0 m/s; the faster arrives 5.5 min = 330 s earlier. Solving s/1.5 - s/2 = 330 gives s/6 = 330, so AB = 1980 m.
E11 is 150 GPa and nu12 is 0.286. Both follow the parallel rule of mixtures: multiply each phase property by its volume fraction and add the two contributions.
The first integral is (1/2)(x')^2 - 0.005cos(x) = C, and the exact solution needs elliptic integrals. Small angles give x = C1cos(wt) + C2sin(wt) with w = 0.0707.
Reduce the constants first, then use the standard form 1/(a^2 - u^2), whose antiderivative is a logarithm. The integral is 7.209 and dividing by 3.6 gives t = 2.00 s.
The equivalent equation is V = m/d. Multiply both sides by V to free it from the denominator, then divide by d; the answer dm = V has the units the wrong way round.
F_B = 1300 N at 62.2 degrees. Forcing the horizontal components to cancel and the vertical components to total 1500 N turns one vector condition into two scalar equations.
E11 = 225.4 GPa and sigma11 = 2428.1 MPa. The rule of mixtures weights fibre and matrix properties by volume fraction: 0.66(340) + 0.34(3.05) for stiffness.
An object in uniform straight-line motion covers 180 km in 6 h, so v = s/t = 180/6 = 30 km/h. Unit conversion shows this equals about 8.33 m/s, not 30 m/s.
Laminate 1 exceeds Laminate 2 by 35.75 GPa. What matters is the fraction of 0 degree plies, 50 percent against 20 percent, not the total number of layers.
The correct root is about 947 K, not the 1729 K a linear estimate suggests. The T⁴ radiation term is far from negligible, so the quartic must be solved numerically.
A pedestrian walks 780 m at a constant 2 m/s. Rearranging v = s/t gives t = 780/2 = 390 s, and 390 s converts to exactly 6 minutes 30 seconds.
Convert 67 percent by weight to Vf = 0.5357, then apply the inverse rule of mixtures. Both transverse properties end up close to the weak matrix, not the stiff fibre.
E22 is 5.2 GPa. Transverse loading puts fibre and matrix in series, so the compliances add: 1/E22 = Vf/Ef + Vm/Em, and the soft matrix dominates the result.
The net work is about 142.5 J. Only the horizontal component 87 cos 33 = 72.96 N drives the box, and friction removes 62 N of it over the 13 m path.