Let his initial speed be V. Then for the last 80 miles he drove with V-15 mph speed. It took him 80/(V-15) hours to complete last 80 miles. The first 300-80=220 miles he drove with speed V and it took him 220/V hours to complete that part of the trip. Total time is 6 hours. Thus,   220/V+80/(V-15)=6   Or...

Let us have ladder with mass M and length L, standing against the wall at an angle θ with respect to ground. A person with mass m climbs distance x along the ladder before ladder starts sliding. The distance x can be found from the equilibrium conditions. We will assume FW to be the normal...

a) In the no-slipping case, energy conservation tells us:   mgh=mv2/2+Iω2/2; here ω=v/R, R is the radius of the marble, I=(2/5)mR2; Plugging in all the data into the first equation yields:   mgh=(7/10)mv2; From this we obtain:   h=(7/10)(v2/g); h=0...

Use energy conservation. Potential energy transforms into rotational energy.   mgh=Iω2/2; Here I is the moment of inertia with respect to the axis passing through the point at the rim, where the rope leaves the disk. If you draw a picture, you will realize that the disk at every instant...

The field on the axis is directed perpendicular to the disk and is given by:   E=∫035 2π*ρ*x/(x2+y2)3/2*y*dx; Here ρ=7.90*10-3 C/m2, charge density, x is the distance from the axis to a point on the disc, y is the distance from the disc to the point on the axis where the...

Since perimeter is the sum of all side lengths, it scales just as the sides themselves. So two perimeters scale as 2:3. If 56 cm is the perimeter of a small triangle, the perimeter of the large one is (3/2)*56=3*28=84 cm.

Since (a+βb) and (a-βb) are perpendicular, it means that (a+βb)•(a-βb)=0 or   a•a-β2*(b•b)=|a|2-β2*|b|2=0   Substituting values for |a| and |b| into the equation, we obtain:   9-25β2=0 or β=±3/5

-2(2m+1)2-17=11 ⇔ -2(2m+1)2=28 ⇔ (2m+1)2=-14 ⇔ 2m+1=±i√14, where i=√(-1);    Finally,    m=-½±(i/2)√14

I would like to add to answers something for you to think about. Why the sky is not violet? It is true that the shorter the wavelength, the stronger the scattering is. Then the sky shall be violet, since the violet light has shorter wavelength than the blue light. So why not?

Factor -54. It is -9*6 or 9*(-6). Then check the sum of factors, it shall come to -3, opposite to the coefficient in front of x. -9+6=-3, so roots are x1=-9 and x2=6. The factoring is given by:   (x-x1)(x-x2), in your case,  (x+9)(x-6)

For direct variation y=kx, where k is the number. If y=5 for x=10, then k=y/x=5/10=1/2. So we have:   y=1/2*x; For y=15 x=2*y=2*15=30   Answer:30

I would try to see which interval/intervals the roots may lie into. First, I would see what I get if I plug in x=0, I will get -2; If I plug in -1, I will get zero, which means x=-1 is the root (lucky me!). It turns out that x=1 is also the root. Thus you can now divide your polynomial by (x-1)*(x+1)=x2-1...

Centrifugal force is given by:   F=mω2R, where m is the mass of the hand, R is the radius of the circle the hand makes, ω is the angular frequency in rad/s. Mass of one hand is 1.25% of the body mass, that is 50*0.0125=0.625 kg; radius equals to 0.8 m (half the distance between...

a=dv/dt; Let us differentiate v(x):   dv/dt=dv/dx*dx/dt=dv/dx*v, since dx/dt is velocity itself. I used a chain rule here, as you may notice.   dv/dx is given by:   dv/dx=2x+1; Thus we obtain:   a=dv/dt=(2x+1)*v(x)=(2x+1)*(x2+x)=2x3+3x2+x   Now...

Use work-energy theorem. It states that the change in kinetic energy equals to the work done on the object.   Work done on the object when moved from x=1 to x=3 is:   W=∫13(x3-3x)dx=[x4/4-3/2*x2]|13=81/4-27/2-1/4+3/2=8 J.   Since the body was at rest at...

Since box is not moving, the forces are balanced. There are three forces: gravity, normal force, and horizontal force F.   1) Let x-axis be parallel to the incline. In this case, component of force F along this axis is F*cos(60). Component of the gravity force along the same axis is...

You have two places and any letter can be at any place, thus, you have N=42=16 codes.   AA AB AC AD BB BA BC BD CA CB CC CD DA DB DC DD