The image produced by a concave mirror of 5.15 cm radius of curvature is erect and has its size doubled compared to the real object. Where is the image (in cm)?
M=h'/h =-q/p = 2
Image is virtual,upright and enlarged but ...
I still haven't figured out how to do this simple problem1 respuestaPhysicshace 1 década
The space shuttle, with an initial mass M = 2.41E+6 kg, is launched from the surface of the earth with an initial net acceleration a = 26.1 m/s2. The rate of fuel consumption is R = 6.90E+3 kg/s. The shuttle reaches outer space with a velocity of vo = 4632 m/s, and a mass of Mo = 1.45E+6 kg.
How much fuel must be burned after this time to reach a velocity vf = 5041 m/s?
I have tried it so many times , I'm kind of confused if I'm getting Ve correctly...
A circle of radius is dropped into the parabola . If too large, the circle will not fall all the way to the bottom, if is sufficiently small, the circle will touch the parabola at its vertex. Find the largest value of r so that the circle will touch the vertex of the parabola
A car with a mass of 1328 kg is coasting in neutral on a straight , level road. It slows down, and its speed as a afunciton of time is given by the equation:
v(t)=26.8 - 0.310t + 0.0210t^2
At time 14.0 s , the speed, as given by the above equation ,is 22.87 m/s.Calculate the power which the engine must deliver(to compensate for air resistance and rolling resistance) in order to maintain that speed.
What I did ...
p=F*v p=ma*v . I did the derivative of V(t) to get a but I got the wrong answer .Also when it says "maintain that speed" that's it mean constant speed, then a would be 0 ?
A block of mass m = 2.10 kg rests on the left edge of a block of larger mass M = 8.15 kg. The coefficient of kinetic friction between the two blocks is 0.330, and the surface on which the 8.15 kg block rests is frictionless. A constant horizontal force of magnitude F = 17.80 N is applied to the 2.10 kg block, setting it in motion
a). If the distance L that the leading edge of the smaller block travels on the larger block is 3.20 m, how long will it take before this block reaches the right side of the 8.15 kg block
b).how far does the 8.15 kg block move in the process?
a.)so I tried Fnet=17.80-umg a=Fnet/m and then t=sqrt(2L/a)
but it's wrong.1 respuestaPhysicshace 1 década
o Volcanoes on the Earth eject rocks at speeds of up to 250 m/s. Consider a 1250 m high volcano which ejects rocks in all directions. What is the maximum height above sea level reached by the rocks?
- 1 respuestaOtros - Salud y Bellezahace 1 década