What happens to the resistance of a wire as it gets wider

A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with rig

aleksandr82 [10.1K]

Answer:

The rate of change of the shadow length of a person is 2.692 ft/s

Solution:

As per the question:

Height of a person, H = 20 ft

Height of a person, h = 7 ft

Rate = 5 ft/s

Now,

From Fig.1:

b = person's distance from the lamp post

a = shadow length

Also, from the similarity of the triangles, we can write:

$\frac{a + b}{20} = \frac{a}{7}$

$a = \frac{7}{13}b$

Differentiating the above eqn w.r.t t:

$\frac{da}{dt} = \frac{7}{13}.\frac{db}{dt}$

Now, we know that:

Rate = $\frac{db}{dt} = 5\ ft/s$

Thus

$\frac{da}{dt} = \frac{7}{13}.\times 5 = 2.692\ ft/s$

5 0

3 years ago

A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

7 0

3 years ago

To run a centrifuge correctly, you should:

ser-zykov [4K]

Balance tubes by spacing them equally around the centrifuge and Always balance tubes with other tubes containing a same volume of liquid are right.
If you don't space them out equally, you will have a lot of broken glass to clean up...trust me. The same thing can happen if you don't have equal amounts of liquid in each tube, but it doesn't have to be exactly the same in every one.

7 0

3 years ago

Dadas dos sustancias con distinta densidad, las cuales se mezclan, formarán una nueva sustancia con una densidad.....

zhannawk [14.2K]

Answer:

Explanation:

Equal to the sum of the densities of both substances

Equal to the difference in the densities of both substances

of which I cannot predict anything about its value

of an intermediate value to both substances which will depend on which substance is in greater quantity

8 0

2 years ago

Block 1, of mass m1

musickatia [10]

Answer:

M2=0.52kg

Explanation:

thats rightt

7 0

2 years ago