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3 years ago

6

You are out on a hike, and you notice that the line of telephone wires you have been following along a slope is crooked. Two of

the posts are much farther down slope then the others. What has likely occurred on this hillside?
Explain.

2 answers:

ivolga24 [154]3 years ago

3 0

Answer: a creep

Explanation:

horsena [70]3 years ago

3 0

Answer:

This is an example of the creep

Explanation:

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A car experiences a centripetal acceleration of 4.4 m/s ^2 as ur rounds a corner with a speed of 15 m/s. What is the radius of t

damaskus [11]

The calculation of the centripetal acceleration of an object following a circular path is based on the equation,

a = v² / r

where a is the acceleration, v is the velocity, and r is the radius.

Substituting the known values from the given above,

4.4 m/s² = (15 m/s)² / r

The value of r from the equation is 51.14 m.

Answer: 51.14 m

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3 years ago

Assuming no air resistance, how far will a 0.0010 kg raindrop have fallen when it hits the ground 30.0 s later. Assume g = 9.8 m

Serga [27]

D = (1/2)·at²
where d is the distance fallen, a is the acceleration (g in this problem), and t is the time
d = (1/2)·(9.8 m/s²)·(30 s)² = (1/2)·(9.8)·(900) m
d = 4410 m
The answer is b) 4410 m

Note: the mass of the raindrop is irrelevant since the acceleration due to gravity is independent of mass. (Galileo's Leaning Tower of Pisa experiment)

6 0

3 years ago

Read 2 more answers

The cylinder valve is open and the gas is collected and atmospheric pressure

jok3333 [9.3K]

Less gas will be collected because some of the gases will escape from the open cylinder valve.

Cylinders used to store carbon dioxide will have thicker walls than those of butane because of higher pressures.

<h3>What are compressed gases?</h3>

Compressed gases are gases which are compressed under high pressure in gas cylinders.

Cylinder valves are used to reduce the pressure of the compressed gases and in the process, some of the gas molecules escape.

Since the cylinder valve is open and the gas is collected at atmospheric pressure, less gas will be collected because some of the gases will escape.

Since, the carbon dioxide not liquefy under pressure compared to butane, the cylinders used to store carbon dioxide will have thicker walls than those of butane.

Learn more about compressed gases at: brainly.com/question/518065

4 0

2 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago

A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1

pav-90 [236]

I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
$F_a=F_cf-F_g$
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
$F_a=m\frac{v^2}{r}-mg=179 $N$
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
$F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N$
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
$F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}$

6 0

3 years ago