1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
tatiyna
3 years ago
9

How high is the average telephone pole

Physics
2 answers:
kenny6666 [7]3 years ago
7 0

The average telephone pole stands about 40ft in the air.

Ksivusya [100]3 years ago
7 0

Well the first dude isn't wrong it is 40ft....But it's taller than me

You might be interested in
What word is modified by an adjectival phrase in the following sentence? The savings to the project manager at NASA were substan
DedPeter [7]
I am pretty sure that the<span> word which is modified by an adjectival phrase in the following sentence is </span>being revealed by the second options from the box represented above : b. project.  I choose this one as a correct answer because it is clear that this word is <span>modified by the phrase :"to the project manager". Hope it will help!</span>
8 0
4 years ago
is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke’s
yulyashka [42]

Answer:

a) 156960 N/m

b) Yes the player can play on the team.

Explanation:

F = Force

a = Acceleration due to gravity = 9.81 m/s²

m = Mass

k = Spring constant

x = Deformation length of spring

F = ma

From Hooke's law

F=kx\\\Rightarrow ma=kx\\\Rightarrow k=\frac{ma}{x}\\\Rightarrow k=\frac{120\times 9.81}{0.0075}\\\Rightarrow k=156960\ N/m

The effective spring constant is 156960 N/m

When x = 0.48 cm = 0.0048 m

m=\frac{kx}{a}\\\Rightarrow m=\frac{156960\times 0.0048}{9.81}\\\Rightarrow m=76.8\ kg

The mass of the player is 76.8 kg which is less than the required mass limit of the players' which is 85 kg. So, the player is eligible to play.

5 0
3 years ago
A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l
artcher [175]

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, \lambda_1=8\times 10^{-6}\ C

linear charge density of the other charge distribution, \lambda_2=-6\times 10^{-6}\ C

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

E=\frac{\lambda}{2\pi.r.\epsilon_0}

where:

\lambda= linear charge density

r = radial distance from the center of wire

\epsilon_0= permittivity of free space

Therefore,

E_1=E_2

\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}

\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}

\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}

x=0.0629\ m

∴at y=6.29 cm the charge of the two distribution will be equal.

9 0
3 years ago
Please Help <br><br> Conservation of energy
Sonja [21]

Answer:

a principle stating that energy cannot be created or destroyed, but can be altered from one form to another.

Explanation:

8 0
3 years ago
A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s2. Pa
ValentinkaMS [17]

Answer:

45 s .

Explanation:

The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .

displacement s = ?

acceleration a = 1 m /s²

Final speed v = 5 m/s

initial speed u = 0

v² = u² + 2as

5² = 0 + 2 x 1 x s

s = 12.5  m

B)  Let time of acceleration or deceleration be t

v = u + a t

5 = 0 + 1 t

t = 5 s

Similarly displacement during deceleration = 12.5 m

Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) =  175 m .

velocity of uniform motion = 5 m /s

time during which there was uniform velocity = 175 / 5 = 35 s

Total time = 5 + 35 + 5 = 45 s .

4 0
3 years ago
Other questions:
  • An 8kg soccer ball starts at rest and is kicked 31.5 meters until it reaches a final velocity of 21m/s in 3 seconds. Assuming th
    9·1 answer
  • Scientist accept ideas of the past and do not try to revise or change them
    13·1 answer
  • The voltage entering a transformer’s primary winding is 120 volts. The primary winding is wrapped around the iron core 10 times.
    13·2 answers
  • Compared to its weight on Earth, a 10-kg object on the moon will weigh Question 9 options: less. the same amount. more.
    6·2 answers
  • A heavy ball with a weight of 100 NN is hung from the ceiling of a lecture hall on a 4.4-mm-long rope. The ball is pulled to one
    15·1 answer
  • If a bar magnet is suspended at the center on a string and allowed to swing freely, in what direction will its south pole point
    6·1 answer
  • A circular loop of wire of area 25 cm2 lies in the plane of the paper. An increasing magnetic field B is coming out of the paper
    13·1 answer
  • Assertion: When I P+ Q I = I P- QI, then P must be perpendicular to Q. Reason: The relation will hold even when Q is a null vect
    7·1 answer
  • 10points asap <br><br> A force of 30 N acts upon a 7 kg block. Calculate its acceleration.
    11·1 answer
  • A car starts from rest and accelerates uniformly over a time of 7.25 seconds for a distance of 210 m. Determine the acceleration
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!