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ollegr [7]
2 years ago
11

Utility poles are to be set every 30 meters. How many poles will be set in one mile if one was set at the beginning of the mile)

?
Physics
2 answers:
maksim [4K]2 years ago
7 0

Answer:

54 poles

Explanation:

1600 / 30 = 53.333 (only 53 poles)

note: there was one set at the beginning, so you would have to add 1  = 54

Zepler [3.9K]2 years ago
4 0

Answer:

53

Explanation:

Because there are 1609.34 meters in a mile. 1609.34÷30=53.64 but because you put one at the beginning of the mile it will stay 53 and not round up to 54

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An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

Charge on electron, q=1.6\times 10^{-19}\ C    

Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}

v = 341250  m/s

or

v=3.41\times 10^5\ m/s

So, the speed of the electron is 3.41\times 10^5\ m/s

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0
2 years ago
In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce
noname [10]

Answer:

14.7 m/s

Explanation:

a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²

v₀ = initial speed of the driver = 0 m/s

v = final speed of the driver after 30 ms

t = time interval for which the acceleration is experienced = 30 ms = 0.030 s

Using the equation

v = v₀ + a t

Inserting the values

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v = 14.7 m/s

6 0
2 years ago
We know that every object exerts an attraction on every other object and the heavier the object ___ the attraction.
hichkok12 [17]
I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question? 
6 0
1 year ago
Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a
Alchen [17]

The height of the roof is <u>3.57m</u>

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

s_2=\frac{1}{2} g(4t)^2=8gt^2\\  s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2

The length of the window s is given by,

s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2

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The height of the roof h is the distance traveled by the first drop and is given by,

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the height of the roof is 3.57 m



8 0
2 years ago
Read 2 more answers
On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s
xenn [34]
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.

First of all, we need to calculate the speed of sound at temperature of T=31^{\circ}C:
v=(331+0.60 T) m/s = (331+0.6 \cdot 31) m/s = 349.6 m/s

The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
v= \frac{2L}{t}
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
L= \frac{vt}{2}= \frac{(349.6 m/s)/4.60 s)}{2}=  804.1 m
6 0
2 years ago
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