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kumpel [21]
3 years ago
11

g At some point the road makes a right turn with a radius of 117 m. If the posted speed limit along this part of the highway is

25.1 m/s, how much should Raquel bank the turn so that a vehicle traveling at the posted speed limit can make the turn without relying on the frictional force between the tires and the road
Physics
1 answer:
user100 [1]3 years ago
6 0

Answer:

Ф = 28.9°

Explanation:

given:

radius (r) = 117m

velocity (v) = 25.1 m/s

required: angle Ф

Ф = inv tan (v² / (r * g))      we know that g = 9.8

Ф = inv tan (25.1² / (117 * 9.8))

Ф = 28.9°

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Karen runs sets in basketball practice. She starts from a line runs 2.0 m, returns to the line, runs 4.0 m, to the line, runs 6.
Aleonysh [2.5K]

D. distance = 23 m, displacement = + 1 m

Explanation:

Let's remind the difference between distance and displacement:

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- displacement is a vector connecting the starting point and the final point of a motion, so its magnitude is given by the length of this vector, and its direction is given by the direction of this vector.

In this case, the distance covered by Karen is given by the sum of all its movements:

distance = 2.0 m + 2.0 m+4.0 m+4.0 m +6.0 + 5.0 m=23.0 m

The displacement instead is given by the difference between the final point (1.0 m in front of the starting line) and the starting point (the starting line, 0 m):

displacement = +1.0 m-0 m=+1 m

8 0
2 years ago
The spectra of most stars are dark-line spectra because ________.
pav-90 [236]
Because dark line spectra result from passing white light through ionized gasses and plasmas, which is what the atmosphere of stars are made of.  These frequencies are scattered by the star's atmosphere as it leaves the surface (photosphere) of the star, and don't make it to earth.
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3 years ago
What is the period of a sound wave having a frequency of 340 Hz
sertanlavr [38]

Answer:

lambda = 343 m/s divided by 340 Hz = 1.009 seconds

Hope it helps and have a wonderful day!

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2 years ago
A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c
den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

We know that, the time period of a simple pendulum of length 'L' is given as:

T=2\pi\sqrt{\frac{L}{g}}-------------- (1)

So, for the new length, the time period is given as:

T_1=2\pi\sqrt{\frac{L_1}{g}}------------ (2)

Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

Now, plug in the given values and calculate 'L'. This gives,

L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

Therefore, the original length of the simple pendulum is 2.97 m

4 0
2 years ago
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