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3 years ago

8

The speed limmit on an interstate highway is posted at 75mi/h. What is the speed in kilometers per hour? In feet per second?

1 answer:

jok3333 [9.3K]3 years ago

4 0

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

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The resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is:a) 180 d

r-ruslan [8.4K]

Answer:

0 degrees

Explanation:

Let $F_1\ and\ F_2$ are two forces. The resultant of two forces acting on the same point is given by :

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos\theta}$

Where $\theta$ is the angle between two forces

When $\theta=0$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos(0)}$

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2}$

When $\theta=90^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(90)}$

$F_R=\sqrt{F_1^2+F_2^2}$

When $\theta=180^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(180)}$

$F_R=\sqrt{F_1^2+F_2^2-2F_1F_2}$

It is clear that the resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is 0 degrees. Hence, this is the required solution.

8 0

3 years ago

Which one will it be

liq [111]

Answer:

none

Explanation:

it's to high up to be affected by the gravity

3 0

3 years ago

Read 2 more answers

What is the displacement of the GBS cross-country team if they begin at the school, run 10 miles, and finish back at the school?

olga2289 [7]

Answer:

0

Explanation:

8 0

3 years ago

) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

$\omega = 2\pi f$

And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s$

(d) $2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s$

In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

7 0

3 years ago

Explain why the<br> ball looks red<br> under normal<br> light.

allochka39001 [22]

That’s the color of the ball?

6 0

3 years ago