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kolbaska11 [484]
3 years ago
15

In which decade was Sputnik I launched? A. 1930-1939 B. 1950-1959 C. 1960-1969 D. 1980-1989

Physics
1 answer:
Olenka [21]3 years ago
6 0
B) 1950-1959
Sputnik I was launched on October 4, 1957
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The speed limit on an interstate highway is posted at 75mi/h. What is the speed in kilometers per hour? In feet per second?
Tatiana [17]

Answer:

See explanation

Explanation:

The conversion factor from miles per hour to kilometers per hour is 1.609344.

Hence;

To convert 75mi/h to kilometers per hour,

75mi/h * 1.609344 = 120.7 kilometers per hour

b)

To convert 75mi/h to feet per second

Since

1 mile per hour = 1.46667 foot per second

75mi/h = 75 * 1.46667 = 110 feet  per second

5 0
3 years ago
Hlp
Nadusha1986 [10]

Answer:

i think its C. cools

Explanation:

7 0
3 years ago
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A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa
andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

\omega=\frac{v}{r}

where

\omega is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

\omega=\frac{v}{r}

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s

(c) 5550 m

The maximum playing time of the CD is

t =74.0 min \cdot 60 s/min = 4,440 s

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) -6.4\cdot 10^{-3} rad/s^2

The angular acceleration of the CD is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 21.6 rad/s is the final angular speed (when the CD is scanned at the outermost part)

\omega_i = 50.0 rad/s is the initial angular speed (when the CD is scanned at the innermost part)

t=4440 s is the time elapsed

Substituting into the equation, we find

\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2

5 0
3 years ago
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PLEASE ANSWERRRRR ASAPPPPPP
Bad White [126]
D Valence
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^^^answer
5 0
3 years ago
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FOR FLVS STUDENTS ONLY///NO BIT LINKS, SCAM LINKS, OR GIBBERSH
Marysya12 [62]

Answer:

whats thw question tho

Explanation:

6 0
3 years ago
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