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pashok25 [27]
3 years ago
7

NEED HELP ASAP!!!!! IN MIDDLE OF TEST RN HELP PLZ

Physics
2 answers:
svlad2 [7]3 years ago
4 0
ONEEEEEEEEEEEEEEEEEEEEEEEEE
igomit [66]3 years ago
4 0
The answer should be number one
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The front and rear sprockets on a bicycle have radii of 8.40 and 4.91 cm, respectively. The angular speed of the front sprocket
Rudik [331]

Explanation:

It is given that,

Radius of the front sprockets, r_f=8.4\ cm=0.084\ m

Radius of the rear sprockets, r_r=4.91\ cm=0.0491\ m

The angular speed of the front sprocket is 12.3 rad/s, \omega_f=12.3\ rad/s

(a) Linear speed of the front sprockets, v_f=r_f\times \omega

v_f=0.084\times 12.3    

v_f=1.0332\ m/s

v_f=103.32\ cm/s

Linear speed of the rear sprockets, v_r=r_r\times \omega

v_r=0.0491\times 12.3    

v_r=0.60393\ m/s

v_r=60.393\ cm/s

(b) Let a_r is the centripetal acceleration of the chain as it passes around the rear sprocket.

a_r=\dfrac{v_r^2}{r_r}

a_r=\dfrac{(60.393)^2}{0.0491}

a_r=74283.39\ m/s^2

a_r=7428339\ cm/s^2

`Hence, this is the required solution.

8 0
3 years ago
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What are the first three harmonics of a note produced on a 0.31 m long violin string if the waves on this string have a speed of
bezimeni [28]
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4 0
2 years ago
Which of these determines an objects ability to sink or float in water?
Zigmanuir [339]

Answer:

C. Density

Explanation:

8 0
3 years ago
Peering through a telescope could be considered Step 1 of the scientific method.<br> True<br> False
aleksklad [387]

Answer:

<em><u>True:</u></em> because when u look thru a telescope you are making an observation

Explanation:

4 0
2 years ago
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Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of
max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0
3 years ago
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