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Georgia [21]
3 years ago
15

The figure shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a frictionless pulle

y
and pulls with a force of magnitude 91.7 N on the wall to which it is attached. The tensions in the shorter cords are T1 = 59.2 N, T2 =
33.4 N, and T3 = 9.65 N. What are the masses of (a) disk A. (b) disk B, (c) disk C, and (d) disk D?
11
79
Physics
1 answer:
garik1379 [7]3 years ago
7 0

Answer:

A) m = F/a = 91.7/9.81 = 9.35 kg

B) m = F/a = 59.2/9.81 = 6.03 kg

C) m = F/a = 33.4/9.81 = 3.40 kg

D) m = F/a = 9.65/9.81 = 0.984 kg

Explanation:

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A steel cable has a cross-sectional area 2.54 10-3 m2 and is kept under a tension of 1.01 104 N. The density of steel is 7860 kg
ElenaW [278]

Answer:

The speed is equals to 22.49 m/s

Explanation:

Given Data:

Area = A=2.54*10^-^3m^2\\Force = F = 1.01*10^4N\\density = p = 7860 kg/m^3

Required:

Speed of Traverse wave = c =?

Solution:

As we know that

p=m/V\\\\ p=m/(L*A)\\p*A=m/L

Now the equation for speed of traverse wave is calculated through:

\sqrt \frac{F*L}{m}\\

=\sqrt\frac{F}{m/L} \\\sqrt{} \frac{F}{p*A}

Substituting the values

\sqrt\frac{1.01*10^4}{7860*2.54*10^-^3}  \\

=22.49 m/s

4 0
3 years ago
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Nata [24]
When the child is moving fastest
6 0
3 years ago
Read 2 more answers
Write the correct abbreviation for each metric unit
Romashka-Z-Leto [24]
Meter - m
Kilometer - km
Hectometer- hm
Dekameter - dam
Decimeter - dm
Centimeter - cm
Millimeter - mm
Micrometer - μm
5 0
3 years ago
A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of
nlexa [21]

Answer:

a_c=1.44\ m/s^2

Explanation:

<u>Centripetal Acceleration</u>

It's the acceleration that an object has when traveling on a circular path to take into consideration the constant change of velocity it must have in order to keep going in the circular path.

Being v the tangent speed, and r the radius of curvature of the circle, then the centripetal acceleration is given by

\displaystyle a_c=\frac{v^2}{r}

We can compute the value of v by using the distance and the time taken to travel:

\displaystyle v=\frac{x}{t}=\frac{200\ m}{26.4\ s}

v=7.58\ m/s

Now we calculate the centripetal acceleration

\displaystyle a_c=\frac{7.58^2}{40}=1.44\ m/s^2

\boxed{a_c=1.44\ m/s^2}

4 0
3 years ago
Take the following measured gauge (or gage) pressures and convert them to absolute pressures in both kPa and psi units for an am
mamaluj [8]

Answer:

Explanation:

Given

ambient Pressure =98.10 kPa

(a)gauge pressure 152 kPa

we know

Absolute pressure=gauge pressure+Vacuum  Pressure

P_{abs}=152+98.10=250.1 kPa or 36.27 psi

(b)P_{gauge}=67.5 Torr or 8.99 kpa

as 1 Torr is 0.133 kPa

P_{abs}=8.99+98.10=107.09 kPa or 15.53 psi

(c)P_{vaccum}=0.1 bar or 10 kPa

Thus absolute pressure=98.10-10=88.10 kPa or 12.77 psi

as 1 kPa is equal to 0.145 psi

(d)P_{vaccum}=0.84 atm  or 85.113 kPa

as 1 atm is equal to 101.325 kPa

P_{abs}=98.10-85.11=12.99 kPa or 1.88 psi

6 0
3 years ago
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