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MrRissso [65]
2 years ago
6

John is pushing (horizontally) on a 100 kg (Fw = 980N) bench with an unknown force. If the coefficient of static friction is 0.2

0 and the acceleration is 1.84m/s/s, what is the applied force?
Physics
1 answer:
atroni [7]2 years ago
4 0

Answer:

1176N

Explanation:

Given parameters:

Mass of bench  = 100kg

Forward force  = 980N

Coefficient of static friction = 0.2

Acceleration  = 1.84m/s²

Unknown:

Magnitude of the applied force  =?

Solution:

Frictional force is a force that opposes motion, for a body to move, the applied force must be greater than the frictional force;

The Force applied;

  Force applied  = Frictional force + Weight of the body

   Frictional force  = umg

     u is the coefficient of static friction

      m is the mass

        g is the acceleration due to gravity

    Force applied  = 0.2 x 100 x 9.8 + 980 = 1176N

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A sample of a diatomic ideal gas has pressure P and volume V. When the gas is warmed, its pressure triples and its volume double
Andreyy89

Answer:

Amount of  Energy transferred =8.5PV

Explanation:

Given:

Initial volume=V

Initial pressure=P

Final volume=2V

Final pressure=3P

Now w know that the Energy transferred in constant pressure pressure is given by

E_1=nc_pdT\\\\E_1=n\times\dfrac{7R}{2}dT\\E_1=3.5(nRdT)\\E_1=3.5(V(2P-P))\\E_1=3.5PV

Now the Energy transferred in constant volume process is given by

E_2=nc_vdT\\\\E_2=n\times\dfrac{5R}{2}dT\\E_1=2.5(nRdT)\\E_1=2.5(V(3V-V))\\E_1=5PV

The total Energy transferred is given by

E_{total}=E_1+E_2\\E{total}=3.5PV+5PV\\=8.5PV

3 0
3 years ago
Read 2 more answers
a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second
lawyer [7]

Answer:

a) Yes, there are 10^8 shakes in a second (1 s \frac{1 shake}{10^{-8}s}=10^8 shake) in a year there are 31,536,000 seconds... that is 3.1536x10^7 (1year\frac{365 day}{1 year} \frac{24 h}{1 day} \frac{60 min}{1 h}\frac{60s}{1min}=3.1536 *10^7)

b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s

That is about 2 and half hours.

3 0
3 years ago
Im confused on number one
mr Goodwill [35]

Explanation:

v=?, u=0, a=?, S=22m.

Using the formula, S=ut+½at²

22={0×5}+(½.a.5²)

22=½.a.5²

a=44/25 = 1.76m/s².

Therefore, net force = work done = ma = 48×1.76 = 84.48N.

therefore, power = work done/time = 84.48/5 = 16.896W.

hope this helps you.

5 0
2 years ago
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