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lakkis [162]
2 years ago
7

The hammer of an electric bell strikes 1440 times in a minute and does a work of 0.2 J per strike. what is the power of hammer i

n S.I.system?
Physics
2 answers:
crimeas [40]2 years ago
7 0
First we need to convert the frequency to the correct SI units (seconds)

Frequency of bell = 1440/60 = 24 Hertz (Strikes per second)

Power is the rate of work being done or the work done per second.

P = E/t = (24 * 0.2)/1 = 4.8 Watt
Kitty [74]2 years ago
4 0
Hello there.

The hammer of an electric bell strikes 1440 times in a minute and does a work of 0.2 J per strike. what is the power of hammer in S.I.system?

4.8 Watt
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mrs_skeptik [129]
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If we pick T_R=T whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that T_L=-T

8 0
2 years ago
Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4
kati45 [8]

Answer:

velocity  = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = \frac{\pi }{4} × (4.50×10^{-2})²  × velocity

solve it we get

velocity  = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = \frac{\pi }{4} × 3 ×  (4.50×10^{-2})²  × velocity

velocity = 52.4 m/s

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Brass is an alloy made from copper and zinc a 0.59 kg brass sample at 98.0 is dropped into 2.80 kg of water at 5.0 c if the equi
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The specific heat capacity of brass would be ranked between 0 and infinity
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2 years ago
Read 2 more answers
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

3 0
2 years ago
Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights
Arlecino [84]

Answer:

               275 kPa

Explanation:

             mass of the gas=m=1.5 kg

             initial volume if the gas=V₁=0.04 m³

             initial pressure of the gas= P₁=550 kPa

as the condition is given final volume is double the initial volume

             V₂=final volume

             V₂=2 V₁

As the temperature is constant

             T₁=T₂=T

\frac{P1V1}{T1}=\frac{P2 V2}{T2}

putting the values in the equation.

\frac{P1V1}{T1}=\frac{P2 *2V1}{T2}

P₂=\frac{P1}{2}

P₂=\frac{550}{2}

P₂=275 kPa

So the final pressure of the gas is 275 kPa.

           

3 0
2 years ago
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