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SVETLANKA909090 [29]
3 years ago
7

Spring has an unstretched length of 0.40 meters. The spring is stretched to a length of 0.60 meters when a 10.-newton weight is

hung motionless from one end. The spring constant of this spring is:

Physics
1 answer:
Nonamiya [84]3 years ago
7 0

Answer: The spring constant is 50N/m

Explanation: Please see the attachment below

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A large cylindrical tank contains 0.750 cubic meters of nitrogen gas at 27 degrees celsius and 1.5 e5 pa absolute pressure. the
k0ka [10]
<span>3.36x10^5 Pascals The ideal gas law is PV=nRT where P = Pressure V = Volume n = number of moles of gas particles R = Ideal gas constant T = Absolute temperature Since n and R will remain constant, let's divide both sides of the equation by T, getting PV=nRT PV/T=nR Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation P1V1/T1 = P2V2/T2 where P1, V1, T1 = Initial pressure, volume, temperature P2, V2, T2 = Final pressure, volume, temperature Now convert the temperatures to absolute temperature by adding 273.15 to both of them. T1 = 27 + 273.15 = 300.15 T2 = 157 + 273.15 = 430.15 Substitute the known values into the equation 1.5E5*0.75/300.15 = P2*0.48/430.15 And solve for P2 1.5E5*0.75/300.15 = P2*0.48/430.15 430.15 * 1.5E5*0.75/300.15 = P2*0.48 64522500*0.75/300.15 = P2*0.48 48391875/300.15 = P2*0.48 161225.6372 = P2*0.48 161225.6372/0.48 = P2 335886.7441 = P2 Rounding to 3 significant figures gives 3.36x10^5 Pascals. (technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>
8 0
3 years ago
A block of 7.80 kg kept on an inclined plane just begins to slide at an angle of inclination of 35.0°. Once it has been set into
Bas_tet [7]

Answer:

The answer is....i am sorry idk

8 0
3 years ago
g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
Elza [17]

Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

                 = 0.031 \ \ rem

4 0
3 years ago
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