Answer:
1.927 m/s^2
Explanation:
period = 2 pi sqrt ( l/g)
3.2 = 2 pi sqrt (.5/g) =1.927 m/s^2
OA the speed will increase four times
Answer:
Temperature of the hot reservoir is 1540K
Explanation:
![E= 1- \frac{T_{c}}[tex]{T_h}=308+{T_c}\\Efficiency of a carnot engine is given by the aboveTc=temperature of the cold reservoirTh= temperature of the hot reservoirK=273+ 35 (convert 35°C to kelvin)K=308k{T_h}={T_c}+308-----------------------(equation 1)20%=1-{T_c}/{T_h}](https://tex.z-dn.net/?f=E%3D%201-%20%5Cfrac%7BT_%7Bc%7D%7D%5Btex%5D%7BT_h%7D%3D308%2B%7BT_c%7D%5C%5C%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EEfficiency%20of%20a%20carnot%20engine%20is%20given%20by%20the%20above%3C%2Fp%3E%3Cp%3ETc%3Dtemperature%20of%20the%20cold%20reservoir%3C%2Fp%3E%3Cp%3ETh%3D%20temperature%20of%20the%20hot%20reservoir%3C%2Fp%3E%3Cp%3EK%3D273%2B%2035%20%20%28convert%20%2035%C2%B0C%20to%20kelvin%29%3C%2Fp%3E%3Cp%3EK%3D308k%3C%2Fp%3E%3Cp%3E%3Cstrong%3E%7BT_h%7D%3D%7BT_c%7D%2B308-----------------------%28equation%20%201%29%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3E20%25%3D1-%7BT_c%7D%2F%7BT_h%7D)
0.2=({T_c}+308-{T_c})/{T_c}+308
.2({T_c}+61.6=308
0.2{T_c}=246.4
{T_c}=1232
recall from equation 1
{T_h}=308+1232
{T_h}=1540K
The heat flows into the sample in the segment <u> BC </u> will yield the value of the heat of fusion of this substance.
<h3>Heat of fusion:</h3>
The heat of the fusion of water is unusually high. The amount of heat required to convert 1 g of solid into a liquid while maintaining the same temperature is known as the heat of fusion. It is a latent heat as well and is occasionally referred to as the latent heat of fusion. Since water only freezes at a single temperature (0 °C), it only has one value, which is 79.71 cal/g, or the rounded amount of 80 cal/g.
In order to control frost, water fuses at a high temperature. The water used for irrigation is frequently consistently above freezing when it is extracted from the ground.
Learn more about heat of fusion here:
brainly.com/question/13726231
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<span>Force F = 280 N
Angle with the ground = 40 degrees
Weight of the Lawnmower = 350 N
Velocity is constant so Acceleration is 0
So Forward force Ff = F cos theta = 280 cos40
Frictional force with resists to back Fb = (u x Force from pressure) + vertical component of Force, where u is the coefficient of friction.
Fb = (u x m x g) + (u x 280sin40)
AS Ff = Fb => 280 cos40 = u x ((m x g) + 280sin40)
u = 280 cos40 / ((350 x 9.81) + 280sin40) = 214.49 / () = 0.405
So the coefficient of friction u = 0.405</span>