Answer:
A) 21.2 kg.m/s at 39.5 degrees from the x-axis
Explanation:
Mass of the smaller piece = 200g = 200/1000 = 0.2 kg
Mass of the bigger piece = 300g = 300/1000 = 0.3 kg
Velocity of the small piece = 82 m/s
Velocity of the bigger piece = 45 m/s
Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s
Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s
since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems
Resultant momentum² = 16.4² + 13.5² = 451.21
Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis ( tan^-1 (13.5 / 16.4)
<span>The property that matter has that energy does not is that matter has size, shape and occupies space. Matter also has inertia. Energy does not have any of these.</span>
Answer:
5.3 cm
Explanation:
This question is an illustration of real and apparent distance.
From the question, we have the following given parameters
Real Distance, R = 8.0cm
Refractive Index, μ = 1.5
Required
Determine the apparent distance (A)
The relationship between R, A and μ is:
μ = R/A
i.e.
Refractive Index = Real Distance ÷ Apparent Distance
Substitute values in the above formula
1.5 = 8/A
Multiply both sides by A
1.5 * A = A * 8/A
1.5A = 8
Divide both side by 1.5
1.5A/1.5 = 8/1.5
A = 8/1.5
A = 5.3cm
Hence, the letters would appear at a distance of 5.3cm
Answer:
Q = 160.36[kJ], is the heat lost.
Explanation:
This is a thermodynamic problem where we can find the latent heat of vaporization, at a constant temperature of 100 [°C].
We know that for steam at 100[C], the enthalpy is
![h_{gas} = 2675.6[kJ/kg]](https://tex.z-dn.net/?f=h_%7Bgas%7D%20%3D%202675.6%5BkJ%2Fkg%5D)
For liquid water at 100[C], the enthalpy is
![h_{water} = 419.17[kJ/kg]](https://tex.z-dn.net/?f=h_%7Bwater%7D%20%3D%20419.17%5BkJ%2Fkg%5D)
Therefore
![h_{g-w} = 2675.6-419.17 = 2256.43[\frac{kJ}{kg} ]](https://tex.z-dn.net/?f=h_%7Bg-w%7D%20%3D%202675.6-419.17%20%3D%202256.43%5B%5Cfrac%7BkJ%7D%7Bkg%7D%20%5D)
The amout of heat is given by:
![Q=h_{g-w}*m\\ where:\\m = mass = 0.07107[kg] = 71.01[g]\\Q = heat [kJ]\\Q =2256.43*0.07107\\Q=160.36[kJ]](https://tex.z-dn.net/?f=Q%3Dh_%7Bg-w%7D%2Am%5C%5C%20where%3A%5C%5Cm%20%3D%20mass%20%3D%200.07107%5Bkg%5D%20%3D%2071.01%5Bg%5D%5C%5CQ%20%3D%20heat%20%5BkJ%5D%5C%5CQ%20%3D2256.43%2A0.07107%5C%5CQ%3D160.36%5BkJ%5D)
Responder:
39200 J
Explicación:
Dado:
Masa de Tamara (m) = 50 kg
Altura a la que se encuentra Tamara (h) = 80 m
Aceleración debido a la gravedad (g) = 9.8 m / s²
La energía potencial de un objeto de masa 'm' ubicada a una altura 'h' sobre el suelo se da como:
Ahora, conecte los valores dados y resuelva la energía potencial. Esto da,
Por lo tanto, la energía potencial de Tamara ubicada a una altura de 80 m es 39200 J.
