Answer:
592000 J
Explanation:
We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:
1 Pa = 1 Kg/ms²
Therefore,
3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²
Next, we shall determine the workdone.
Workdone is given by the following equation:
Workdone (Wd) = pressure (P) × change in volume (ΔV)
Wd = PΔV
With the above formula, the work done can be obtained as follow:
Pressure (P) = 3.7×10⁵ Kg/ms²
Change in volume (ΔV) = 1.6 m³
Workdone (Wd) =?
Wd = PΔV
Wd = 3.7×10⁵ × 1.6
Wd = 592000 Kgm²/s²
Finally, we shall convert 592000 Kgm²/s² to Joule (J). This can be obtained as follow:
1 Kgm²/s² = 1 J
Therefore,
592000 Kgm²/s² = 592000 J
Therefore, the Workdone is 592000 J.
When frequency increases more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases. The opposite is also true—as frequency decreases, wavelength increases.
Answer:
except ii and iii
Explanation:
The angle of reflection is the angle to the normal the white rays strikes the water surface and it is the incidence angle. Since the white light is moving from less dense medium to a denser medium or a medium with a higher refractive index; the angle of refraction will be less than 30 degrees. Total internal reflection cannot occur because the white light is traveling from a less dense medium to a denser medium.
Kinetic of automobile
Mass m = 1,250 Kg; V = 11 m/s
Formula: K.E = 1/2 mV²
K.E = 1/2(1,250 Kg)(11 m/s)²
K.E = 75,625 J
Speed required for insect to have the same kinetic energy as automobile
Mass of insect = 0.72 g convert to Kg m = 7.2 x 10⁻⁴ Kg
K.E = 1/2 mV² Derive V =?
V = 2 K.E/m
V = √2(75,625 J)/7.2 x 10⁻4 Kg
V = √2.1 x 10⁸ m²/s²
V = 14,491.34 m/s (velocity of insect)
Answer:
Work done in all the three cases will be the same.
Explanation:
1) The free falling body has only one force acting on it, the gravitational force. The work done on the body = mgH (Gravitational potential energy)
2) There are two forces acting on the body going down on a frictionless inclined plane - gravity and the normal force. The gravitational potential energy will be the same. The work done due to the normal force is zero, since the direction of the force is perpendicular to the displacement. Hence, total work done on the body = mgH
3) In the case of the body swinging on the end of a string, the change in gravitational potential enrgy will once again be the same since difference in height is H. The additional force on the body is the tension due to the string. But the work done due to this force is <em>zero, </em>since the displacement of the body is perpendicular to the tension. Therefore, the total work done on the body is once again mgH.
