**Given Information: **

Power of bulb = w = 25 W
atts

distance = d = 9.5 cm = 0.095 m

**Required Information: **

Radiation Pressure = ?

**Answer:
**

Radiation Pressure =7.34x10⁻⁷ N/m²

**Explanation:
**

We know that radiation pressure is given by

P = I/c

Where I is the intensity of radiation and is given by

I = w/4πd²

Where w is the power of the bulb in watts and d is the distance from the center of the bulb.

So the radiation pressure becomes

P = w/c4πd²

Where c = 3x10⁸ m/s is the speed of light

P = 25/(3x10⁸*4*π*0.095²)

P = 7.34x10⁻⁷ N/m²

Therefore, the radiation pressure due to a 25 W bulb at a distance of 9.5 cm from the center of the bulb is 7.34x10⁻⁷ N/m²

<span>As a crate slides down an inclined rough surface, its total mechanical energy decreases.

Since the surface of the incline is rough, friction is robbing mechanical energy from the crate all the way down, and blowing it away as heat into the surrounding air.</span>

(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:

And so, the kinetic energy of the object is

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is

And so the new kinetic energy is

So, the work done on the object is the variation of kinetic energy of the object:

<h2>

</h2><h3>kinetic energy is given as</h3>

KE = (0.5) m v²

given that : v = speed of the bottle in each case = 4 m/s when m = 0.125 kg

KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J