A frictionless inclined plane is 8.0 m long and rests on a wall that is 2.0 m high. How much force is needed to push a block of

Kinetic energy refers to the energy an object possesses while in motion, whereas thermal energy corresponds to heat energy. In situations where heat increases in materials, such as a solid turning into a liquid, the molecules start to move more rapidly, resulting in a rise in kinetic energy.

The electric flux through the cylindrical surface surrounding the infinite charged wire is given by the formula ∅E = E x 2πrl. To analyze this, we consider an infinitely long straight wire with a uniform linear charge density of λ Cm⁻¹. The electric field at a distance r from this charge can be evaluated using a cylindrical Gaussian surface of radius r and length l, oriented along the wire. Only the curved surface of the cylinder contributes to the total flux since the other surfaces are perpendicular to E.

Answer:

57.94°

Explanation:

We understand that the formula for flux is

where Ф represents flux

           E indicates electric field

           S denotes surface area

        θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78

                          E=

                          S=

                         

 =  

 =0.5306

 θ=57.94°

Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension present in rope = 550 N

The total torques will be

The worker is positioned at 2.5 m

Answer:

A. Yes, the ball clears the crossbar by 2.83 meters

Explanation:

This situation pertains to projectile motion.

The horizontal velocity component of the ball is calculated as 26 cos 35 = 21.3 m/s

The vertical velocity component is 26 sin 35 = 14.9 m/s

The time taken to travel the horizontal distance to the goalpost, which is 54.9 m, is:

= distance / horizontal speed

= 54.9 / 21.3

= 2.577 seconds.

The vertical distance achieved during this time is:

h = ut - 1/2 gt², where u is the initial vertical velocity, and t = 2.577 seconds.

h = 14.9 x 2.577 - 0.5 x 9.8 x (2.577)²

= 38.39 - 32.54

= 5.85 m

Thus, the ball surpasses the crossbar by 5.85 - 3.05 = 2.8 m