a mass of 25kg on a ramp at an angle of inclination 30 degrees to the horizontal, is oulled with a force of 75 kg an angle of 20 degrees to the ramp what is the frictional force if it is pulled up the ramp at constant speed

Decompose the forces into components that are parallel or perpendicular to the plane. Then the fricitional force (magnitude f ) is only included in the net force parallel to the surface. The other two are the parallel components of the box's weight (mag. w ) and the pulling force (mag. p)

The box is moving at constant speed, so there is no acceleration and the box is in equilibrium. By Newton's second law,

∑ F = p cos(20º) - w sin(30º) - f = 0

We're given p = 75 N, and

w = m g = (25 kg) (9.80 m/s²)

so

f = (75 N) cos(20º) - (25 kg) (9.80 m/s²) sin(30º)

f ≈ -52 N

That the magnitude is negative makes me think there's a mistake in the question.

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 =

x = 3, y = 216 =

x = 4, y = 1296 =

x = 5, y = 7776 =

so, the geometric series obtained is .

thus, the common ratio is given by = = 6.

hence, the common ratio is 6.