What is the approximate distance from the center of the milky way galaxy to the sun? select one:

Anyone help me please?

choli [55]

Answer:

I believe that its B my apologies if its wrong/

Explanation:

3 0

3 years ago

So far, you’ve been working with an "ideal" pulley system. How do you think real pulley systems are different, and how would tha

almond37 [142]

Answer:

In an ideal pulley system is assumed as a perfect system, and the efficiency of the pulley system is taken as 100% such that there are no losses of the energy input to the system through the system's component

However, in a real pulley system, there are several means through which energy is lost from the system through friction, which is converted into heat, sound, as well as other forms of energy

Given that the mechanical advantage = Force output/(Force input), and that the input force is known, the energy loss comes from the output force which is then reduced, and therefore, the Actual Mechanical Advantage (AMA) is less than the Ideal Mechanical Advantage of an "ideal" pulley system

The relationship between the actual and ideal mechanical advantage is given by the efficiency of the pulley system as follows;

$Efficiency \, \% = \dfrac{AMA}{IMA} \times 100$

Explanation:

8 0

3 years ago

The lower the angle of the slope, ________ the acceleration along the ramp, therefore, the speed at the bottom of a slope will b

pogonyaev

Answer:

Lower

gsintheta (gsinθ)

Explanation:

The sum of forces resolved parallel to the inclined plane is given by;

F - mgsinθ = 0

ma - mgsinθ = 0

ma = mgsinθ

a = gsinθ

Acceleration is proportional to angle of inclination, thus the lower the angle of the slope, lower the acceleration along the ramp.

therefore, the speed at the bottom of a slope will be lower, (velocity is directly proportional to acceleration) and, consequently, the control will be better.

The acceleration along the ramp, is gsintheta (gsinθ)

3 0

3 years ago

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The

IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s$