Answer:
Work done, W = 1786.17J
Explanation:
The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "
Mass of a painter, m = 75 kg
He climbs 2.75-m ladder that is leaning against a vertical wall.
The ladder makes an angle of 30 degrees with the wall.
We need to find the work done by the gravity on the painter.
The angle between the weight of the painter and the displacement is :
θ = 180 - 30
= 150°
The work done by the gravity is given by :
Hence, the required work done is 1786.17 J.
Answer:
V = 10 km / 1 hr = 10 km/hr
V = -10 j km / hr if one were to use i, j, k as unit vectors with the usual orientation
Answer:
<h3>0.445</h3>
Explanation:
In friction, the coefficient of friction formula is expressed as;
Ff is the frictional force = Wsinθ
R is the reaction = Wcosθ
Substitute inti the equation;
Given
θ = 24°
Hence the coefficient of kinetic friction between the box and the ramp is 0.445
I think its Mercury because it's the closest to the sun.
Explanation:
The given data is as follows.
Velocity of bullet, 
= 814.8 m/s
Observer distance from marksman, d = 24.7 m
Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.
t = 
(velocity in air = 343 m/s)
= 0.072 sec
Now, before the observer hears the report the distance traveled by the bullet is as follows.
= 
= 58.66
= 59 (approx)
Thus, we can conclude that each bullet will travel a distance of 59 m.
