I notice that even though we're working with frames of reference
here, you never said which frame the '5 km/hr' is measured in.
In fact ! You didn't even say which frame the '12 km/hr' of his
bike is measured in.
So there are several different ways this could go. I'll do it the way
I THINK you meant it, but that doesn't guarantee anything.
-- Simon is riding his bike at 12 km/hr relative to the sidewalk,
away from Keesha.
-- He throws a ball at Keesha, at 5 km/hr relative to his own face.
-- Keesha sees the ball approaching her at (12 - 5) = 7 km/hr
relative to the ground and to her.
Answer:
Net forces which pushes the window is 30342.78 N.
Explanation:
Given:
Dimension of the office window.
Length of the window = 
m
Width of the window = 
m
Area of the window = 
Difference in air pressure = Inside pressure - Outside pressure
= 
atm = 
atm
Conversion of the pressure in its SI unit.
⇒ 
atm = 
Pa
⇒ atm = 
Pa
We have to find the net force.
We know,
⇒ Pressure = Force/Area
⇒ 
⇒ 
⇒ Plugging the values.
⇒ 
⇒ 
Newton (N)
So,
The net forces which pushes the window is 30342.78 N.
Answer:The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.
Explanation:
From the picture, I see that you had no trouble at all with #4.
Well, #5, 6, and 7 are easily handled in exactly the same way.
Just as you did with #4, please sketch these on paper
as I walk you through the solutions. That'll help you see
immediately what's going on.
#5.b).
Traveling east at 3 m/s for 4 seconds,
he covers (3 m/s) x (4 sec) = 12 meters.
Traveling south at 5 m/s for 2 seconds,
he covers (5 m/s) x (2 sec) = 10 meters.
The total distance he covers is (12m + 10m) = 22 meters.
#5.c).
Average speed (scalar)
= (distance covered)/(time to cover the distance)
= (22 meters)/(6 sec) = 3-2/3 m/s .
#5.d).
Displacement (vector)
= distance between the start-point and the end-point,
regardless of the route traveled,
in the direction from the start-point to the end-point.
Distance from the start-point to the end-point =
√(12² + 10²) = √(144 + 100) = √(244) = 15.62 meters
in the direction of arctan(10/12) south of east
= 39.8° south of east.
#5.e).
Average velocity (vector) =
(displacement vector) / (time)
= 15.62 meters directed 39.8° south of east / 6 seconds
= 2.603 m/s directed 39.8° south of east.
#6).
Magnitude = √(5.2² + 2.1²) = √(27.04 + 4.41) = √31.45 = 5.608 km.
Direction = arctan(5.2/2.1) south of east
= 68° south of east = 158° bearing .
#7).
Magnitude = √(39² + 57²) = √(1521 + 3249) = √( 4770)
= 69.07 m/s .
Direction = arctan (57/39) south of west
= 55.6° south of west
Bearing = 214.4°
Compass: 0.65° past "southwest by south".
I'm grateful for the privilege and opportunity to practice my math,
and I shall cherish the bounty of 5 points that came with it.
