Answer:
The angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is <u>10°.</u>
Explanation:
Given:
Mass of the driver is, 
Radius of circular turn is, 
Linear speed of the car is, 
Since, the car makes a circular turn, the driver experiences a centripetal force radially inward towards the center of the circular turn. Also, the driver experiences a downward force due to her weight. Therefore, two forces act on the driver which are at right angles to each other.
The forces are:
1. Weight = 
2. Centripetal force, 'F', which is given as:
Now, the angle of the net force acting on the driver with respect to the vertical is given by the tan ratio of the centripetal force (Horizontal force) and the weight (Vertical force) and is shown in the triangle below. Thus,
°
Therefore, the angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.
Answer:Although any object in motion through space (for example a thrown baseball, kicked football, fired bullet, thrown arrow, stone released from catapult) are projectiles, they are commonly found in warfare and sports.
Explanation:
An air mass is a large pocket of air with a uniform temperature and humidity
Answer:
a) 
s
b) 3.41 mm
Explanation:
a)
We take the speed of light, c = 
m/s and the refractive index of glass as 1.517.
Speed = distance/time
Time = distance/speed
Refractive index, n = speed of light in vacuum / speed of light in medium
b)
We take the refractive index of water as 1.333.
Speed in water = speed in vacuum / refractive index of water
Distance = speed * time
d = 3.41 mm
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
