If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula
Where | A | is the magnitude of the vector and 
is the angle that it forms with the x axis in the opposite direction to the hands of the clock.
In this problem we know the value of Ax and Ay and we need the angle .
Vector A is in the 4th quadrant
So:
So:
So:
= -47.28 ° +360° = 313 °
= 313 °
Option 4.
Answer:a) P = Po + rho×h×g
b) P = 5.4 × 10^9 pa
c) F = P/A = (Po + rho×h×g)/A
d) 1.174×10^11N
Explanation: Using the formula
P = Po + rho×h×g
P = 1.0 x 10^5 + 1000 × 5.5 × 9.81
P = 5.4 × 10^9pa
The magnitude of the force exerted by water on the top of the person's head F at the depth h in terms of P
F = P/A = (Po + rho×h×g)/A
Using the above formula
Where A = 0.046m^2
F = P/ A = 5.4×10^9/0.046
F = 1.174×10^11N
Answer:
9.62 minutes 0r 0.16 of an hour
Explanation:
Speed = distance/time
300mph = 48.1 m/t
xt
300t = 48.1
÷300
t = 48.1/300
t = 0.16033333333 hr
0.16033333333 x 60 = 9.62 minutes
60 minutes in an hour
9.62/60= 0.16033333333 hr
So, around 10 minutes.
Hope this helps!
Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.
