Nikolas, the fire extinguisher, and the skateboard have a combined mass of 50 kg. What force would the fire extinguisher have to
1 answer:
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Answer:
(a) 1.414 km
(b) 1.06 m/s
Explanation:
(a) For John:
Distance = 1 km north and then 1 km east
Speed = 1.5 m/s
total distance traveled = 1 + 1 = 2 km = 2000 m
Time taken to travel = Distance / speed
t = 2000 / 1.5 = 1333.3 seconds
Displacement =
(b) For jane :
Time is same as john = 1333.33 second
Distance = 1.414 km = 1414 m
Speed = distance / time = 1414 / 1333.33 = 1.06 m/s
Answer:
a) vd = 47.88 m/s
b) θ = 80.9°
c) t = 6.8 s
Explanation:
In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.
One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.
(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.
One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m
The other side of the triangle is c2 = 230m - 3m = 227 m
Then, the hypotenuse is:
(1)
Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:
(2)
yo: initial height = 230m
vo: initial vertical speed of the mouse = 0m/s
g: gravitational acceleration = -9.8m/s^2
y: final height of the mouse = 3 m
You replace the values of the parameters in (2) and solve for t:
The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.
You use the value of h and 4.8s to find the diving speed of the hawk:
The diving speed of the Hawk is 47.88m/s
(b) The angle is given by:
Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°
(c) The mouse is falling down during 6.8 s
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x m/s)
f= 3 x / 2.4
f=1.25 x hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x => 800 x
s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x
x 3 x
)/2
=120m