Answer:
1.6 m
Explanation:
Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.
The time for landing should be calculated by using the second equation of motion formula
h = Ut + 1/2gt^2
Let U = 0
0.5 = 1/2 × 9.8 × t^2
0.5 = 4.9t^2
t^2 = 0.5 / 4.9
t^2 = 0.102
t = 0.32 s
The target should be placed so that the toy car lands on it at:
Distance = 5 × 0.32
distance = 1.597 m
Distance = 1.6 m
Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.
Answer:
21.67 rad/s²
208.36538 N
Explanation:
= Final angular velocity =
= Initial angular velocity = 78 rad/s
= Angular acceleration
= Angle of rotation
t = Time taken
r = Radius = 0.13
I = Moment of inertia = 1.25 kgm²
From equation of rotational motion
The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²
Torque is given by
Frictional force is given by
The magnitude of the force of friction applied by the brake shoe is 208.36538 N
Answer: 43.58 min
Explanation:
Knowing the volume of a rectangular object is length x width x height, we have two volumes:
And we know it takes a time of 4 minutes to fill .
If we want to know how long will it take the same hose to fill another tank with volume , we can use the <u>Rule of three</u>, which is a mathematical rule to find out an amount that is with another quantity given in the same relation as other two also known:
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Finally:
This is the concept of trigonometry, the direction of the John will be given by:
tan theta=opposite/adjacent
suppose:
a=theta
opposite=4 mph
adjacent=5.5 mph
tan a=4/5.5
a=tan^-1(4/5.5)
a=36
therefore the boat canoe will be moving 36 degrees from East, his direction will be 126 degrees south east