Explanation:
a) 7.5= 111.1×2°= 0.1111×2^3
which can also be written as
(1/2+1/4+1/8+1/16)×8
sign of mantissa:=0
Mantissa(9 bits): 111100000
sign of exponent: 0
Exponent(5 bits): 0011
the final for this is:011110000000011
b) -20.25= -10100.01×2^0= -0.1010001×2^5
sign of mantissa: 1
Mantissa(9 bits): 101000100
sign of exponent: 0
Exponent(5 bits): 00101
the final for this is:1101000100000101
c)-1/64= -.000001×2^0= -0.1×2^{-5}
sign of mantissa: 1
Mantissa(9 bits): 100000000
sign of exponent: 0
Exponent(5 bits): 00101
the final for this is:1100000000100101
Answer: 16.3 seconds
Explanation: Given that the
Initial velocity U = 80 ft/s
Let's first calculate the maximum height reached by using third equation of motion.
V^2 = U^2 - 2gH
Where V = final velocity and H = maximum height.
Since the toy is moving against the gravity, g will be negative.
At maximum height, V = 0
0 = 80^2 - 2 × 9.81 × H
6400 = 19.62H
H = 6400/19.62
H = 326.2
Let's us second equation of motion to find time.
H = Ut - 1/2gt^2
Let assume that the ball is dropped from the maximum height. Then,
U = 0. The equation will be reduced to
H = 1/2gt^2
326.2 = 1/2 × 9.81 × t^2
326.2 = 4.905t^2
t^2 = 326.2/4.905
t = sqrt( 66.5 )
t = 8.15 seconds
The time it will take for the rocket to return to ground level will be 2t.
That is, 2 × 8.15 = 16.3 seconds
Answer:
W = 1222.4 J = 1.22 KJ
Explanation:
The work done on an object is the product of the force applied on it and the displacement it covers as a result of this force. It must be noted that the component of displacement in the direction of force should only be used. Hence, the work can be calculated as:
W = F d Cosθ
where,
W = Work Done = ?
F = Force Applied = 64 N
d = Distance Covered by Box = 19.1 m
θ = Angle between force and displacement = 0°
Therefore,
W = (64 N)(19.1 m)Cos 0°
<u>W = 1222.4 J = 1.22 KJ</u>
