Answer:
14 m/s
Explanation:
The following data were obtained from the question:
Mass = 50 kg
Initial velocity (u) = 0 m/s
Height (h) = 10 m
Acceleration due to gravity (g) = 9.8 m/s²
Final velocity (v) =?
The velocity (v) with which the person hit the water can be obtained as shown below:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 10)
v² = 0 + 196
v² = 196
Take the square root of both side
v = √196
v = 14 m/s
Therefore, he will hit the water with a speed of 14 m/s
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
We Know, F = m*a
Here, m = 0.25 Kg
a = 196 m/s²
Substitute it into the expression,
F = 0.25*196
F = 49 N
So, option B is your final answer.
Hope this helps!
An educated guess pretty much
Answer: SG = 2.67
Specific gravity of the sand is 2.67
Explanation:
Specific gravity = density of material/density of water
Given;
Mass of sand m = 100g
Volume of sand = volume of water displaced
Vs = 537.5cm^3 - 500 cm^3
Vs = 37.5cm^3
Density of sand = m/Vs = 100g/37.5 cm^3
Ds = 2.67g/cm^3
Density of water Dw = 1.00 g/cm^3
Therefore, the specific gravity of sand is
SG = Ds/Dw
SG = (2.67g/cm^3)/(1.00g/cm^3)
SG = 2.67
Specific gravity of the sand is 2.67
