Answer: 2.55meter
Explanation: Using the second equation of motion.
S{hieght} = U*t + {g*t²}/2
Where U is initial velocity =0m/s
g is acceleration due to gravity 10m/s²
t is time 1secs
So we have,
hieght = 0 + {g*t²}/2
hieght = {10*(1)²}/2
Total hieght travelled is 10/2
Which is 5 meter.
But we are asked to find the hieght above the window which as a hieght of 2.45meter.
So,
hieght above window would be
{5 - 2.45}meter
Which is 2.55 meter.
Answer:
Explanation:
Kinetic energy of block will be converted into heat energy by friction .
Heat energy produced = 1/2 m v²
= .5 x 4.8 x 1.2²
= 3.456 J
85% of energy is converted into heat energy , so heat energy produced
= .85 x 3.456 = 2.9376 J .
If Q heat is given to m mass of object having s as specific heat and Δt is increase in temperature
Q = msΔt
specific heat of iron s = 462 J / kg C
Putting the values ,
2.9376 = 4.8 x 462 x Δt
Δt = 13.24 x 10⁻⁴ ⁰C.
Answer:
The velocity at impact is 222.2 m/s
Explanation:
The given information are;
The time at which the woman heard the sound of the coin after dropping it into the wishing well = 30 seconds
The average speed of sound in air = 344 m/s
The time in which the dime travel = t₁
The time in which the sound travel = t₂
1/2 × 9.8 × t₁² = 344 × t₂
t₁ + t₂ = 30
4.9·t₁² = 344 × (30 - t₁)
4.9·t₁² + 344·t₁ - 10320 = 0
Which gives -344±
t₁ = 22.676 seconds or -92.9 seconds
Therefore the correct natural time is t₁ = 22.676 seconds
The velocity at impact, v = g×t₁ = 9.8 × 22.676 = 222.2 m/s
The velocity at impact= 222.2 m/s.
The perpendicular distance from the equilibrium line to the peak or trough
Answer:
TATTCATTCATTA—TGATTT—ATTCG
Explanation:
A mutation is a permanent change in the nucleotide sequence of DNA. A mutation occurs during replication or recombination. It may be due to base substitutions, deletions and insertions. As per the question, DNA segment forms encodes for the enzyme pepsin is CATTGTTA.
Option TATTCATTCATTA—TGATTT—ATTCG is the correct answer. In the DNA segment which encodes pepsin, a purine base (G) guanine is replaced by another purine (A) adenine. This type of mutation is called a transition type point mutation.
Due to base substitution, the mutated segment CATTCATTA will nor encode pepsin.