The initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
CALCULATE INITIAL VELOCITY:
The initial velocity of the car can be calculated by using one of the equation of motion as follows:
V = u + at
Where;
- V = final velocity (m/s)
- u = initial velocity (m/s)
- a = acceleration due to gravity (m/s²)
- t = time (s)
According to this question, a car accelerates at a constant rate of 3 m/s² for 5 seconds. If it reaches a velocity of 27 m/s, its initial velocity is calculated as follows:
u = v - at
u = 27 - 3(5)
u = 27 - 15
u = 12m/s.
Therefore, the initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
Learn more about motion at: brainly.com/question/974124
Answer:
A) 12.57 m
B) 5 RPM
C) 3.142 m/s
Explanation:
A) Distance covered in 1 Revolution:
The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:
s = rθ
where,
s = distance covered = ?
r = radius of circle = 2 m
θ = Angle = 2π radians (For 1 complete Revolution)
Therefore,
s = (2 m)(2π radians)
<u>s = 12.57 m</u>
B) Angular Speed:
The formula for angular speed is given as:
ω = θ/t
where,
ω = angular speed = ?
θ = angular distance covered = 15 revolutions
t = time taken = 3 min
Therefore,
ω = 15 rev/3 min
<u>ω = 5 RPM</u>
C) Linear Speed:
The formula that gives the the linear speed of an object moving in a circular path is given as:
v = rω
where,
v = linear speed = ?
r = radius = 2 m
ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s
Therefore,
v = (2 m)(1.571 rad/s)
<u>v = 3.142 m/s</u>
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(amount of heat)Q = ? , (Mass) m= 4 g , ΔT = T f - T i = 180 c° - 20 °c = 160 °c ,
Ce = 0.093 cal/g. °c
Q = m C ΔT
Q = 4 g × 0.093 cal/g.c° × ( 180 °c- 20 °c )
Q= 4×0.093 × 160
Q = 59.52 cal
I hope I helped you^_^
Answer:
Hello! Your answer is BELOW
Explanation:
1.About 91.754% of all iron is iron-56. Of all nuclides, iron-56 has the lowest mass per nucleon. With 8.8 MeV binding energy per nucleon, iron-56 is one of the most tightly bound nuclei.
2.The atomic weight of lead is quite variable in nature because the three heaviest isotopes are the stable end-products of the radioactive decay of uranium (238U to 206Pb and 235U to 207Pb) and thorium (232Th to 208Pb).
3.Mass defect for uranium-238 is 3.983 × 10-25 kg.
4.Energy and Mass Are Relative
The equation E = mc^2 states that the amount of energy possessed by an object is equal to its mass multiplied by the square of the speed of light.
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