Answer:
The final velocity of the car is 26.65 m/s.
Explanation:
Given;
acceleration of the racecar, a = 6.5 m/s²
initial velocity of the car, u = 0
time of motion, t = 4.1 s
The final velocity of the car is given by;
v = u + at
where;
v is the final velocity of the car
suvstitute the givens
v = 0 + (6.5)(4.1)
v = 26.65 m/s.
Therefore, the final velocity of the car is 26.65 m/s.
Answer:
If one plays ice hockey he should must have necessary equipment to support his sport. These equipment include Ice Skates,Helmet with Cage and Mouth-guard:, Hockey stick, Hockey pants,Hockey gloves, shoulder pads, elbow pads, Shin Guard:Neck guard and Jockstrap (men) or Pelvic protector (women)
Explanation:
Answer:
Explanation:
Let T be the tension .
Applying newton's second law on the downward movement of the bucket
mg - T = ma
On the drum , a torque of TR will be acting which will create an angular acceleration of α in it . If I be the moment of inertia of the drum
TR = Iα
TR = Ia/ R
T = Ia/ R²
Replacing this value of T in the other equation
mg - T = ma
mg - Ia/ R² = ma
mg = Ia/ R² +ma
a ( I/ R² +m)= mg
a = mg / ( I/ R² +m)
mg - T = ma
mg - ma = T
mg - m x mg / ( I/ R² +m) = T
mg - m²g / ( I/ R² +m ) = T
mg - mg / ( 1 + I / m R² ) = T
b ) T = Ia/ R²
I = TR² / a
c ) Moment of inertia of hollow cylinder
I = 1/2 M ( R² - R² / 4 )
= 3/4 x 1/2 MR²
= 3/8 MR²
I / R² = 3/8 M
a = mg / ( I/ R² +m)
a = mg / ( 3/8 M + m )
T = Ia/ R²
= 3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²
= 
Answer:
ntergalactic space takes up most of the volume of the universe, but even galaxies and star systems consist almost entirely of empty space.
Explanation:
so to be exact galixes do make up most of the volume of space because galaxies are what make up space!
By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet