Given:
The initial velocity of the object, v=30 m/s
a_t=0
a_c≠0
The time period is Δt.
To find:
The right conclusion among the given choices.
Explanation:
a_t represents the tangential accleration on the object and a_c represents the centripetal acceleration on the object.
The centripetal acceleration is the acceleration that keeps the object in its circular path. The centripetal force only changes the direction of the velocity and not the magnitude.
Thus the magnitude of the velocity of the object remains the same after a time interval of Δt. But the direction of the velocity of the object will be changed and will be unknown after Δt seconds.
Final answer:
Thus the object will be traveling at 30 m/s in some unknown direction.
Therefore, the correct answer is option a.
Answer:
Addition reactions with benzenes lead to the loss of aromaticity.
Benzene and its derivatives undergo a type of substitution reaction in which a hydrogen atom is replaced by a substituent, but the stable aromatic benzene ring is regenerated at the end of the mechanism.
Benzene and its derivatives tend to undergo electrophilic aromatic substitution reactions.
Explanation:
<span>The initial velocity of the bike was 1.67 (vf)m/s. This is found by evaluating 7.5/4.5 which yields the velocity per unit of time which is equivalent to initial velocity.</span>
A) the periodic time is given by the equation;
T= 2π√(L/g)
For the frequency will be obtained by 1/T (Hz)
T = 2 × 3.14 √ (0.66/9.81)
= 6.28 × √0.0673
= 1.6289 Seconds
Frequency = 1/T = f = 1/1.6289
thus; frequency = 0.614 Hz
b) The vertical distance, the height is given by
h= 0.66 cos 12
h = 0.65 m
Vertical fall at the lowest point = 0.66 - 0.65 = 0.01 m
Applying conservation of energy
energy lost (MgΔh) = KE gained (1/2mv²)
mgh = 1/2mv²
v² = 2gΔh = 2×9.81 × 0.01
= 0.1962
v = 0.443 m/s
c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
Thus total energy is equal to;
E = 1/2mv²
= 1/2 × 0.310 × 0.443²
= 0.0304 J
Answer:
The acceleration is 
Explanation:
From the question we are told that
The lift up speed is 
The distance covered for the take off run is 
Generally from kinematic equation we have that
Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest
So
=>
