The minimum initial speed of the dart so that the combination makes a complete circular loop after the collision is 58.5 m/s.
<h3>Minimum speed for the object not fall out of the circle</h3>
The minimum speed if given by tension in the wire;
T + mg = ma
T + mg = m(v²)/R
tension must be zero for the object not fall
0 + mg = mv²/R
v = √(Rg)
<h3>Final speed of the two mass after collision</h3>
Use the principle of conservation of energy
K.Ef = K.Ei + P.E
¹/₂mvf² = ¹/₂mv² + mg(2R)
¹/₂vf² = ¹/₂v² + g(2R)
¹/₂vf² = ¹/₂(Rg) + g(2R)
vf² = Rg + 4Rg
vf² = 5Rg
vf = √(5Rg)
vf = √(5 x 2.8 x 9.8)
vf = 11.7 m/s
<h3>Initial speed of the dart</h3>
Apply principle of conservation of linear momentum for inelastic collision;
5v = vf(20 + 5)
5v = 11.7(25)
5v = 292.5
v = 58.5 m/s
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Answer:
2ms-¹ means that the body under consideration moves 2m in a second, and may be it will continue to move 2m in every 1 second, if there's no external unbalanced force acting on that body (those forces do include frictional forces). mark its brainlist plz. Kaneppeleqw and 6 more users found this answer helpful. Thanks 3.
Well, Break the problem up into parts.
For speeding up:
The car accelerates at 4 m/s^2 for 3 seconds. Multiplying these values tells you the car reaches a speed of 12 m/s.
Vf^2 = Vi^2 + 2a(Xf - Xi)
12^2 = 0 + 2(4)(Xf - 0)
144 = 8 Xf
Xf = 18 m
For coasting:
The car is at the 12 m/s and does this for 2 seconds.
x = vt = (12)(2) = 24 m
For slowing down:
The car decelerates at 3 m/s^2 to come to a stop at the next sign. From 12 m/s, this would take 12/3 seconds or 4 s.
Vf^2 = Vi^2 + 2a(Xf - Xi)
0 = 12^2 + (2)(-3)(Xf - 0)
-144 = -6 Xf
Xf = 24 m
Summing the distances: Xtotal = 18 + 24 + 24 = 66 m
Answer:
1. Velocity = 75.39 m/s
2. Air resistance
3. Mass of the coin
Explanation:
1. Given that the height of the building is 290 m. From Newton's third law of motion,
= 
+ 2gh
where: V is the final velocity of the coin, U is the initial velocity, g is the force of gravity and h is the height.
Since the coin was dropped from a height, its initial velocity is zero. The force of gravity is taken as 9.8 m/
, so that:
= 0 + 2 x 9.8 x 290
= 5684
V = 
= 75.3923
The velocity with which the coin hits the ground is 75.39 m/s.
2. Air resistance: During the free fall of the coin, the direction of wind flow could either cause an increase or decrease the value predicted.
ii. Mass of the coin: This can also affect the predicted value.
Answer:
Answer. to final velocity 'v' =10.9 m/s in time 't' = 2.37 secs. So acceleration = -7.09 m/sec^2 or, decceleration is 7.09 m/sec^2
Explanation:
