1. A jet lands on an aircraft carrier at a speed of 140 mi/h [ 1.00 mile = 1.61 km]
1 answer:
Answer:
Acceleration, a = -31.29 m/s²
Explanation:
It is given that, Initial speed of the aircraft, u = 140 mi/h =
62.58 m/s
Finally, it stops, v = 0
Time taken, t = 2 s
Let a is the acceleration of the aircraft. We know that the rate of change of velocity is called acceleration of the object. It is given by :
a = t
0-62.58 2 a =
a = -31.29 m/s²
So, the acceleration of the aircraft is 31.29 m/s²
and the car is decelerating. Hence, this is the required solution.
(i found this answer becuase i could do the math)
Acceleration, a = -31.29 m/s²
Explanation:
It is given that, Initial speed of the aircraft, u = 140 mi/h =
62.58 m/s
Finally, it stops, v = 0
Time taken, t = 2 s
Let a is the acceleration of the aircraft. We know that the rate of change of velocity is called acceleration of the object. It is given by :
a = t
0-62.58 2 a =
a = -31.29 m/s²
So, the acceleration of the aircraft is 31.29 m/s²
and the car is decelerating. Hence, this is the required solution.
(i found this answer becuase i could do the math)
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Answer:
The right answer is:
(a) 63.83 kg
(b) 0.725 m/s
Explanation:
The given query seems to be incomplete. Below is the attachment of the full question is attached.
The given values are:
T = 3 sec
k = 280 N/m
(a)
The mass of the string will be:
⇒
or,
⇒
On substituting the values, we get
⇒
⇒
⇒
(b)
The speed of the string will be:
⇒
then,
⇒
On substituting the values, we get
⇒
⇒
⇒
⇒
For 3 kg mass: <em>F = =
=
= 4
/
</em>
For 6 kg mass: <em>F = =
=
= 2
/
</em>
Hope this helped, have a great day!~