Answer:
a. The speed is 2.39 m/s
b. The acceleration of the block is 10.2
Explanation:
First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:
Δx=
W_{ext}=20.4 Nm
To determine, the acceleration we solve the following equation for a:
Hey I wanna was irie wiiw you a good night time to come home I wanna was your gonna day you were a
Answer:
Option D. 4.4 m/s²
Explanation:
The following data were obtained from the question:
Velocity (v) = 21 m/s
Radius (r) = 100 m
Centripetal acceleration (a) =.?
The centripetal acceleration of the car can be obtained as follow:
Centripetal acceleration (a) = Velocity square (v²) / radius (r)
a = v²/r
a = 21²/100
a = 441/100
a = 4.41 ≈ 4.4 m/s²
Therefore, the centripetal acceleration of the car is 4.4 m/s².
Answer:Solution
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Given:u=−10cm f=20cm (convex lens)
To find: v and image's nature.
Solution: From lens formula
v
1
−
u
1
=
f
1
v
1
+
10
1
=
20
1
Explanation:
Answer:
Explanation:
Let T be the tension
For linear motion of hoop downwards
mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .
For rotational motion of hoop
Torque by tension
T x R , R is radius of hoop.
Angular acceleration be α,
Linear acceleration a = α R
So TR = I α
= I a / R
a = TR² / I
Putting this value in earlier relation
mg -T = m TR² / I
mg = T ( 1 + m R² / I )
T = mg / ( 1 + m R² / I )
mg / ( 1 + R² / k² )
Tension is less than mg or weight because denominator of the expression is more than 1.
