First, let's put 22 km/h in m/s:
Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by
The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:
Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:
The actual position of the object is <span>at a great distance, effectively infinite. The other options given in the question are not at all correct. The correct option among all the options that are given in the question is the last option or option "D". I hope that this answer has actually come to your great help.</span>
Answer:
v_g,i = 1.208 m/s
Explanation:
We are given;
Mass of girl; m_g = 47.2 kg
Mass of plank; m_p = 177 kg
Let the velocity of girl to ice be v_g,i
Let the velocity of plank to ice be v_p,i
Since the velocity of the girl is 1.53 m/s relative to the plank, then;
v_g,i + v_p,i = 1.53
From conservation of momentum;
m_g × v_g,i = m_p × v_p,i
Thus;
47.2(v_g,i) = 177(v_p,i)
Dividing both sides by 47.2 gives;
v_g,i = 3.75(v_p,i)
v_pi = (v_g,i)/3.75
Thus, from v_g,i + v_p,i = 1.53, we have;
v_g,i + ((v_g,i)/3.75) = 1.53
v_g,i(1 + 1/3.75) = 1.53
1.267v_g,i = 1.53
v_g,i = 1.53/1.267
v_g,i = 1.208 m/s
Explanation:
Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path. ... Waves are generated by vibrating sources, their amplitude being proportional to the amplitude of the source.
Answer:
beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.
To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm
Solution:
As per the given criteria,
the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)
(a) lens is a convex lens with
focal length, f=20cm
object distance, u=12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
20
1
+
12
1
⟹
v
1
=
60
3+5
⟹v=7.5cm
Hence the image formed is real, at 7.5cm from the lens on its right side.
(b) lens is a concave lens with
focal length, f=−16cm
object distance, 12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
−16
1
+
12
1
⟹
v
1
=
48
−3+4
⟹v=48m
Hence the image formed is real, at 48 cm from the lens on the right side.
