Answer:
6.18 m/s
Explanation:
Roller skate collision
The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;
x-axis component form (+x east);
+ 
+ 
= 
+
Ф
60 ·8 + 0 = (60 + 80)
Ф
480 = 140
Ф................. (I)
y-axis component form (+y north);
+ 
+ 
= 
+ 
Ф
0 + 80.9 = (60 + 80)
Ф
720=
140Ф
140Vf=
Ф......................................(2)
Substituting (2) into (1) to give the angle;
480 = 720tan Ф
Ф = arctan(0.67) =33.69°.......................(3)
Evaluating (1) with (3) gives the velocity magnitude
480 = 140Vfsin 33.69°
Vf=6.18 m/s
note 1:
This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.
Answer:
The speed with which the baseball leaves the hand = 20.58 m/s
Explanation:
The time take to reach highest height during a projectile's flight is given by
t = (u sin θ)/g
u = initial velocity of the baseball = ?
θ = angle of throw above the horizontal
g = acceleration due to gravity = 9.8 m/s²
1.05 = (u sin 30)/9.8
u = (1.05 × 9.8)/0.5
u = 20.58 m/s
Answer:
Explanation:
Image of distant object will be made at far point or at 52.5 so
object distance u = infinity
image distance v = - 52.5 cm
focal length required = f
Lens formula
1 / v - 1 / u = 1 / f
1 / - 52.5 - 0 = 1 / f
f = -52.5 cm
= -.525 m
Power P = 1 / f = - 1 / .525
= - 1.90
now , for eye with glass we shall find new near point .
v = ?
u = - 17.2 cm
f = - 52.5 cm
1 / v - 1 / u = 1 / f
1 / v + 1 / 17.2 = - 1 / 52.5
1 / v = - 1 / 17.2 - 1 / 52.5
= - .05813 - .019
= - .07713
u = - 12.96 cm
so new near point will be 12.96 cm
Answer:
Explanation:
Impulse = change in momentum
mv - mu , v and u are final and initial velocity during impact at surface
For downward motion of baseball
v² = u² + 2gh₁
= 2 x 9.8 x 2.25
v = 6.64 m / s
It becomes initial velocity during impact .
For body going upwards
v² = u² - 2gh₂
u² = 2 x 9.8 x 1.38
u = 5.2 m / s
This becomes final velocity after impact
change in momentum
m ( final velocity - initial velocity )
.49 ( 5.2 - 6.64 )
= .7056 N.s.
Impulse by floor in upward direction
= .7056 N.s
Answer:
The frequency of the signal is 2 GHz
Explanation:
Given;
period of the clock signal, T = 500 ps = 500 x 10⁻¹² s
the frequency of the signal is given by;
F = 2 GHz
Therefore, the frequency of the signal is 2 x 10⁹ Hz or 2 GHz
