Given that,
Puck slide total Deltax = 12m
Puck and board Mk = 0.10
Find the initial speed = ?
We know that,
Vf^2-Vi^2 = 2a Deltax
-Vi^2 = -2MkgDeltax ............(1)
then,
fk = -Mk mg = ma
a = -Mkg .........(2)
From equation (1),
Vi^2 = 2MkgDeltax
Vi = âš2MkgDeltax [g=9.8m/s^2]
Vi = âš2(0.10)(0.98)(12)
Vi = âš0.2(9.8)(12)
Vi = âš1.92(12)
Vi = âš23.52
Vi = 4.8 m/s
Therefore the initial speed of the puck Vi = 4.8 m/s
Answer:
assuming ur speaking about chemistry
c=frequency×wavelength
or
energy=frequency×h
Answer:
0.34s, 8.5m,31.89m
Explanation:
The above motion defines a projectile motion.
Now the athletes lands on a cliff 30° to the horizontal this means the velocity at that point would be 25m/s cos30°
Now from Newton's law of motion.
The body would be decelerating so,
V = u - gt
Where u is initial velocity and v is final velocity. g is acceleration of free fall due to gravity.
Hence,
V-U/ -g = t
Hence 25cos30 - 25/ -9.8 = 0.34s.
2.Now the length of the jump is defined as the total horizontal distance which is marked off by the horizontal velocity and time taken for take off and landing.
Hence Distance,S = u × t
25 ×0.34 =8.5m.
3. The maximum height is defined that at that point the Final velocity is 0m/s
Now the initial velocity is 25m/s
From Newton's law that;
V2= U2 -2gH; where U and V are initial and final velocity and H is height.
Hence H = V2-U2/-2g
=(0)^2- (25)^2/ -2×9.8
= -625/-19.6 =31.89m
Answer: Please see the attachments below
Explanation: Please see the attachments below
