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
The formula to find the magnitude of velocity is:
▲v= ▲M/m
▲v=Velocity Change
▲M=Momentum Change
m=Mass
Plug in the information;
▲v=345/5
The answer is 69
Answer:
Approximate escape speed = 45.3 km/s
Explanation:
Escape speed
Here we have
Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²
R = 1 AU = 1.496 × 10¹¹ m
M = 2.3 × 10³⁰ kg
Substituting
Approximate escape speed = 45.3 km/s
Answer:
The fractional Intensity 
= 0.0146
Given:
wavelength of the light, 
slit and screen separation difference, D = 130 cm = 1.3 m
distance of the point from the center of the principal maximum, y = 4.10 mm = 0.041 m
slit width, d = 0.420 mm = 
Solution:
To calculate the fractional intensity, we use the given formula:
(1)
For very small angle:
(2)
where
Using eqn (2):
Now, using eqn (1):
Answer:
A) The ball will roll forever in a straight path.
