Answer: A projectile which is fired horizontally is being constantly acted upon by acceleration due to gravity, acting vertically downwards. Hence, it does not follow a straight line path. Also Why a projectile fixed along the horizontal not follow a straight line path? Because the projectile fired horizontally is constantly acts upon by acceleration due to gravity acting vertically downwards.
Explanation:
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Answer: 0.86 × 10^14
Explanation:
Given the following :
Radius of proton = 1.2 × 10-15 m
Radius of hydrogen atom = 5.3 × 10-11 m
Density of proton could be calculated thus:
Mass of proton = 1.67 × 10^-27 kg
Using the formula :
(4/3) × pi × r^3
(4/3) × 3.142 × (1.2 × 10^-15)^3 = 7.24 × 10^-45
Density = mass / volume
Density = (1.67 × 10^-27) / ( 7.24 × 10^-45)
= 0.2306 × 10^18
Density of hydrogen atom:
Mass of hydrogen atom= 1.67 × 10^-27 kg
Using the formula :
(4/3) × pi × r^3
(4/3) × 3.142 × (5.3 × 10^-11)^3 = 6.24 × 10^-31
Density = mass / volume
Density = (1.67 × 10^-27) / ( 6.24 × 10^-31)
= 0.2676 × 10^4
Ratio is thus:
Density of proton / density of hydrogen atom
0.2306 × 10^18 / 0.2676 × 10^4 = 0.8617 × 10^14
Explanation:
The time period rotation of the bob is t and the tension in the thread is T. ... A simple pendulum of length. ... Balancing the forces in Horizontal and vertical direction: ... Circular motion 2.
1 answer
Answer:
0.54454
104.00902 N
Explanation:
m = Mass of wheel = 100 kg
r = Radius = 0.52 m
t = Time taken = 6 seconds
= Final angular velocity
= Initial angular velocity
= Angular acceleration
Mass of inertia is given by

Angular acceleration is given by

Equation of rotational motion

The coefficient of friction is 0.54454
At r = 0.25 m

The force needed to stop the wheel is 104.00902 N