Answer: yes u can use a 6.75 when measuring the pencil
Answer:
The velocity of the ball is 3.52 m/s.
Explanation:
A projectile is any object that moves under the influence of gravity and momentum only. Examples are; a thrown ball, a fired bullet, a kicked ball, thrown javelin, etc.
Given that the ball was thrown vertically upward on the top of a skyscraper of height 61.9 m. So that the velocity can be determined by;
u = 
Where: u is the velocity of the object, H is the height and g is the gravitational force on the object. Given that: H = 61.9 m and g = 10 m/
, then;
u = 
= 
u = 3.5185
The velocity of the ball is 3.52 m/s.
Answer
Given,
Magnetic field, B = 0.0000193 T
speed, v = 121 m/s
mass of electron, m = 9.11 x 10⁻³¹ Kg
charge of electron, q = 1.6 x 10⁻¹⁹ C
radius of the electron path, r = ?


r = 3.64 x 10⁻⁵ m
We know frequency is inverse of time period
d = v t



t = 1.889 x 10⁻⁶ s.
now, frequency



If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
Mexico Mexico language to is the answer to the question