Explanation:
initial velocity U = 20m/s
Final velocity V = 35m/s
time = 15.0 secs
change in velocity = 35 - 15
= 20m/s
acceleration a = change in velocity/time V/t
a = (35-20)/15
a= 15/15
Hence, your acceleration is 1m/s^2
Explanation:
Activation energy and reaction rate
The activation energy of a chemical reaction is closely related to its rate. Specifically, the higher the activation energy, the slower the chemical reaction will be. ... The released energy helps other fuel molecules get over the energy barrier as well, leading to a chain reaction.
Answer:
Explained below
Explanation:
Newton's first law of motion: This law states that an object will remain at rest or continue in constant motion except it's acted upon by an external force. In projectile motion, the horizontal component of velocity will remain unchanged because we ignore air resistance since no force is acting in that horizontal direction.
Newton's second law of motion: This law states that force is the product of mass and acceleration. In projectile the force acts downwards, thus f = mg.
But g = a since internal forces will cancel out.
Thus, F = ma
Answer:
1. Luminosity
2.Apparent brightness
Explanation:
There are two factors on which brightness of star appear to be in the sky
The two factors are
1. Luminosity
2.Apparent brightness
1.Luminosity :It is defined as the total energy emitted by the object in a given time.Luminosity vary with the distance of observer from the star.Luminosity is a intrinsic property which depends on the fundamental chemical composition and structure of the material.Luminosity is depends on the size of star.Lager the star luminosity will be more.
2.Apparent brightness: It is defined as how bright a star appears from an observer on the earth and the amount of starlight reaching the earth.if the distance is large then the brightness decreases.When the distance of star from us small then the brightness of star increases.Distance is inversely proportional to brightness of the star.
Answer
given,
v = 128 ft/s
angle made with horizontal = 30°
now,
horizontal component of velocity
vx = v cos θ = 128 x cos 30° = 110.85 ft/s
vertical component of velocity
vy = v sin θ = 128 x sin 30° = 64 m/s
time taken to strike the ground
using equation of motion
v = u + at
0 =-64 -32 x t
t = 2 s
total time of flight is equal to
T = 2 t = 2 x 2 = 4 s
b) maximum height
using equation of motion
v² = u² + 2 a h
0 = 64² - 2 x 32 x h
64 h = 64²
h = 64 ft
c) range
R = v_x × time of flight
R = 110.85 × 4
R = 443.4 ft
