True
The half-life isn’t applicable to a first order reaction because it does not rely on the concentration of reactant present. However the 2nd order reaction is dependent on the concentration of the reactant present.
The relationship between the half life and the reactant is an inverse one.
The half life is usually reduced or shortened with an increase in the concentration and vice versa.
Answer:
v = 7.67 m/s
Explanation:
Given data:
horizontal distance 11.98 m
Acceleration due to gravity 9.8 m/s^2
Assuming initial velocity is zero
we know that
solving for t
we have
substituing all value for time t
t = 1.56 s
we know that speed is given as
v = 7.67 m/s
The correct answer to the question is : B) The weight of the water, and C) The height of the water.
EXPLANATION :
Before coming into any conclusion, first we have to understand potential energy of a body.
The potential energy of a body due to its position from ground is known as gravitational potential energy.
The gravitational potential energy is calculated as -
Potential energy P.E = mgh
Here, m is the mass of the body, and g is the acceleration due to gravity.
h stands for the height of the body from the ground.
We know that weight of a body is equal to the product of mass with acceleration due to gravity.
Hence, weight W = mg
Hence, potential energy is written as P.E = weight × height.
Hence, potential energy depends on the weight and height of the water.
Let <em>F</em> be the magnitude of the force applied to the cart, <em>m</em> the mass of the cart, and <em>a</em> the acceleration it undergoes. After time <em>t</em>, the cart accelerates from rest <em>v</em>₀ = 0 to a final velocity <em>v</em>. By Newton's second law, the first push applies an acceleration of
<em>F</em> = <em>m a</em> → <em>a</em> = <em>F </em>/ <em>m</em>
so that the cart's final speed is
<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>
<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>
<em />
If we force is halved, so is the accleration:
<em>a</em> = <em>F</em> / <em>m</em> → <em>a</em>/2 = <em>F</em> / (2<em>m</em>)
So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give
(<em>F</em> / (2<em>m</em>)) (2<em>t</em>) = (<em>F</em> / <em>m</em>) <em>t</em> = <em>v</em>
