Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.
In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.
Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.
We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.
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>
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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>
Mass (m) and volume (v)
the equation is d = m/v
The atom becomes positively charged.
EXPLANATION :
When all the electrons are removed from the atom, the atom contains only protons and neutrons. The neutron is a neutral nucleon. ...
Hence, the correct answer is that atom will be positively charged.
Answer:
9.8 × 10⁴Pa
Explanation:
Given:
Velocity V₁ = 12m/s
Pressure P₁ = 3 × 10⁴ Pa
From continuity equation we have
ρA₁V₁ = ρA₂V₂
A₁V₁ = A₂V₂
making V₂ the subject of the equation;
the pipe is widened to twice its original radius,
r₂ = 2r₁
then the cross-sectional area A₂ = 4A₁
⇒ 
This implies that the water speed will drop by a factor of 
because of the increase the pipe cross-sectional area.
The Bernoulli Equation;
Energy per unit volume before = Energy per unit volume after
p₁ + 
ρV₁² + ρgh₁ = p₂ + ρV₂² + ρgh₂
Total pressure is constant and 
= P = ρV₂²ρV²
p₁ + ρV₁² = p₂ + ρV₂²
Making p₂ the subject of the equation above;
p₂ = p₁ + ρV₁² - ρV₂²
But 
so,
p₂ = p₁ + ρV₁² - ρ
p₂ = 3.0 x 10⁴ + ( × 1000 × 12²) - ( × 1000 × 12²/4² )
P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³
P₂ = 9.79 × 10⁴Pa
P₂ = 9.8 × 10⁴Pa
