B explanation : they are both filled to the same pint
Answer:
1170 m
Explanation:
Given:
a = 3.30 m/s²
v₀ = 0 m/s
v = 88.0 m/s
x₀ = 0 m
Find:
x
v² = v₀² + 2a(x - x₀)
(88.0 m/s)² = (0 m/s)² + 2 (3.30 m/s²) (x - 0 m)
x = 1173.33 m
Rounded to 3 sig-figs, the runway must be at least 1170 meters long.
Answer:
<em>The work done by the car is 363 kJ</em>
Explanation:
Work : Work is said to be done when a Force moves an object through a certain distance. Work and Energy are interchangeable because they have the same unit. The unit of work is Joules (J).
Mathematically work done can be expressed as,
E = W = 1/2mv²
W = 1/2mv²................................ Equation 1
Where E = Energy, W = work done, m = mass of the car, v = velocity of the car
<em>Given: m=1500 kg, v=22 m/s</em>
<em>Substituting these values into equation 1</em>
<em>W = 1/2(1500)(22)²</em>
<em>W = 750 × 484</em>
<em>W = 363000 J</em>
<em>W = 363 kJ</em>
<em>Thus the work done by the car is 363 kJ</em>
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )
