Answer:
V is approximately = 23m/s
Explanation:
Kinetic energy = ½ mv²
Where m= mass = 0.450kg
V= velocity =?
K. E = 119J
Therefore
K. E = ½ mv²
Input values given
119= ½ × 0.450 × v²
Multiply both sides by 2
119 ×2 = 2 × 1/2 × 0.450 × v²
238= 0.450v²
Divide both sides by 0.450
238/0.450 = 0.450v²/0.450
v² = 528.89
Square root both sides
Sq rt v² = sq rt 528.89
V = 22.998m/s
V is approximately = 23m/s
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Answer:
E = k*Q₁/R₁² V/m
V = k*Q₁/R₁ Volt
Explanation:
Given:
- Charge distributed on the sphere is Q₁
- The radius of sphere is R₁
- The electric potential at infinity is 0
Find:
What is the electric field at the surface of the sphere?E.
What is the electric potential at the surface of the sphere?V
Solution:
- The 3 dimensional space around a charge(source) in which its effects is felt is known in the electric field.
- The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.
- If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by
F = k*Q₁/R₁²
- Then the electric field at that point is
E = F/1
E = k*Q₁/R₁² V/m
- The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.
- Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation
V = k*Q₁/R₁ Volt
Answer:
15m
Explanation:
Hello! first to solve this problem we must find that so much that the tree grew in the 15 years this is achieved by dividing the height of the tree before and after
the tree grew 10 times its initial length in 15 years, so to find how tall the nail is, we multiply this factor by 1.5m
X=(1.5m)(10)=15m
the nail is 15 meters above ground level
Answer:
Qa/Qb = k/2×2k = 1/4
Explanation:
a situation is series and b situation is parallel.
let conductance of each plate = k
so net conductance in series = k/2
net conductance in parallel = 2k
so Qa/Qb = k/2×2k = 1/4
<span>x=((12.3/100)m)cos[(1.26s^−1)t]
v= dx/dt = -</span><span>((12.3/100)*1.26)sin[(1.26s^−1)t]
v=</span>-((12.3/100)*1.26)sin[(1.26s^−1)t]=-((12.3/100)*1.26)sin[(1.26s^−1)*(0.815)]
v=<span>
<span>-0.13261622 m/s
</span></span>the object moving at 0.13 m/s <span>at time t=0.815 s</span>
