Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
The Impulse delivered to the baseball is 89 kgm/s.
To solve the problem above, we use the formula of impulse.
⇒ Formula:
- I = m(v-u)................. Equation 1
Where:
- I = Impulse delivered to the baseball
- m = mass of the baseball
- v = Final velocity of the baseball
- u = initial speed of the baseball
From the question,
⇒ Given:
- m = 0.8 kg
- u = 67 m/s
- v = -44 m/s
⇒ Substitute these values into equation 1
- I = 0.8(-44-67)
- I = 0.8(-111)
- I = -88.8
- I ≈ -89 kgm/s
Note: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.
Hence, The Impulse delivered to the baseball is 89 kgm/s.
Learn more about impulse here: brainly.com/question/7973509
I don't know if you need to complete this question or do it otherwise, however, I managed to find on the Internet on several places this completion of your sentence:
<span>Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of q = 2 x 10</span>⁶ W/m³.
I'm not sure whether that is the answer you were looking for, but that's what I found.
Answer:
5.66 × 10⁻²³ m/s
Explanation:
If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.
Since initial momentum = final momentum,
mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.
My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?
So mv₁ + M × 0 = m × 0 + MV₂
mv₁ = MV₂
V₂ = mv₁/M = 54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s
Mechanical energy equals the sum of potential and kinetic energy. During the process, all PE converts into KE, assuming air resistance is neglected. So, the mechanical energy does not change and is equal to the initial potential energy.
ME
=mgh
=0.005 x 9.81 x 3
=0.147J
