Physics, 11.12.2019 03:31, dommalb

A100 ton ship traveling at 1 miles per hour relative hits a tennis ball sized one pound piece of iron. what is the force generated and what would happen to the ship?

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answered: trammellb91

The force generated in 1 second of impact is, F = 201161.37 N

The ship slightly slows down and moves forward with velocity V = 9.9 x 10¹⁰ kg miles/hr

Explanation:

Given data,

The mass of the ship, M = 100 ton

= 100,000 kg

The velocity of the ship, V = 1,000,000 miles/hr

The mass of the iron piece, m = 1 lb

= 0.45 kg

The velocity of the iron piece, v = 0

The momentum of the ship,

P = MV

= 100,000 kg x 1,000,000 miles/ hr

= 1 x 10¹¹ kg miles/ hr

The momentum of the iron piece, p = 0

According to the conservation of momentum,

MV + mv = U(M + m)

$U=\frac{MV+mv}{M+m}$

$U=\frac{1\times 10^{11} +0}{100000+0.45}$

U = 999,995.5 miles/hr

Therefore the momentum of the ship after impact,

P₁ = MU

= 100,000 kg x 999,995.5 miles/hr

= 9.9 x 10¹⁰ kg miles/hr

Therefore, the change in the momentum of the ship,

ΔP = P - P₁

= 1 x 10¹¹ - 9.9 x 10¹⁰

= 449997.98 kg miles/hr

= 201161.37 kg m/s

Let the time period of impact be 1 s

Therefore, the force exerted by the ship on the iron is,

F = ΔP / t

= 201161.37 / 1

= 201161.37 N

Hence, the force generated in 1 second of impact is, F = 201161.37 N

The ship slightly slows down and moves forward with velocity V = 9.9 x 10¹⁰ kg miles/hr

answered: Guest

its c its is a mixture of gases

answered: Guest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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