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2 years ago

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A 15-g bullet moving at 300 m/s passes through a 2.0 cm thick sheet of foam plastic and emerges with a speed of 90 m/s. Let's as

sume that the speed of the bullet changes takes place uniformly (uniform deceleration). Find the change in momentum of the bullet. Find the time the bullet is in contact with the plastic, Delta t. Find the average force impeded by the bullet's motion through the plastic.

1 answer:

Shkiper50 [21]2 years ago

8 0

Answer:

Explanation:

a) Change in momentum, Δp = mΔv = m(v - u) = (15 * 10-3) * (90 - 300) = -3.15 kg-m2

b) Acceleration of the bullet, a = (v2 - u2) / 2s = (902 - 3002) / (2 * 0.02) = -2047500 m/s2

So, the bullet is in contact with the plastic for the time, $\bigtriangleupt = \frac{(v - u)}{a} =\frac{(90 - 300)}{(-2047500)} = 1.03 \times 10^{-4} s$

c) Average force, $F_{avg} =\frac{\bigtriangleup p}{\bigtriangleup t} =\frac{(-3.15)}{(1.03 \times 10^{-4})} = 30.6 kN$

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A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T

atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.$

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

$\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.$

<em>where</em>,

k is the Coulomb's constant, having value = $\dfrac{1}{4\pi \epsilon_o}$

$\epsilon_o$ is the electrical permittivity of free space.

Also, r>>s, therefore, $\rm r^2+s^2\approx r^2.$

$\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.$

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

$\rm F=qE$

Therefore, the electric force on the charge Q due to the dipole is given by

$\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.$

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = $\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole is given by

$\rm \tau = Fs\ \sin\theta$

Since the charge Q is placed in the plane that bisects the dipole, therefore, $\theta = 90^\circ$ .

$\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.$

4 0

2 years ago

The atmosphere pressure can support mercury in a tube, which the upper end is closed, up to 0.76 meter. If the mercury is replac

Leni [432]

Answer:

Maximum height the atmosphere pressure can support the

water=10.336 m

Explanation:

We know that ,

$Pressure = h\cdot\rho\cdot g$

Case 1 - Mercury in the tube

$Density\ of\ mercury =\rho_1\\and\ height\ attained\ for\ mercury\ column = h_1$

Case 2 - Water in the tube

$Density\ of\ water =\rho_2\\and\ height\ attained\ for\ water\ column = h_2$

Since atmospheric pressure is same

. $P=h_1\cdot\rho_1\cdot g = h_2\cdot\rho_2\cdot g$

or, $h_2=\frac{h_1\rho_1}{\rho_2}$

$Given\ h_1= 0.76\ m,\rho_1=13.6\cdot\rho_2$

∴ $h_2=0.76\cdot13.6=10.336\ m$

Hence height of the water column =10.336 m

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2 years ago

Water, with a density of 1000 kg/m3, flows out of a spigot, through a hose, and out a nozzle into the air. The hose has an inner

stepan [7]

Answer:

P₁ = 2.3506 10⁵ Pa

Explanation:

For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

A₁ v₁ = A₂ v₂

Let's look for the areas

r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm

r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm

A₁ = π r₁²

A₁ = π 1.125²

A₁ = 3,976 cm²

A₂ = π r₂²

A₂ = π 0.1²

A₂ = 0.0452 cm²

Now with the continuity equation we can look for the speed of water inside the hose

v₁ = v₂ A₂ / A₁

v₁ = 11.2 0.0452 / 3.976

v₁ = 0.1273 m / s

Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)

P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂

P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂

Let's calculate

P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25

P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴

P₁ = 2.3506 10⁵ Pa

7 0

2 years ago

A battery has some internal resistance. Can the potential difference across the terminals of the battery be equal to its emf.

yarga [219]

Answer:

yes, the potential difference across the terminals of the battery can be equal to its emf.

Explanation:

when the current in the battery is zero, meaning the current though, and hence the potential drop across the internal resistance is zero. This only happens when there is no load placed on the battery.

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2 years ago

Follow these steps to solve this problem: Two identical loudspeakers, speaker 1 and speaker 2, are 2.0 m apart and are emitting

horsena [70]

Answer:

Explanation:

Wave length of sound from each of the speakers = 340 / 1700 = .2 m = 20 cm

Distance between first speaker and the given point = 4 m.

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Path difference = 4.472 - 4 = .4722 m.

Path difference / wave length = 0.4772 / 0.2 = 2.386

This is a fractional integer which is neither an odd nor an even multiple of half wavelength. Hence this point of neither a perfect constructive nor a perfect destructive interference.

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2 years ago