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3 years ago

A 10 g bullet accelerates to 400 m/s after being fired from a gun with a mass of 2.0 kg.

Physics

1 answer:

Gala2k [10]3 years ago

3 0

Answer:

See the answers below.

Explanation:

Momentum is defined as the product of mass by velocity, and can be calculated by means of the following expression.

$P=m*v$

where:

P = Momentum [kg*m/s]

m = mass = 10 [g] = 0.01 [kg]

v = velocity = 400 [m/s]

$P=0.01*400\\P=4[kg*m/s]$

ii)

The momentum of the gun is equal to zero, because it does not move before being fired, the weapon only moves after having fired the weapon.

$P_{gun}=0$

iii)

Since the momentum is conserved before and after the shot, the same momentum given to the bullet is equal to the momentum received by the gun.

$P=m*v_{recoil}$

$v_{recoil}=P/m\\v_{recoil}= 4/2\\v_{recoil}=2[m/s]$

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Gravity: The force due to gravity is F=mg where g=9.80 m/s². A. Find the force due to gravity on a 41.63-kg object. B. The force

olga_2 [115]

A. F=mg
= 41.63kg x 9.8m/s²
= 407.974
= 408N (3 significant figure)
B. F=mg
632N= y x 9.8m/s²
y= 632 ÷ 9.8
y = 64.48
y= 64.5kg (3 significant figure)

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3 years ago

Two ice boats (one of mass m, one of mass 2m) hold a race on a frictionless, horizontal, frozen lake. Bath ice boats start at re

Elenna [48]

Answer:

They both cross the finish line with the same kinetic energy

Explanation:

Same force, same displacement, so same KE at the location of the finish line. They don’t cross the line at the same time, but that was not the question!

∆KA= (mA/2)vF,A2

∆KA=∆KBso vF,B/vF,A= (MA/MB)1/2∆KB= (mB/2)vF,B2

Lighter boat goes faster so reaches finish line 1st

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3 years ago

Explain why elements in the same family have similar physical and chemical properties

zhenek [66]

They have similar physical and chemical properties because of thier valence electrons

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3 years ago

Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$

Power factor is given by

$F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802$

The power factor is 0.28802

The average power to the circuit is given by

$P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W$

The average power to the circuit is 2.57162 W

Power to resistor

$P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W$

Power to resistor is 14.28 W

Power to inductor

$P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W$

Power to the inductor is 53.55 W

Power to the capacitor

$P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W$

The power to the capacitor is 6.07142 W

8 0

3 years ago

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$

Substitute the given values to find "terminal speed"

$\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}$

$\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}$

$\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}$

$\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}$

The “terminal speed” of the ball bearing is 5.609 m/s

7 0

3 years ago