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2 years ago

11

a bobsled has a momentum of 1500 kgm/s to the south. Air resistance reduces its momentum to 750 kgm/s to the south. What impul

se is applied to the bobsled by the air resistance?

2 answers:

Elza [17]2 years ago

8 0

Answer:

750 kg•m/s north

Explanation:

kherson [118]2 years ago

7 0

The impulse is exactly the change in momentum,
and they have the same units.

The impulse imparted to the bobsled by the air
resistance is

(750 - 1500) = -750 kg-m/s.

It's negative because it reduced the bobsled's momentum.

If the impulse had come from a rocket engine on the back of
the bobsled, then it would have been a positive impulse.

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In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li

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Answer:

Total energy saving will be 0.8 KWH

Explanation:

We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW

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So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW

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2 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

balandron [24]

Answer with Explanation:

We are given that

Radius of solid core wire=r=2.28 mm= $2.28\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of each strand of thin wire=r'=0.456 mm= $0.456\times 10^{-3} m$

Current density of each wire= $J=3750 A/m^2$

a.Area = $\pi r^2$

Where $\pi=3.14$

Using the formula

Cross section area of copper wire has solid core = $3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2$

Current density = $J=\frac{I}{A}$

Using the formula

$3750=\frac{I}{16.3\times 10^{-6}}$

$I=3750\times 16.3\times 10^{-6}=0.061 A$

Total number of strands=19

Area of strand wire= $A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2$

$J'=\frac{I'}{A'}$

$3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}$

$I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2$

$I'=0.047 A$

b.Resistivity of copper wire= $\rho=1.69\times 10^{-8}\Omega-m$

Length of each wire =6.25 m

Resistance, R= $\frac{\rho l}{A}$

Using the formula

Resistance of solid core wire= $R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega$

Resistance of strand wire= $R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega$

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laila [671]

Answer:

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Explanation:

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v = 0.027 c

v = 0.027 3 10⁸

v = 8.1 10⁶ m / s

for this part we can use the conservation of mechanical energy

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Em₀ = U = q V

final point. Electrons with maximum speed

Em_f = K = ½ m v2

Em₀ = Em_{f}

e V = ½ m v²

V = ½ m v² / e

let's calculate

V = ½ 9.1 10⁻³¹ (8.1 10⁶)² / 1.6 10⁻¹⁹

V = 1.866 10² V

V = 1866 V

b) if this acceleration protons is the mass of the proton is m_{p} = 1.67 10-27

V = ½ 1.67 10⁻²⁷ (8.1 10⁶)² / 1.6 10⁻¹⁹

V = 3.424 10⁵ V

V = 342402 V

c) this potential difference should give the protons the same speed as the electrons

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Answer:

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Therefore, we can write:

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$q_2=-1\cdot 10^{-5} C$ is the charge on the bag

r is the separation betwen the bag and the balloon

$m=0.02 g=2\cdot 10^{-5} kg$ is the mass of the bag

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Solving for r, we find the distance at which the bag must be held:

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