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2 years ago

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While a car is stopped at a traffic light in a storm, raindrops strike the roof of the car. The area of the roof is 5.0 m2. Each

raindrop has a mass of 3.7 ✕ 10−4 kg and speed of 2.5 m/s before impact and is at rest after the impact. If, on average at a given time, 150 raindrops strike each square meter, what is the impulse of the rain striking the car?

1 answer:

julsineya [31]2 years ago

5 0

Answer:

J = 0.693 N.s

Explanation:

The impulse of one single drop is given by:

J1 = m*(Vf - Vo) where Vf = 0

$J1 = -9.25*10^{-4}N.s$

The magnitude of the total impulse will be:

Jt = J1 * 150 * 5

Jt = 0.693 N.s

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Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V

Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

$Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter$

$Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter$

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

$R=\dfrac{\rho\times l}{A}$

Where,

R = Resistance

l = length

A = Area of cross section = πr²

$\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter$

Substituting the values we get

$R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}$

$R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}$

$R=\dfrac{1\times 10^{2}}{3.23}$

$R=30.95\ ohm$

Now by Ohm's Law,

$V= I\times R$

Substituting the values we get

$I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere$

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

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2 years ago

A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and betwe

natta225 [31]

First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.

Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.

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2 years ago

A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a

aleksley [76]

Answer:

T₂ =602 °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

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P₁ = 100 + 100 KPa

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The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

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The temperature at condition 2 = T₂

We know that

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Three identical train cars, coupled together, are rolling east at speed v0. A fourth car traveling east at 2v0 catches up with t

aliina [53]

Answer: $v_o$

Explanation:

It is given that three cars has same mass m with speed $v_o$

suppose rest two cars also has same mass m

As there is no external force therefore momentum is conserved

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$P_i=3mv_0+m(2v_0)+m\times 0$

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where v=final velocity

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$v=v_o$

thus final velocity is $v_o$

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Answer:

Vectors have both magnitude and direction.

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