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Alina [70]
4 years ago
13

a 1150 kg car is on a 8.70 hill. using x-y axis tilted down the plane, what is the x-component of the normal force(unit=N)

Physics
1 answer:
rodikova [14]4 years ago
3 0

The x-component of the normal force is equal to <u>1706.45 N.</u>

Why?

To solve the problem, and since there is no additional information, we can safely assume that the x-axis is parallalel to the hill surface and the y-axis is perpendicular to the x-axis. Knowing that, we can calculate the components of the normal force (or weight for this case), using the following formulas:

N_{x}=W*Sin(\alpha)=mg*Sin(\alpha)\\\\N_{y}=W*Cos(\alpha)=mg*Cos(\alpha)

Now, using the given information, we have:

mass=m=1150Kg\\\alpha=8.70\°\\g=9.81\frac{m}{s^{2}}

Calculating, we have:

N_{x}=mg*Sin(\alpha)

N_{x}=1150Kg*9.81\frac{m}{s^{2}}*Sin(8.70\°)\\\\N_{x}=11281.5\frac{Kg.m}{s^{2} }*Sin(8.70\°)=1706.45\frac{Kg.m}{s^{2} }=1706.45.23N

Hence, we have that the x-component of the normal force is equal to  <u>1706.45 N.</u>

Have a nice day!

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A voltage source is connected across two 1 kohm resistors in parallel. The current flowing through the voltage source is measure
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Answer:

<em>1 Volt</em>

<em></em>

Explanation:

The resistors have resistance R = 1 kΩ each = 1 x 10^3 Ω each

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\frac{1}{R_{T}} = \frac{1}{R_{1}}  + \frac{1}{R_{2}}

\frac{1}{R_{T}} = 1/1 kΩ  + 1/1 kΩ

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From V = IR

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substituting, we have

V = 500 x 2 x 10^-3 =<em> 1 V   This is the voltage across each resistor</em>

7 0
3 years ago
The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb
balu736 [363]

Answer:

16.8 lb is the force on the brake pad of one wheel.

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Area of the brakes = A_2=1.8 inches^2

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

\frac{F_1}{A_2}=\frac{F_2}{A_2}

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A rotating wheel with diameter 0.800 m is speeding up with constant angular acceleration. The speed of a point on the rim of the
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Answer:

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radius of the wheel, r = 0.4 m

initial velocity of the wheel, u = 3 m/s

final velocity of the wheel, v = 6 m/s

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v = ωr

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θ is the angular rotation (rad)

θ = Number of revolutions x 2π rad/rev

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Therefore, the angular acceleration of the wheel is 3.357rad/s²

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