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Alina [70]
2 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]2 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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EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

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where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

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8 0
2 years ago
g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the
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Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

R = \frac{u^{2}Sin2\theta }{g}

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

R = \frac{u^{2}Sin60 }{9.8}

R = 0.088u^2

If the angle of projection is 45 degree.

R = \frac{u^{2}Sin90 }{g}

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igor_vitrenko [27]

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3 0
1 year ago
A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
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Answer:

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Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

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E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

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4 0
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If the 20% of the total energy is used

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