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3 years ago

Which fact is NOT true about gravity?

Physics

2 answers:

Arturiano [62]3 years ago

5 0

The correct answer is B.

Gre4nikov [31]3 years ago

5 0

B i think thats what i put

You might be interested in

For sprinters running at 12 m/s around a curved track of radius 26 m, how much greater (as a percentage) is the average total fo

Snezhnost [94]

Answer:

114.86%

Explanation:

In both cases, there is a vertical force equal to the sprinter's weight:

Fy = mg

When running in a circle, there is an additional centripetal force:

Fx = mv²/r

The net force is found with Pythagorean theorem:

F² = Fx² + Fy²

F² = (mv²/r)² + (mg)²

F² = m² ((v²/r)² + g²)

F = m √((v²/r)² + g²)

Compared to just the vertical force:

F / Fy

m √((v²/r)² + g²) / mg

√((v²/r)² + g²) / g

Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:

√((12²/26)² + 9.8²) / 9.8

1.1486

The force is about 114.86% greater (round as needed).

5 0

3 years ago

The Golden Gate Bridge in San Francisco has a main span of length 1.28 km, one of the longest in the world. Imagine that a steel

Valentin [98]

Answer:

361.46 N

Explanation:

$\alpha$ = Coefficient of thermal expansion = $11\times 10^{-6}\ /^{\circ}C$

Y = Young's modulus for steel = $20\times 10^{10}\ Pa$

A = Area = $3.1\times 10^{-6}\ m^2$

$L_0$ = Original length = 1.28 km

$\Delta T$ = Change in temperature = $45-(-10)$

Length contraction is given by

$\Delta L=\alpha L_0\Delta T$

Also,

$\Delta L=\dfrac{L_0T}{YA}$

$\alpha L_0\Delta T=\dfrac{L_0T}{YA}\\\Rightarrow T=\alpha \Delta TYA\\\Rightarrow T=11\times 10^{-6}\times (43-(-10))\times 20\times 10^{10} \times 3.1\times 10^{-6}\\\Rightarrow T=361.46\ N$

The tension in the wire is 361.46 N

4 0

3 years ago

In your own words describe what free fall is

nekit [7.7K]

Answer:

u are falling

Explanation:

In Newtonian physics, free fall is any motion of a body where gravity is the only force acting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it.

6 0

3 years ago

Read 2 more answers

A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3

Zinaida [17]

Answer:

a. $\bar{d}=4.34 cm$

b. $\sigma=0.023 cm$

c. $\rho=(0.0089\pm 0.00058) kg/cm^{3}$

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

$\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}$

$x_{i}$ is values of each diameter
N is the total number of values. N=6

$\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}$

$\bar{d}=4.34 cm$

b) The standard deviation equations is:

$\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}$

If we put all this values in that equation we will get:

$\sigma=0.023 cm$

Then the mean diameter will be:

$\bar{d}=(4.34\pm 0.023)cm$

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: $V=\pi R^{2}h$

Then:

$\rho=\frac{m}{\pi R^{2}h}$

Using the values that we have, we can calculate the value of density:

$\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}$

We need to use propagation of error to find the error of the density.

$\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}$

δm is the error of the mass value.
δd is the error of the diameter value.
δh is the error of the length value.

Let's find each partial derivative:

1. $\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089$

2. $\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004$

3. $\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071$

Therefore:

$\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}$

$\delta\rho=0.00058$

So the density is:

$\rho=(0.0089\pm 0.00058) kg/cm^{3}$

I hope it helps you!

3 0

3 years ago

How does sound intensity differ from loudness?

Alchen [17]

Loudness refers to how loud or soft a sound seems to a listener. The loudness of sound is determined, in turn, by the intensity, or amount of energy, in sound waves. The unit of intensity is the decibel (dB). As decibel levels get higher, sound waves have greater intensity and sounds are louder.

3 0

3 years ago