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3 years ago

1.) if you are sitting still are you accelerating? How do you know?

Physics

1 answer:

castortr0y [4]3 years ago

5 0

yes. gravity is working on you and everything on you.

also earth is rotating

no not in respect ofimmediate surroundings

object could move at constant velocity = no change in either magnitude or direction

change of position "proves" this

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A hammer strikes one end of a thick iron rail of length 8.80 m. A microphone located at the opposite end of the rail detects two

stepladder [879]

Answer:

ΔT = 0.02412 s

Explanation:

We will simply calculate the time for both the waves to travel through rail distance.

FOR THE TRAVELING THROUGH RAIL:

$T_{rail} = \frac{Distance}{Speed\ of\ Sound\ in\ Rail}\\\\T_{rail} = \frac{8.8\ m}{5950\ m/s}\\\\T_{rail} = 0.00148\ s$

FOR THE WAVE TRAVELING THROUGH AIR:

$T_{air} = \frac{Distance}{Speed\ of\ Sound\ in\ Air}\\\\T_{air} = \frac{8.8\ m}{343\ m/s}\\\\T_{air} = 0.0256\ s$

The separation in time between two pulses can now be given as follows:

$\Delta T = T_{air}-T_{rail} \\\Delta T = 0.0256\ s - 0.00148\ s\\$

3 0

3 years ago

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul

GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that for spherical shell and pure rolling conditions

$v = R \omega$

$I = \frac{2}{3}mR^2$

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh$

$\frac{5}{6}mv^2 = mgh$

$h = \frac{5v^2}{6g}$

$h = \frac{5(8^2)}{6(9.81)} = 5.44 m$

Part b)

If ball is not rolling and just sliding over the hill then in that case

$\frac{1}{2}mv^2 = mgh$

$h = \frac{v^2}{2g}$

$h = \frac{8^2}{2(9.81)} = 3.16 m$

3 0

3 years ago

If a single circular loop of wire carries a current of 61 A and produces a magnetic field at its center with a magnitude of 1.70

Schach [20]

Answer:

The radius is $R = 0.22 5 \ m$

Explanation:

From the question we are told that

The current is $I = 61 \ A$

The magnetic field is $B = 1.70 *10^{-4} \ T$

Generally the magnetic field produced by a current carrying conductor is mathematically represented as

$B = \frac{\mu_o * I}{2 * R }$

=> $R = \frac{\mu_o * I }{ 2 * B }$

Here $\mu_o$ is the permeability of free space with value $\mu_o = 4\pi * 10^{-7} N/A^2$

=> $R = \frac{ 4\pi * 10^{-7} * 61 }{ 2 * 1.70 *10^{-4} }$

=> $R = 0.22 5 \ m$

8 0

3 years ago

By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect

sergejj [24]

Answer:

$Weight\ loss=1.6321N$

Explanation:

From the question we are told that:

Weight $W=85.9kg$

Altitude $h= 6.33 km$

Let

Radius of Earth $r=6380km$

Gravity $g=9.8m/s^2$

Generally the equation for Gravity at altitude is mathematically given by

$g_s=9.8(\frac{6380}{6380+6.33})^2$

$g_s=9.781m/s^2$

Therefore

Weight at sea level

$W_s=9.8*85.9$

$W_s=841.82N$

Weight at 6.33 altitude

$W_a=9.781*85.9$

$W_a=840.2N$

Therefore

$Weight loss=W_s-W_b$

$Weight loss=841.82-840.2$

$Weight loss=1.6321N$

3 0

3 years ago

Rocks are classified as sedimentary igneous or metamorphic primarily based upon differences in their

yanalaym [24]

The differences in their formations. Igneous is formed by cooled and crystallized rock, sedimentary is formed by sediment compaction and cementation, and metamorphic is formed by intense heat and pressure.

3 0

2 years ago

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