All categories

3 years ago

Do subscripts and coefficient matter?

Physics

1 answer:

Serga [27]3 years ago

4 0

UHHHHHH yes...... well depends on what the question is asking but more times then not yes they matter.

You might be interested in

As a train accelerates uniformly, it passes successive 5-kilometer markers while traveling at velocities 10 m/s and 25 m/s. what

gregori [183]

<span>122.0 km/hr. First letâ€™s make sure all of our units are in the base meter form: i.e. convert 5km to 5000m. (We will convert back to km later). The first thing to do is look at the equation relating velocity, acceleration, and distance: Vf^2 = Vi^2 + 2*a*d, where Vf is final velocity, Vi is initial velocity, a is acceleration, and d is distance. 25^2 = 10^2 + 2*a*5000 =?> 625 = 100 +10000a => a= 0.0525m/s^2. Now that we have acceleration, we can use the same equation again with different numbers.: Vf^2 = Vi^2 + 2*a*d = 25^2 + 2*0. 0525m*5000 = 625 + 525 =1150 => Vf^2 = 1150 => 33.9m/s. Convert to km/hour: 33.9m/s * 1km/1000m *60s/1min * 60min/ 1 hr = 122.0 km/hr.</span>

8 0

4 years ago

How does light and heat energy from the sun reach earth?

defon

Explanation:

uuubbv. very ecrcvtyfyhc g you f gg and you are the one to give me the chance for a little bit of time and effort into this is 6.0 and if

6 0

3 years ago

A 8.0-g bullet with an initial velocity of 115.0 m/s lodges in a block of wood and comes to rest at a

Gnesinka [82]

A. When your fire a bullet into the air, it typically takes between 20 and 90 second for it to come down,depending on the angle it was fired at,its muzzle velocity and caliber.

3 0

3 years ago

Whenever a charged particle encounters a magnetic field, the particle tends to spiral around the magnetic field lines. as it doe

fomenos

The electromagnetic radiation produced when a charged particle moves spirally in a magnetic field is called <span>synchrotron radiation. This phenomenon occurs due to the fact the particle is accelerated radially (due to the presence of the magnetic field), and every charged particle when it is accelerated emits electromagnetic radiation.</span>

4 0

3 years ago

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

Nesterboy [21]

Answer:

Compression distance: $d \approx 0.102\,m$

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy ( $K$ ), in joules, at the expense of elastic potential energy ( $U$ ), in joules, and overcoming work losses due to friction ( $W_{l}$ ), in joules:

$K + W_{l} = U$ (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$ (2)

Where:

$m$ - Mass of the block, in kilograms.

$v$ - Final velocity of the block, in meters per second.

$\mu$ - KInetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Width of the rough patch, in meters.

$k$ - Spring constant, in newtons per meter.

$d$ - Compression distance, in meters.

If we know that $m = 1.2\,kg$ , $v = 2.3\,\frac{m}{s}$ , $\mu = 0.44$ , $g = 9.807\,\frac{m}{s^{2}}$ , $s = 0.05\,m$ and $k = 730\,\frac{N}{m}$ , then the compression distance of the spring is:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$

$m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}$

$d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }$

$d \approx 0.102\,m$

4 0

3 years ago