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3 years ago

When there is no number in front of a chemical formula in a chemical equation, what number is understood?

1 answer:

5 0

The number in front is the number of molecules (or atoms) taking part in the (balanced) chemical reaction equation.

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A train slows down as it rounds a sharp horizontal turn, going from 88.0 km/h to 52.0 km/h in the 18.0 s that it takes to round

zmey [24]

Answer:

The acceleration at the moment the train speed reaches 52 kilometers per hour is approximately 1.826 meters per square second.

Explanation:

According to Rotational Physics, the total acceleration of the train rounding the horizontal turn is a combination of tangential ( $a_{t}$ ) and radial accelerations ( $a_{r}$ ), measured in meters per square second. The former one represents the change in the magnitude of the velocity, whereas the latter one represents the change in its direction. By definition of magnitude and Pythagorean Theorem we get that magnitude of total acceleration ( $a$ ), measured in meters per square second, is:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (Eq. 1)

Magnitudes of tangential and radial accelerations are determined by using the following formulas:

$a_{t} = \frac{v_{f}-v_{o}}{t}$ (Eq. 1)

$a_{r} = \frac{v_{f}^{2}}{R}$ (Eq. 2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds, measured in meters per second.

$t$ - Time, measured in seconds.

$R$ - Radius, measured in meters.

If we know that $v_{o} = 24.444\,\frac{m}{s}$ , $v_{f} = 14.444\,\frac{m}{s}$ , $t = 18\,s$ and $R = 120\,m$ , then the magnitude of the total acceleration when the train speed reaches 52 kilometers per hour is:

$a_{t} = \frac{14.444\,\frac{m}{s}-24.444\,\frac{m}{s} }{18\,s}$

$a_{t} = -0.556\,\frac{m}{s^{2}}$

$a_{r} = \frac{\left(14.444\,\frac{m}{s} \right)^{2}}{120\,m}$

$a_{r} = 1.739\,\frac{m}{s^{2}}$

$a = \sqrt{\left(-0.556\,\frac{m}{s^{2}} \right)^{2}+\left(1.739\,\frac{m}{s^{2}} \right)^{2}}$

$a \approx 1.826\,\frac{m}{s^{2}}$

The acceleration at the moment the train speed reaches 52 kilometers per hour is approximately 1.826 meters per square second.

8 0

3 years ago

Dr. gavin decides that instead of conducting a 2 ´ 4 independent-groups factorial design, he is going to conduct a 2 ´ 4 within-

vampirchik [111]

If Dr. Gavin decides that instead of conducting a 2 ´ 4 independent-groups factorial design, and he is going to conduct a 2 ´ 4 within-subjects factorial design, then the things that will change are the various independent groups which are involved.

8 0

3 years ago

What happens to the pressure in a tire if air is slowly leaking out of the tire? explain your answer?

creativ13 [48]

It's pretty simple. When air is leaking out of a tire, like a tiny hole or something, the pressure in the tire decreases, because without air in the tire, there is no pressure.

3 0

3 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

3 years ago

What is angle of dip

DENIUS [597]

Angle of Dip is the angle in the vertical plane aligned with magnetic north (the magnetic meridian) between the local magnetic field and the horizontal.

8 0

3 years ago