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3 years ago

8

What do I do if someone is choking me with both of their hands, inside of the classroom and the teacher isn't around? How do I g

et out of their clutches.

1 answer:

guajiro [1.7K]3 years ago

6 0

Answer:

If you are laying down then nee them in the X.

If you are standing up you gan take them by surprise and give them a hook punch.

Explanation:

Self defenc is very important.

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Check my work please

katrin [286]

We can use the ideal gas equation which is expressed as PV = nRT. At a constant volume and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:

T1/P1 = T2/P2

P2 = T2 x P1 / T1

P2 = 273 x 340 / 713

<span>P2 = 130 kPa</span>

6 0

3 years ago

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0

3 years ago

Which three quantities can be used to calculate acceleration?

ikadub [295]

The correct option will be
D. Time, initial velocity and final velocity
The Formula can be written as,
Acceleration=Final velocity-Initial Velocity/Time

5 0

2 years ago

A wave has a wavelength of 5m and frequency of 5 Hz. What is the speed of the wave?

Anna71 [15]

If you multiply m (the unit for wavelength) with 1s (the unit for frequency), you will get m/s, the unit for speed. Now multiply! 25 m/s is your final answer!

3 0

2 years ago

An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0∘C to 25.0∘C at constant pressure. The gas do

ipn [44]

Answer:

(a) ΔU=747J

(b) γ=1.3

Explanation:

For (a) change in internal energy

According to first law of thermodynamics the change in internal energy is given as

ΔU=Q-W

Substitute the given values

ΔU=970J-223J

ΔU=747J

For(b) γ for the gas.

We can calculate γ by ratio of heat capacities of the gas

γ=Cp/Cv

Where Cp is the molar heat capacity at constant pressure

Cv is the molar heat capacity at constant volume

To calculate γ we first need to find Cp and Cv

So

For Cp

As we know

Q=nCpΔT

Cp=(Q/nΔT)

$C_{p}=\frac{970J}{1.75mol*(25^{o}C-10^{o}C )}\\C_{p}=37J/mol.K$

From relation of Cv and Cp we know that

Cp=Cv+R

Where R is gas constant equals to 8.314J/mol.K

So

$C_{v}=C_{p}-R\\C_{v}=37-8.314\\C_{v}=28.687J/mol.K\\$

So

γ=Cp/Cv

γ=[(37J/mol.K) / (28.687J/mol.K)]

γ=1.3

4 0

3 years ago