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2 years ago

10

A 5.00 L air sample at a temperature of -50 °C has a pressure of 107 kPa. What will be the new pressure if the temperature is ra

ised to 102 °C and the volume expands to 7.00 L?

1 answer:

Serjik [45]2 years ago

4 0

Its simple use formuila ,
PV=nRT
n,R is constant as the both have same moles.
so,
(p1v1)/T1 = (p2v2)/T2
so, 128.53338kpa

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Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

<h3>a.</h3>

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

so

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

<h3>b.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

<h3>c.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

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2 years ago

The magnetic field in the region between the poles of an electromagnet is uniform at any time, (1 point) but is increasing at th

olga nikolaevna [1]

Answer:

B)

The magnitude of induced emf in the conducting loop is 0.99 mV.

Explanation:

Rate of increase in magnetic field per unit time = 0.090 T/s

Area of the conducting loop = 110 cm^2 = 0.0110 m^2

Electromagnetic induction is the production of an emf or voltage in a coil of wire due to a changing magnetic field through the coil.

Induced e.m.f is given as:

EMF = (-N*change in magnetic field/time)*Area

EMF = rate of change of magnetic field per unit time * Area

EMF = 0.090 * 0.0110

EMF = 0.00099 V

EMF = 0.99 mV

5 0

2 years ago

A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge

Digiron [165]

Answer:

F = 2.49 x 10⁻⁹ N

Explanation:

The electrostatic force between two charged bodies is given by Colomb's Law:

$F = \frac{kq_1q_2}{r^2}\\$

where,

F = Electrostatic Force = ?

k = colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

q₂ = second charge = 1.4 C

r = distace between charges = 0.9 m

Therefore,

$F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.4\ C)}{(0.9\ m)^2}$

<u>F = 2.49 x 10⁻⁹ N</u>

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2 years ago

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Yuliya22 [10]

Instantaneous speed?

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2 years ago

Use the dimensional analysis and check the correctness of given equation:-<br> PV= nRT

Harman [31]

PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

$\\ \tt\bull\leadsto PV$

$\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]$

$\\ \tt\bull\leadsto [ML^5T^{-2}]$

RHS

$\\ \tt\bull\leadsto nRT$

$\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]$

$\\ \tt\bull\leadsto [ML^5T^{-2}]$

LHS=RHS

hence verified

6 0

2 years ago