8. A point of Charge +3.0 X 10^-7 Coulomb is placed 2.0 X 10^-2 from a second

When a certain gas under a pressure of 4.65 106 Pa at 21.0°C is allowed to expand to 3.00 times its original volume, its final p

Marina CMI [18]

Answer:

-72.0°C

Explanation:

PV = nRT

Since n, number of moles, is constant:

PV / T = PV / T

(4.65×10⁶ Pa) V / (21 + 273.15) K = (1.06×10⁶ Pa) (3V) / T

T = 201.16 K

T = -72.0°C

7 0

4 years ago

A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of

SCORPION-xisa [38]

Answer:

a).β=0.53 $x10^{-3}$ T

a).β=0.40 $x10^{-4}$ T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

$\beta =\frac{u_{o}*I*N}{2\pi*r}$

a).

$N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A} \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T$

b).

The distance is the radius add the cross section so:

$r_{1}=15cm+5cm\\r_{1}=20cm$

$r_{1} =20cm*\frac{1m}{100cm}=0.20m$

$\beta =\frac{u_{o}*I*N}{2\pi*r1}$

$\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T$

3 0

3 years ago

Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha

Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .

Let the electric field intensity due to $Q_2$ be + $E_2$ and that due to $Q_3$ be - $E_3$ since the charge is negative. Hence at the origin;

$+E_2-E_3=0..................(1)$

From equation (1) above, we obtain the following;

$E_2=E_3.................(2)$

From Coulomb's law the following relationship holds;

$+E_2=\frac{kQ_2}{r_2^2}\\$

$-E_3=\frac{kQ_3}{r_3^2}$

where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

$\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)$

k can cancel out from both side of equation (3), so that we finally obtain the following;

$\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)$

Given;

$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

$Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C$

6 0

3 years ago

Rita is a salon owner. She notices that her salon charged one of her clients, Linda, extra for a service that the clent did not

Aloiza [94]

Answer:

C hope it helps

call the client and inform her that she was incorrectly charged.

7 0

3 years ago

Studying this brochure from nasa, which explains more detail the instruments carried by the Juno spacecraft which scientific act

adoni [48]

Answer:

Juno scientific payload includes:

A gravity/radio science system (Gravity Science)
A six-wavelength microwave radiometer for atmospheric sounding and composition (MWR)
A vector magnetometer (MAG)
Plasma and energetic particle detectors (JADE and JEDI)
A radio/plasma wave experiment (Waves)
An ultraviolet imager/spectrometer (UVS)
An infrared imager/spectrometer (JIRAM)

Explanation:

Each mission of NASA has a specific set of instruments that it uses to perform scientific experiments on the desired heavenly body. In case of Juno, the mission for Jupiter has a series of instruments that would study domains of gravitational forces, magnetic effect, particle detection, radiation detection, UV/IR imaging, and plasma experiments.

3 0

3 years ago