1. Explain why when objects become charged it is elec-

if the root mean square speed of a gas particle is 200 m/s at a temperature of 400k, at what approximate temperature will urms w

jekas [21]

Answer:

The correct solution is "1230 K".

Explanation:

The given values are:

$(V_{rms})_1= 200 \ m/sec$

$(V_{rms})_2= 350 \ m/sec$

$T_1=400 \ K$

As we know,

⇒ $V_{rms} \propto \sqrt{T}$

or,

⇒ $\frac{(V_{rms})_1}{(V_{rms})_2} =\sqrt{\frac{T_1}{T_2} }$

On substituting the values, we get

⇒ $\frac{200}{350} =\sqrt{\frac{400}{T_2} }$

⇒ $T_2=1230 \ K$

5 0

3 years ago

A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field. What is

Alexeev081 [22]

Answer:

Explanation:

In the region of potential difference , protons will acquire velocity which can be found as follows

1/2 m v² = Ve , V is potential diff , v is velocity attained , m is mass of proton , e is charge on proton

v² = 2Ve / m

For circular path of proton in a magnetic field

mv² / r = Bev r is radius of circular path.

r = mv / Be

r² = m²v² / B²e²

= 2Ve/m x m²/ B²e²

= 2Vm / B²e

= 2 x 1000 x 1.67 x 10⁻²⁷ /( .04² x 1.6 x 10⁻¹⁹ )

= .013

r = .114 m

= 11.4 cm

8 0

3 years ago

There are two balloons of charges +3.37 x 10-6 C and –8.21 x 10-6 C. The distance between the two balloons is 2.00 m. Determine

andrew-mc [135]

F = 0.06N. Since the charges has different signs the force of atraction between them is 0.06N.

In order to solve this exercise we have to use Coulomb's Law equation which says that the magnitude of each of the electric forces with which two point charges at rest interact is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them. Given by the equation:

$F = k\frac{|q_{1}q_{2}|}{d^{2}}$ . Where k is the Coulomb's Constant $k = 9x10^{9}\frac{Nm^{2} }{C^{2} }$ , q1 and q2 are the charges value in Coulomb (C), and d is the distance between charges in meters (m).

There are two balloons of charges +3.37 x 10-6 C and –8.21 x 10-6 C. The distance between the two balloons is 2.00 m. Calculate the force between the two ballons.

$F = 9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{|(3.37x10^{-6}C)(-8.21x10^{-6}C)|}{(2.00m)^{2}}\\F = 9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{|-2.77x10^{-11}C^{2} )|}{4.00m^{2}}\\F = 9x10^{9}\frac{Nm^{2} }{C^{2} }(6.92x10^{-12})\frac{C^{2} }{m^{2}}\\F = 0.06N$

4 0

3 years ago

what will be the final temperature is 0.05kg water at 0 degree centigrade is added to 0.25kg pf water at 90 degree centigrade sp

scoray [572]

Q in = Q out

0.25 x 4200 x (90-t)=0.05 x 4200 x (t-0)

1050 (90-t)=210t

94500-1050t=210t

94500=1260t

t=75°

7 0

2 years ago

. If the student had a mass of 88.6 kg, what was his weight?

Andre45 [30]

<span>F = m*g = 88.6kg * 9.81 m/s^2 = 869 N (3 s.f.)

The equation above is Newton's 2nd Law of Motion where g is the acceleration due to gravity near the Earth's surface.</span><span>
</span>

8 0

3 years ago

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