Which force is stronger electrical or gravitational?

A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0 °C. During the summer, the temperature in the storage a

VMariaS [17]

Answer:

5.63 atm

Explanation:

P1 = 5 atm

P2 = ?

T1 = 5 °C = 5 + 273 = 278 K

T2 = 40 °C = 273 + 40 = 313 K

Let the pressure is P2.

Use

Pressure is directly proportional to the absolute temperature as the volume remains constant.

$\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$

$\frac{5}{278}=\frac{P_{2}}{313}$

P2 = 5.63 atm

Thus, the pressure is 5.63 atm.

3 0

3 years ago

A current of 5 A exists in a copper wire of length 50 m and diameter of 2.5 mm when applying a

taurus [48]

Answer:

a) 1273.23 A/m^2

b) 7.19*10^-5 m/s

c) 236881.7 Ohms

Explanation:

(a) To find the current density you use the following formula:

$J=\frac{I}{A}=\frac{I}{\pi r^2}$

I: current in the wire

A: cross area of the wire

r: radius of the wire

$J=\frac{5A}{\pi(1.25*10^{-3}m)^2}=1273.23\frac{A}{m^2}$

(b) The electron drift speed is given by:

$v_d=\frac{I}{nqA}=\frac{I}{nq\pi r^2}$

n: number of conduction electrons per m^3

q: charge of the electron = 1.6*10^-19C

The number of free electrons is calculated by using:

$n=\frac{\rho N_A}{M}\\\\n=\frac{(9*10^{3}kg/m^3)(6.22*10^{23})}{63.54*10^{-3}}=8.5*10^{28}$

Next, you replace the values of the parameters in the equation for vd:

$v_d=\frac{5A}{(8.85*10^{28}m^{-3})(1.6*10^{-19}C)\pi (1.25*10^{-3}m)^2}\\\\v_d=7.19*10^{-5}\frac{m}{s}$

(c) The conductivity is given by:

$\sigma=\frac{L}{RA}$

You first calculate R:

$R=VI=(0.86V)(5A)=4.3\Omega$

Next, replace for sigma:

$\sigma=\frac{50m}{(4.3\Omega)(\pi (1.25*10^{-3}m)^2)}=236881.7\Omega^{-1}m^{-1}$

5 0

3 years ago

Pulling the string on a bow back with a force of 29.0 lb, an archer prepares to shoot an arrow.If the archer pulls in the center

deff fn [24]

Answer:

The tension in the string 33.08 lb.

Explanation:

Step 1

Consider the forces acting where the arrow is held. The string has a tension T on both sides of the arrow inclined 64 degrees to the horizontal.

The sum of all the forces in the x-directions is:

∑F_x = T cos(Ф₁) + T cos(Ф₂)

Since Ф₁ = Ф₂ = 64°, the sum of all the forces in the x-direction will be

∑F_x = 2T cos(64°)

The sum of all the forces in the y-directions is:

∑F_y = T sin(Ф₁) + T sin(Ф₂)

Since Ф₁ = Ф₂ = 64°, the sum of all the forces in the y-direction will be

∑F_y = T sin(64°) - T sin(64°)

= 0

Step 2:

The tension in the string:

The net force acting at the midpoint is 29.0 lb

Thus,

F_net = ∑F_x

F_net = 2T cos(64°)

29.0 = 2T cos(64°)

⇒ T = 29.0 / (2 cos(64°))

T = 33.08 lb

Therefore, the tension in the string 33.08 lb.

6 0

3 years ago

Standing waves on a 1.0m long string that is fixed at both ends are seen at successive frequencies of 24 hz and 36hz. what is th

kicyunya [14]

The multiples of the fundamental frequency are the standing wave frequencies for a string.

Since 24 Hz and 36 Hz are consecutive frequencies, the fundamental frequency must be 36 −12 = 12 Hz. Meaning that the standing wave with frequency 36 Hz is the third harmonic (i.e., m = 3),

The wave speed fulfills the equation f1 = v / 2L where f1 is the fundamental frequency and L is the length of the string

So we can say that:

v = 2Lf1

= 2(1.0)(12)

= 24 m/s

3 0

3 years ago

3. Gravitational potential energy depends on the ____ and _____ of the object

Licemer1 [7]

Answer is height and mass

4 0

3 years ago