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strojnjashka [21]
3 years ago
7

as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p

article has a speed of 230 m/s and moves perpendicular to the magnetic field. Just as the particle enters the magnetic field, an electric field is turned on. What must be the magnitude of the electric field such that the net force on the particle is twice the magnetic force
Physics
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta

where,

E = Magnitude of Electric Field = ?

v = speed of charge = 230 m/s

B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

Therefore,

E = (3)(230\ m/s)(0.61\ T)Sin90^o

<u>E = 420.9 N/C</u>

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8 0
4 years ago
500cm3 of water is to be jeated from room temperature 28oc to 100oc in order to prepare ahot cup pf coffee. What os the minimum
klasskru [66]

Answer:

The minimum heat required is 171626 joules.  

Explanation:

By First Law of Thermodynamics, the amount of heat (Q), in joules, is equal to the change in internal energy of water. The minimum heat required is the sensible heat necessary to heat up the water. That is:

Q = \rho\cdot V \cdot c\cdot \Delta T (1)

Where:

\rho - Density of water, in kilograms per cubic meter.

V - Volume of water, in cubic meters.

c - Specific heat of water, in joules per kilogram-degree Celsius.

\Delta T - Change in temperature, in degrees Celsius.

If we know that \rho = 1000\,\frac{kg}{m^{3}}, V = 5\times 10^{-4}\,m^{3}, c = 4186\,\frac{J}{kg\cdot ^{\circ}C} and \Delta T = 82\,^{\circ}C, then the minimum heat required is:

Q = 171626\,J

The minimum heat required is 171626 joules.  

7 0
3 years ago
The radius of he Earth orbit around the sun (assumed circular) is 1.50 X 10^8km, with T=365d. What is the radial acceleration of
Karolina [17]

Answer:

ar = 5.86*10^-3 m/s^2

Explanation:

In order to calculate the radial acceleration of the Earth, you first take into account the linear speed of the Earth in its orbit.

You use the following formula:

v=\sqrt{\frac{GM_s}{r}}         (1)

G: Cavendish's constant = 6.67*10^-11 m^3 kg^-1 s^-2

Ms: Sun's mass = 1.98*10^30 kg

r: distance between Sun ad Earth = 1.50*10^8 km = 1.50*10^11 m

Furthermore, you take into account that the radial acceleration is given by:

a_r=\frac{v^2}{r}             (2)

You replace the equation (1) into the equation (2) and replace the values of all parameters:

a_r=\frac{1}{r}\frac{GM_s}{r}=\frac{GM_s}{r^2}\\\\a_r=\frac{(6.67*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}{(1.50*10^{11}m)^2}\\\\a_r=5.86*10^{-3}\frac{m}{s^2}

The radial acceleration of the Earth, towards the sun is 5.86*10^-3 m/s^2

7 0
3 years ago
Future space stations could create an artificial gravity by rotating. Consider a cylindrical space station that rotates with a p
aleksandrvk [35]

Answer:

P = 2 pi R / v    period of space station

F / m = v^2 / R    centripetal force per unit of mass

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Also, F / m = 9.8 m/s^2   earth's gravitational attraction

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8 0
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A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is
Tasya [4]

Answer:

C) T

Explanation:

M = Mass attached to an ideal spring

A = Amplitude of the motion

T = Time period of oscillation

k = Spring constant of the spring

A = Amplitude of the motion

Time period of oscillation of the mass attached to the spring is given as

T = 2\pi \sqrt{\frac{M}{k} }

So we see that the time period does not depend on the amplitude. hence the period of oscillation remains the same.

8 0
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