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arlik [135]
3 years ago
13

A proton moves from left to right across the screen at speed v0. Then a uniform magnetic field of strength 0.80 T is directed in

to the screen. What is the direction of the force on the proton?

Physics
1 answer:
Anton [14]3 years ago
8 0

Answer: The force is  directed upward

Explanation: Considering the Lorentz force, given by:

F= qv×B

using the right hand rule and considering the direction of electron velocity and the magnetic field from the figure, the vectorial product gives a  force vector upwards .

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Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y
WITCHER [35]

Question:

A point charge of -2.14uC  is located in the center of a spherical cavity of radius 6.55cm  inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm  from the center of the cavity.  

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 3.65\times 10^5N/C

Explanation:

A point charge ,q = -2.14\times 10^{-6} C is located in the center of a spherical cavity of radius , r =6.55\times 10^{-2}  m inside an insulating spherical charged solid.  

The charge density in the solid , d = 7.35 \times 10^{-4}C/m^3.

Distance from the center of the cavity,R =9.5\times 10^{-2 }m

Volume of shell of charge= V  =(\frac{4\pi}{3})[ R^3 - r^3 ]

Charge on the shell ,Q = V \times d'

Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d

Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}

Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}

Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}

Q =1.7745 \times 10^{-6 }C

Electric field at 9.5\times 10^{-2}m due to shellE1  = \frac{k Q}{R^2}

E1 =  \frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}

E1 =1.769\times 10^6 N/C

Electric field at  9.5\times 10^{-2} due to 'q' at center E2 = \frac{kq}{R^2}

E2 =\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}

E2 =2.134\times 10^6 N/C

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

=[  2.134  - 1.769 ]\times 10^6

= 3.65\times 10^5 N/C

8 0
3 years ago
Some expressways are curved, banked and designed to maximize ___________ at higher speeds.
Liono4ka [1.6K]

Answer:

Safety

Explanation:

Expressways are banked to resist centifugal action

6 0
3 years ago
In as small room fan of rating 50 watt is used for 10hrs 2bulb of rating 10v are used for 8hrs daily.calculate monthly. electric
worty [1.4K]

Answer:

The electric bill for June is Rs198000

Explanation:

Convert volt to watt, but in order to do so I need to know the amps and since it is not provided I converted if the amps was 1.

I multiple 50 with 10 then  with 30 so I know how much watt the fan takes at June.

Since there are 2 light bulb I multiple 10 with 2 than with 8 than with 30.

15000 watts for the fan,

4800 watts for light bulb,

add them and then times it by 10.

Rs198000

4 0
3 years ago
A 55.0-g piece of copper wire is heated, and the temperature of the wire changes from 19.0°C to 86.0°C. The amount of heat absor
Alinara [238K]
The specific heat of a metal or any element or compound can be determined using the formula Cp = delta H / delta T / mass. delta pertains to change. That is change in enthalpy and change in temperature. From the given data, Cp is equal to 343 cal per (86-19) c per 55 grams. This is equal to 0.093 cal / g deg. Celsius
4 0
3 years ago
A mountaintop is a height y above the level ground. A woman measures the angle of elevation of the mountaintop to be θ when she
valentina_108 [34]
Let t = Theta and p = Phi
Tan t = y/x    Then x =y/Tant.
Tant = y/(x-d)  x-d = y/Tanp
y/Tant  - d = y/Tanp
y -d*Tanr = y*Tant/Tanp
y-y*Tant/Tanp  =  d*Tanr
y(1 - Tanr/Tanp = d*Tant

y = d*Tant/(1-Tant/Tanp)
7 0
4 years ago
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