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3 years ago

calculate the electric field acting on a proton so that proton remains stationary in a space a)2.0410^-2 b)1.0210^-7

Physics

1 answer:

Fofino [41]3 years ago

6 0

Answer:

No electric field will exist.

Explanation:

Given the following:

In other to for the proton to remain stationary in space, the electric field acting on the proton will be 0 or no electric field will exist. From the formulae ;

F = qE ; where q = charge of the particle

E = electric field intensity

The electric field force = F

At a stationary position in space, the electric field intensity = 0

F = qE = q * 0 = 0.

Also according to Heisinberg, measuring the momentum and position of a quantum object becomes more and more uncertain when measured simultaneously.

Here, one of position or momentum is know, measuring the electric field becomes very uncertain.

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Q 32.35: The isotope 235U decays by alpha emission with a half-life of 7.0 x 108 y. It also decays (rarely) by spontaneous fissi

Julli [10]

Answer:

rate of fission =5.89*10^3 1\Year

Explanation:

we know that

rate of fission is given as $rate of fission = \frac{0.69}{(T_{1/2})_{fission}} *\frac{ Mass* Avogardo\ number}{Molar\ mass}$

$(T_{1/2})_{fission} = 3*10^{17} y$

mass = 1.0 g

avogardo number = 6.02*10^23

molar mass of isotopes 235U =235

Putting all value to get rate of emission

rate of fission $= \frac{0.69}{3*10^{17}} *\frac{1.0*6.02*10^{23}}{235}$

rate of fission =5.89*10^3 1\Year

5 0

2 years ago

A 0.20 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same exp

KIM [24]

Answer:

The total mechanical energy does not change if the value of the mass is changed. That is, remain the same

Explanation:

The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:

$E=U=\frac{1}{2}kA^2$ (1)

k: spring constant

A: amplitude of the motion = 2.0cm

As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.

Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

Remain the same

8 0

2 years ago

A bullet blasts from the barrel of a gun upward in the vertical direction with an initial speed of 700 m/s . Find the maximum al

ryzh [129]

The maximum altitude that the bullet will reach is the point at which its velocity is zero. The equation that may be used in order to determine the altitude is,

D = ((Vi)2 – (Vf)2)/2g

Where Vi and Vf are the initial and final velocities, respectively. g is the deceleration due to gravity.

Substituting the known values,

D = ((700 m/s)2 – (0 m/s)2) / (2(-9.8 m/s2))

D = 25000 m

Thus, the maximum height is 25000 m.

For the time needed to reach it, we use the equation,

Vf = Vi – (g)(t)

Substituting,

0 = 700 m/s – (-9.8 m/s2)(t)

The value of t is equal to 71.43 s.

6 0

2 years ago

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$

⇒ $Fa*sin(50)=Fb*sin(20)$

⇒ $Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950

⇒ $Fa*cos(20)+Fb*cos(50)=950$

⇒ $2.24*Fb*cos(20)+Fb(50)=950$

⇒ $Fb*(2.24*cos(20)+cos(50))=950$

⇒ $Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒ $Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒ $Fa=2.24*Fb$

$=2.24*345.5$

$=773.93$

Therefore approximately, Fa=774 N and Fb=346 N

5 0

2 years ago

For what value m of the clockwise couple will the horizontal component ax of the pin reaction at a be zero? if a couple of that

sergejj [24]

Thank you for posting your question here at brainly. Below is the answer:

sum of Mc = 0 = -Ay(4.2 + 3cos(59)) + (275)(2.1 + 3cos(59)) + M
- Ay = (M + (275*(2.1 + 3cos(59)))/(4.2 + 3cos(59)) 

sum of Ma = 0 = (-275)(2.1) - Cy(4.2 + 3cos(59)) + M 
- Cy = (M - (275*2.1))/(4.2 + 3cos(59)) 

Ay + Cy = 275 = ((M+1002.41)+(M-577.5))/(5.745) 
= (2M + 424.91)/(5.745) 

M = ((275*5.745) - 424.91)/2 
= 577.483 which rounds off to 577 

Is it maybe supposed to be Ay - Cy = 275

8 0

2 years ago

calculate the electric field acting on a proton so that proton remains stationary in a space a)2.04*10^-2 b)1.02*10^-7

calculate the electric field acting on a proton so that proton remains stationary in a space a)2.0410^-2 b)1.0210^-7