What is the formula for a uniform electric field between two parallel plates?

In a football game, running back is at the 10 yard line and running up the field towards the 50 yard line, and runs for 3 second

lakkis [162]

If he keeps that pace he will be at the 34 yard line

7 0

3 years ago

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

3 years ago

Which statement about the horizontal distance covered by a projectile launched at an angle less than 90° is true?

mylen [45]

<h2>Answer: A</h2>

Explanation because the distance covered is equal time intervals is the same or equal.

5 0

3 years ago

Read 2 more answers

A bucket filled with water has a mass of 54 kg and is hanging from a rope that is wound around a 0.050 m radius stationary cylin

Lubov Fominskaja [6]

Answer:

The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

Explanation:

Given that,

Mass of bucket = 54 kg

Radius = 0.050 m

We need to calculate the magnitude of the torque the bucket produces around the center of the cylinder

Using formula of torque

$\tau=F\times r$

$\tau=mg\times r$

Where, m = mass

g = acceleration due to gravity

r = radius

Put the value into the formula

$\tau=54\times9.8\times0.050$

$\tau=26.46\ N-m$

Hence, The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

3 0

3 years ago

A car move at an initial velocity of 240m and reach at the final velocity of 540m in 8hours. calculate its acceleration.

disa [49]

Answer:

a = 0.01m/s²

Explanation:

V_f = V_0+a*t

V_f = Velocity final

V_0 = Velocity initial

a = acceleration

t = time

a = (V_f-V_0)/t

a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

6 0

2 years ago