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2 years ago

Which of the objects below has the greatest acceleration? *

Physics

1 answer:

lesya692 [45]2 years ago

8 0

$\\ \bull\sf\dashrightarrow F=ma$

$\\ \bull\sf\dashrightarrow a=\dfrac{F}{m}$

Now

$\\ \bull\sf\dashrightarrow a\propto\dfrac{1}{m}$

$\\ \bull\sf\dashrightarrow a\propto F$

Lower mass=Higher acceleration
Lower Force=Lower Acceleration

Option B has lowest mass and highest force hence its correct

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What is needed to create a magnetic field?

lara [203]

Explanation:

A magnetic field can be created by running electricity through a wire. All magnetic fields are created by moving charged particles. Even the magnet on your fridge is magnetic because it contains electrons that are constantly moving around inside

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3 years ago

12. A frain moves from rest to a speed of 25 m/s in 30.0 seconds. What is its acceleration?

ziro4ka [17]

initial velocity=u=0m/s
Final velocity=v=25m/s
Time=t=30s

$\\ \tt\longmapsto Acceleration=\dfrac{v-u}{t}$

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$\\ \tt\longmapsto Acceleration=\dfrac{25}{30}$

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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:

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

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3 years ago

How do you find the capacitance in this?

Lostsunrise [7]

Answer:

Explanation:

parallel capacitances add directly

Series capacitances add by reciprocal of sum of reciprocals.

Ceq = [ C ] + [1 / (1/C + 1/C)] + [1 / (1/C + 1/C + 1/C)]

Ceq = [ C ] + [C / 2] + [C / 3]

Ceq = [ 6C/6 ] + [3C / 6] + [2C / 6]

Ceq = 11C/6

3 0

2 years ago

The electric-power industry is interested in finding a way to store electric energy during times of low demand for use during pe

Stolb23 [73]

Answer: True

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As current passes through an inductor overtime it tends to store current in the form of magnetic field. Therefore the electric-power industry can store energy in large Inductors.

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3 years ago