Identify all the forces acting on a passenger while going through a loop.

Physics

1 answer:

shusha [124]4 years ago

8 0

Answer:

"Gravitational force and the acceleration force" are the forces that are acting on a passenger while going through a loop.

Explanation:

When the passengers reach the top of the loop, The force of the car's acceleration pushes all the passengers off the seat and towards the center of the loop. On the other hand the inertia will try to push the passengers back into the seat. The gravity as well as the acceleration will tend to push the passenger in opposite direction with approximately equal force, which will create a feeling of being mass-less or we can call it weightlessness.

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A string trimmer is a tool for cutting grass and weeds; it utilizes a length of nylon "string" that rotates about an axis perpen

Kaylis [27]

Answer:

r = 22.3 cm

Explanation:

given,

angular speed = 45 rev/s

tangential speed = 63 m/s

length of the rotating string

angular velocity =

ω = 2 π N

ω = 2 π x 45

ω = 282.74 rad/s

we know,

v = r ω

$r = \dfrac{v}{\omega}$

$r = \dfrac{63}{282.74}$

$r =0.223\ m$

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3 0

4 years ago

A proton enters a uniform magnetic field of strength 2 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’

sattari [20]

Answer:

Magnetic force, $F=9.6\times 10^{-17}\ N$

Explanation:

It is given that,

Magnetic field, B = 2 T

Velocity of the proton, v = 300 m/s

Charge on the proton, $q=1.6\times 10^{-19}\ C$

The magnetic field is oriented perpendicular to the proton’s velocity. The magnetic force on the charged particle is given by :

$F=qvB\ sin\theta$

The magnetic field is oriented perpendicular to the proton’s velocity, $\theta=90^{\circ}$

$F=1.6\times 10^{-19}\times 300\times 2$

$F=9.6\times 10^{-17}\ N$

So, the magnitude of the force that the proton experiences while it moves through the magnetic field is $9.6\times 10^{-17}\ N$ . Hence, this is the required solution.