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2 years ago

Claudia throws a baseball to her dog. Which free-body diagram shows the

Physics

1 answer:

chubhunter [2.5K]2 years ago

6 0

Answer:

only the weight of the ball will act on the ball

Explanation: There is no contact force on the ball. Also there is no air resistance on the ball so the friction force on the ball due to air is not shown

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A ladybug starts at the center of a 16.0 in .-diameter turntable and crawls in a straight radial line to the edge. While this is

mafiozo [28]

The problem is about magnitude of the displacement vector of the lady bug and its directions. so the magnitude of the displacement of the lady bug is 16 in / 2 because it started from the center so the magnitude is 8 in. and the direction is the rotation of the turn table which is 60 degrees

3 0

1 year ago

Consider a capacitor made of two rectangular metal plates of length and width , with a very small gap between the plates. There

mezya [45]

Answer:

F= σ² L² /2ε₀

F = (L² ε₀/4π) ΔV² / r⁴

Explanation:

a) For this exercise we can use Coulomb's law

F = - k Q² / r²

where the negative sign indicates that the force is attractive and the value of the charge is equal to the two plates

Capacitance is defined by

C = Q / ΔV

Q = C ΔV

also the capacitance for a parallel plate capacitor is related to its shape

C = ε₀ A / r

we substitute

Q = ε₀ A ΔV / r

we substitute in the force equation

F = k (ε₀ A ΔV / r)² / r²

k = 1 / 4πε₀

F = ε₀ /4π L² ΔV² / r⁴4

F = L² ΔV² ε₀/ (4π r⁴)

F = (L² ε₀/4π) ΔV² / r⁴

b) Another way to solve the exercise is to use the relationship between the force and the electric field

F = q E

where we can calculate the field created by a plane using Gaussian law, where we use a cylinder with a base parallel to the plate as the Gaussian surface

Ф = ∫E .dA = $q_{int}$ / ε₀

the plate have two side

2E A = q_{int} / ε₀

E = σ / 2ε₀

σ = q_{int} / A

substituting in force

F = q σ / 2ε₀

the charge total on the other plate is

q = σ A

q = σ L²

F= σ² L² /2ε₀

4 0

2 years ago

An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$