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Dennis_Churaev [7]

3 years ago

14

A 144-g baseball is dropped from a tree 11.0m above the ground.A.With what speed would it hit the ground if air resistance could

be ignored?Express your answer to three significant figures and include the appropriate units.B.If it actually hits the ground with a speed of 7.50m/s , what is the magnitude of the average force of air resistance exerted on it?Express your answer to three significant figures and include the appropriate units.

1 answer:

nikitadnepr [17]3 years ago

5 0

Answer:

a. 14.68m/s

b. 1.043N

Explanation:

a.

Given

Mass = m = 144g = 0.144kg

Height = h = 11m

Using energy conservation,

½mv² = mgh where g = 9.8m/s² ----- divide both sides by m

½v² = gh ---- divide both sides by ½

v² = 2gh

v = √2gh

v = √(2 * 9.8 * 11)

v = 14.68332387438212m/s

v = 14.68m/s ----- Approximated

b.

v = 7.5m/s

By definition, “the average force of air resistance,”

When work done against air resistance is solved by multiplying height by the average force of resistance ------ by definition

i.e. W = -Fh

Minus is used because F and h points to different direction (when viewed as vectors).

This is proved by:

W = Fhcos(180) because h points toward the direction of motion (downward) while F points opposite.

W = Fh * -1

W = Fh

Using energy conservation,

mgh - Fh = ½mv² ---- Make Fh, the subject of formula

Fh = mgh - ½mv² ---- Divide through by h

F = mg - mv²/2h

F = 0.144 * 9.8 - 0.144 * 7.5²/(2 * 11)

F = 1.043N

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DanielleElmas [232]

Answer:

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Mark all the fermions a) Proton b) Electron c)Anti-top d) Gluon e) Tau Neutrino

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(a) Proton, (b) Electron (c) tau neutrino.

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(1) Quarks: up, down, top, bottom, strange, and charm.

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3 years ago

A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh

daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is $B=2.958\times10^{-5}T$

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

$\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \ \overrightarrow B_s\perp \overrightarrow B_w$

Angle of net magnetic field from axial direction is given by:

$tan\ \theta=\frac{B_w}{B_s}$ ,

Field due to solenoid:

$B_s=\mu_onI_s, \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m$

Field due to wire:

$B_w=\frac{\mu_oI_w}{2\pi r}$

Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

$B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T$

Hence, the magnetic field is $B=2.958\times10^{-5}T$

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3 years ago