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2 years ago

6

A baseball pitcher throws at 95 mile/hr at a hitter 60 feet 6 inches away. How long does it take for the baseball to reach the h

itter?

1 answer:

Dima020 [189]2 years ago

4 0

Answer:

0.43 s

Explanation:

speed (s) = 95 mi/h = 95 x 1600/3600 = 42.5 m/s (converting to m/s)

distance (d) = 60 ft 6 inches = (60 x 0.3048) + (6 x 0.025) = 18.44 m (converting to meters)

speed = distance / time

time = distance / speed = 18.44 / 42.5 = 0.43 s

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Answer:

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2 years ago

The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

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2 years ago

A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coe

Levart [38]

Answer:

Coefficient of static friction will be equal to 0.642

Explanation:

We have given acceleration $a=6.3m/sec^2$

Acceleration due to gravity $g=9.8m/sec^2$

We have to find the coefficient of static friction between truck and a cabinet will

We know that acceleration is equal to $a=\mu g$ , here $\mu$ is coefficient of static friction and g is acceleration due to gravity

So $\mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642$

So coefficient of static friction will be equal to 0.642

3 0

2 years ago

An emf of 28.0 mV is induced in a 501 turn coil when the current is changing at a rate of 12.0 A/s. What is the magnetic flux th

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Answer:

Φ = 5.589×10⁻⁵ Wb

Explanation:

The inductance of a coil is given as

L = e/(di/dt) ..................... Equation 1

Where L = inductance of the coil, e = induced e.m.f, di/dt = rate of change of current in the coil.

Also,

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Where N = number of turns, Φ = magnetic field, i = current.

equating equation 1 and equation 2

e/(di/dt) = NΦ/i

making Φ the subject of the equation,

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Given: e = 28.0 mV = 0.028 V, N = 501 turns, di/dt = 12.0 A/s, i = 4.00 a

Substitute into equation 3,

Φ = (0.028×4)/(12×501)

Φ = 0.112/2004

Φ = 5.589×10⁻⁵ Weber

Φ = 5.589×10⁻⁵ Wb

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2 years ago

A skydiver is falling at constant velocity. If a force of 600 N is pulling down on the skydiver, how much force must be acting u

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2 years ago

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