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Inga [223]
2 years ago
8

The last page of a book is numbered 764. The book is 3.0 cm thick, not including its covers. What is the average thickness (in c

entimeters) of a page in the book, rounded to the proper number of significant figures? A) 0.0039 cm B) 0.072 cm C) 0.0079 cm D) 0.00393 cm E) 0.00785 cm
Physics
1 answer:
Inessa [10]2 years ago
7 0

Answer:

Thickness of each page will be 0.00785 cm

So option (E) will be correct answer

Explanation:

We have given that last page on the book is numbered as 764

So total number of pages in the book =\frac{764}{2}=382pages ( as each page is numbered in both side )

Total thickness of the book = 3 cm

We have to found the thickness of the each page

So thickness of each page will be =\frac{3}{382}=0.00785cm

So option (E) is correct option

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s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

V_L=V_oe^{-\frac{Rt}{L}}

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s

hence, after 1.6s the inductor will have a potential difference of 24V

3 0
2 years ago
A large ebony wood log, totally submerged, is rapidly floating down a flooded river. If the mass of the log is 165 kg, what is t
Sedbober [7]

Answer:

F = 1618.65[N]

Explanation:

To solve this problem we use the following equation that relates the mass, density and volume of the body to the floating force.

We know that the density of wood is equal to 750 [kg/m^3]

density = m / V

where:

m = mass = 165[kg]

V = volume [m^3]

V = m / density

V = 165 / 750

V = 0.22 [m^3]

The floating force is equal to:

F = density * g * V

F = 750*9.81*0.22

F = 1618.65[N]

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2 years ago
José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo
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r = radius of the circle of the ride = 3.00 meters

v = linear speed of the person during the ride = 17.0 m/s

m = mass of the person in angular motion in the ride

L = angular momentum of the person in the ride = 3570 kg m²/s

Angular momentum is given as

L = m v r

inserting the values

3570 kg m²/s = m (17 m/s) (3.00 m)

m = 3570 kg m²/s/(51 m²/s)

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hence the mass comes out to be 7 kg


8 0
2 years ago
An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st
RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

s=ut+\frac{1}{2}at^2

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2

Now we can apply Newton's second law to find the net force on the object:

F=ma=(24 kg)(4 m/s^2)=96 N

This net force is the resultant of the applied force forward (F_a) and the component of the weight acting backward (mg sin \theta), so we can find what is the applied force:

F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N

where

m = 24 kg is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

W=F_a s = (237.6)(18)=4276 J

where s = 18 m is the displacement.

Therefore the average power needed is:

P=\frac{W}{t}=\frac{4276}{3}=1425 W

b)

The instantaneous power at any point of the motion is given by

P=F_av

where

F_a is the force applied

v is the velocity of the object

We already calculated the applied force:

F_a=237.5 N

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

v=u+at=0+(4)(3.0)=12.0 m/s

And therefore, the instantaeous power at 3.0 sec is:

P=Fv=(237.5)(12)=2850 W

Learn more about power:

brainly.com/question/7956557

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