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2 years ago

8

A book is pushed with an initial horizontal velocity of 5.0 meters per second off the top of a 1.19 meter high desk. How far awa

y does the book land?

1 answer:

kipiarov [429]2 years ago

8 0

Answer:

2.45 m

Explanation:

First of all, we have to calculate the time of flight of the book, by using the equation for the vertical motion:

$h=\frac{1}{2}gt^2$

where

h = 1.19 m is the vertical distance covered by the book

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

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.19)}{9.8}}=0.49 s$

Now we can find the horizontal distance covered by the book, which is given by

$d=v_x t$

where

$v_x = 5.0 m/s$ is the horizontal velocity

t = 0.49 s is the time of flight

Substituting,

$d=(5.0)(0.49)=2.45 m$

So the book lands 2.45 m away.

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Which of the following statements cannot be supported by Kepler's laws of planetary motion?

gladu [14]

Answer:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

Explanation:

Kepler's three laws are:

1) The orbits of the planets around the Sun are ellipses, with the Sun at one of the focii

2) A line connecting the Sun with each planet sweeps out equal areas in equal time intervals

3) The cube of the semi-major axis of the orbit of one planet is proportional to the square of its orbital period

There 3 laws help explaining the following statements:

- <em>A planet's distance from the sun will not be the same in six months. --> </em>using the 1st law. In fact, since the orbit is an ellipse (and not a circle), and the Sun is at one of the focii, the distance of the planet from the Sun keeps changing during the year.

-<em> A planet's speed as it moves around the sun will not be the same in six months. -</em>-> using the 2nd law. In fact, since the line connecting the Sun to the planet must cover equal areas in the same time interval, it follows that the speed of the planet cannot be constant during the year (it will be faster when closer to the sun and slower when far from the sun).

- <em>The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets. </em>--> using the 3rd law. In fact, the ratio $\frac{a^3}{T^2}$ (where a is the semi-major axis of the orbit and T the orbital period) is constant and it is the same for every planet orbiting the sun, so by knowing the data of Neptune and the orbital period of Saturn, it is possible to calculate Saturn's average distance.

Instead, the following statement:

<em>The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.</em>

Is not supported by any Kepler's law.

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2 years ago

An activity that is relatively short in time (< 10 seconds) and has few repetitions predominately uses the _____________ ener

tensa zangetsu [6.8K]

An activity that is relatively short in time <10 seconds and has few repetitions predominantly uses the ATP/PC energy system. The cellular respiration procedure that changes food energy into ATP which is a form of energy is largely reliant on oxygen obtainability. During exercise the source and request of oxygen obtainable to muscle is unnatural by period and strength and by the individual’s cardiorespiratory suitability level.
Steps of the ATP-PC system:

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2. Phosphocreatine is then broken down by the enzyme creatine kinase into creatine and phosphate.

3. The energy free in the breakdown of PC permits ADP and Pi to rejoin creating more ATP. This newly made ATP can now be broken down to issue energy to fuel activity.

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2 years ago

Based on the article “Will the real atomic model please stand up?,” why did J.J. Thomson experiment with cathode ray tubes? to s

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Answer:

B.) to determine that electric beams in cathode ray tubes were actually made of particles

Explanation:

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2 years ago

A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a

Yanka [14]

Answer:

Diameter of the cylinder will be $d=2.998\times 10^4m$

Explanation:

We have given young's modulus of steel $E=207GPa=207\times 10^9Pa$

Change in length $\Delta l=0.38mm$

Length of rod $l=500mm$

Load F = 11100 KN

Strain is given by $strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}$

We know that young's modulus $E=\frac{stress}{strain}$

So $207\times 10^9=\frac{stress}{7.6\times 10^{-4}}$

$stress=1573.2\times 10^{-5}N/m^2$

We know that stress $=\frac{force}{artea }$

So $1573.2\times 10^{-5}=\frac{11100\times 1000}{area}$

$area=7.055\times 10^{8}m^2$

So $\frac{\pi }{4}d^2=7.055\times 10^{8}$

$d=2.998\times 10^4m$

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2 years ago

Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V

Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

$Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter$

$Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter$

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

$R=\dfrac{\rho\times l}{A}$

Where,

R = Resistance

l = length

A = Area of cross section = πr²

$\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter$

Substituting the values we get

$R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}$

$R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}$

$R=\dfrac{1\times 10^{2}}{3.23}$

$R=30.95\ ohm$

Now by Ohm's Law,

$V= I\times R$

Substituting the values we get

$I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere$

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0

2 years ago