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3 years ago

9

uppose the bottom block has a mass of 0.4 kg and the top block has a mass of 0.1 kg. What force do you need to exert (in newtons

) to keep the blocks moving at a constant speed of 10 cm/s

1 answer:

notsponge [240]3 years ago

5 0

Answer:

F = 0

Explanation:

Newton's second law is

F = ma

As in this case the two blocks move with constant speed, it implies that the acceleration is zero, therefore the force applied to the system is zero

F = 0

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Why are there zero hours of daylight at the north pole and south pole in the winter

raketka [301]

Answer:

The seasons are caused by the tilt of Earth's axis in relation to the sun. ... During summer, Antarctica is on the side of Earth tilted toward the sun and is in constant sunlight. In the winter, Antarctica is on the side of Earth tilted away from the sun, causing the continent to be dark.

Explanation:

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3 years ago

Sedaia [141]

Answer:

V = 20.5 m/s

Explanation:

Given,

The mass of the cart, m = 6 Kg

The initial speed of the cart, u = 4 m/s

The acceleration of the cart, a = 0.5 m/s²

The time interval of the cart, t = 30 s

The final velocity of the cart is given by the first equation of motion

v = u + at

= 4 + (0.5 x 30)

= 19 m/s

Hence the final velocity of cart at 30 seconds is, v = 19 m/s

The speed of the cart at the end of 3 seconds

V = 19 + (0.5 x 3)

= 20.5 m/s

Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s

6 0

3 years ago

Fill-in-the-blank test questions measure ________; matching concepts with their definitions measures ________.

Ulleksa [173]

Test questions measure recall; matching concepts with their definitions measures recognition.

<u>Explanation: </u>

According to Psychology our brain remembers everything what we learn but the understanding and remembering the right answer for the right question needs training and understanding ability. So in order to enhance the ability of recalling and recognizing among the students, the concept of test questions and matching with definitions are used in curricular activities.

As the students will be learning different terms, definitions, methods and different subjects, they should be able to distinguish among different definitions as well as they should recall the things they have learnt. So the answers for the test questions will help to recall the topics learnt by the students while the matching concept will help the students to recognize each definition with their terms.

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3 years ago

A disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular a

stepan [7]

Answer:

a) α = 1.875 $\frac{rad}{s^{2} }$

b) t = 8 s

Explanation:

Given:

ω1 = 0 $\frac{rad}{s}$

ω2 = 15 $\frac{rad}{s}$

theta (angular displacement) = 60 rad

*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*

a) α = ?

(ω2)^2 = (ω1)^2 + 2α(theta)

$15^{2}$ = $0^{2}$ + 2(α)(60)

225 = 120α

α = 1.875 $\frac{rad}{s^{2} }$

b)

α = (ω2-ω1)/t

t = (ω2-ω1)/α = (15-0)/1.875 = 8

t = 8 s

4 0

3 years ago

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago