The amount of work required for a

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi

vlabodo [156]

Answer:

Smallest drop: Water

Largest drop: Dirt

Explanation:

The heat needed to change the temperature of a sample is:

$Q=cm\Delta T$ (1)

with Q the heat (added(+) or removed(-)), c specific heat, m the mass and $\Delta T$ the change in temperature of the sample. So, if we solve (1) for

Sample A:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{4186*4.0}$

$\Delta T=-\frac{Q}{16744}$

Sample B:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{2700*2.0}$

$\Delta T=-\frac{Q}{5400}$

Sample C:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{1050*9.0}$

$\Delta T=-\frac{Q}{9450}$

Note that the numbers 16744, 5400, 9450 are in the denominator of the expression $-\frac{Q}{cm}$ that gives the drop on temperature. so, if Q is the same for the three samples the smallest denominator gives the largest drop and vice versa.

So, the smallest drop is Sample A and the largest is Sample C.

(Important: The minus sign of $\Delta T$ implies the temperature is dropping)

8 0

3 years ago

A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s

Paraphin [41]

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$

3 0

3 years ago

A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal

zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity $u_{y}=0$

The initial velocity is the horizontal component $u_{x}$

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = $u_{y}$ t + $\frac{1}{2}$ gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , $u_{y}=0$

Substitute these values in the rule

→ -60 = 0 + $\frac{1}{2}$ (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* The ball is in the air for 3.5 seconds 

The initial velocity is the horizontal component $u_{x}$

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = $u_{x}$ t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = $u_{x}$ (3.5)

Divide both sides by 3.5

→ $u_{x}$ = 28.57 m/s

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is $v_{y}$

→ $v_{y}$ = $u_{y}$ + gt

→ $u_{y}$ = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ $v_{y}$ = 0 + (-9.8)(3.5)

→ $v_{y}$ = -34.3 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity v is the resultant vector of $v_{x}$ and $v_{y}$

→ Its magnetude is $v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}$

→ Its direction $tan^{-1}\frac{v_{y}}{v_{x}}$

→ $v_{y}$ = 28.6 , $v_{y}$ = -34.3

Substitute this values in the rules above

→ $v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66$

→ Its direction $tan^{-1}\frac{-34.3}{28.6}=-50.18$

The negative sign means the direction is below the horizontal

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

7 0

3 years ago

what output force is generated when an input force of 630 n is applied to a machine with a mechanical advantage of 3

Marta_Voda [28]

The mechanical advantage is the factor by which
the machine multiplies the input force.

If the MA is 3 and the input force is 630N, then
the output force is

(3) x (630N) = 1,890N

3 0

3 years ago

Read 2 more answers

If the moon is a full moon tonight what will be the moon one week later waxing or waning

tatiyna

-- Starting from nothing (New Moon), the moon's shape grows ('waxes')
for half of the cycle, until it's full, and then it shrinks ('wanes') for the next
half of the cycle.

-- The moon's complete cycle of phases runs 29.53 days . . . roughly
four weeks.

-- So, beginning from New Moon, it spends about two weeks waxing until
it's full, and then another two weeks waning until it's all gone again.

-- After a Full Moon, the moon is waning for the next two weeks. So it's
definitely waning at one week after Full.

5 0

3 years ago