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3 years ago

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Anna pushes a box with a force of 8.00 newtons. She generates a power of 3.00 watts. How much time does it take for Anna to move

the box the distance of 9.72 meters?

2 answers:

QveST [7]3 years ago

8 0

Power is the energy in a system per time. It will have units of Watts which is equal to joules per second. It can be expressed as:

P = E / t

where E = Force x distance

P = Fd / t
t = Fd / P
t = 8 (9.72) / 3.0
t = 25.92 s

Ratling [72]3 years ago

8 0

Answer:

time taken, t = 25.92 seconds

Explanation:

It is given that,

Force acting on the box, F = 8 N

Power generated on the box, P = 3 watts

Distance traveled by the box, d = 9.72 m

We know that the power generated by an object is equal to the total work done divided by total time taken. Its formula is given by :

$P=\dfrac{W}{t}$

Since, W = F d

$P=\dfrac{Fd}{t}$

$t=\dfrac{Fd}{P}$

$t=\dfrac{8\ N\times 9.72\ m}{3\ W}$

t = 25.92 seconds

So, the time taken by Anna to move the box to the distance of 9.72 meters is 25.92 seconds. Hence, this is the required solution.

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An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of

Elan Coil [88]

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

$m=\frac{V}{v}$

$m=m_f+m_g$

$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$

$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$ (taking the value of $v_g$ and $v_g$ at 200°C )

$V=6.304 \times 10^{-3}$

Now quality of vapor

$x=\frac{m_g}{m}$

$=3.377 \times 10^{-3}$

Internal energy at state 1 can be found out by

$u_1=u_f+xu_{fg}$

$=850.65+3.377\times10^{-3}\times 1744.65$

= 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor $v_2=v_g \text { and }\ x=1$

Tank is rigid, so volume of tank is constant.

$v_g=v_2=\frac{V}{m}$

$v_g=\frac{6.304\times 10^{-3}}{1.4}$

$v_g=4.502 \times 10^{-3} \ m^3 /kg$

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$

= 369.11° C

Internal energy at state 2

$u_2=2154.9 \ kJ/kg$

Now power rating of the resistor

$P=\frac{m(u_2-u_1)}{t}$

$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$

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3 years ago

A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140

max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

$u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\$

Using formula:

$v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\$

$= 0.55 \ \frac{m}{sec^2}\\\\$

$F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\$

Calculating the Work by net force

$W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\$

The above work is converted into thermal energy.

Now,

$F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J$

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