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Naddika [18.5K]

2 years ago

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A boy does 465 J of work pulling an empty wagon along level ground with a force of 111 N [31o below the horizontal]. A frictiona

l force of 155 N opposes the motion and is actually slowing the wagon down from an initial high velocity. The distance the wagon travels is

1 answer:

vovikov84 [41]2 years ago

6 0

Can you list the answer's

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At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c

fiasKO [112]

Answer:

497.00977 N

3742514.97005

Explanation:

$\rho$ = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

$F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N$

The drag force on the dolphin's nose is 497.00977 N

at 20°C

$\mu$ = Dynamic viscosity = $1.002\times 10^{-3}\ Pas$

Reynold's Number

$Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005$

The Reynolds number is 3742514.97005

8 0

2 years ago

The circuit in Fig. P4.23 utilizes three identical diodes having IS = 10−14 A. Find the value of the current I required to obtai

Sliva [168]

Answer:

$v = \frac{2 V}{3}= 0.667 v$

Since we have identical diodes we can use the equation:

$I_D =I= I_S e^{\frac{V_D}{V_T}}$

And replacing we have: $I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA$

Since we know that 1 mA is drawn away from the output then the real value for I would be

$I_D = I = 3.86 mA -1 mA= 2.86 mA$

And for this case the value for $v_D$ would be:

$V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V$

And the output votage on this case would be:

$V = 3 V_D = 3 *0.660 V = 1.98 V$

And the net change in the output voltage would be:

$\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V$

Explanation:

For this case we have the figure attached illustrating the problem

We know that the equation for the current in a diode id given by:

$I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}$

For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode $v_1 + v_2 + v_3= 2$ and each voltage is the same v for each diode, so then:

$v = \frac{2 V}{3}= 0.667 v$

Since we have identical diodes we can use the equation:

$I_D =I= I_S e^{\frac{V_D}{V_T}}$

And replacing we have:

$I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA$

Since we know that 1 mA is drawn away from the output then the real value for I would be

$I_D = I = 3.86 mA -1 mA= 2.86 mA$

And for this case the value for $v_D$ would be:

$V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V$

And the output votage on this case would be:

$V = 3 V_D = 3 *0.660 V = 1.98 V$

And the net change in the output voltage would be:

$\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V$

5 0

1 year ago

Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction

gizmo_the_mogwai [7]

<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>

5 0

2 years ago

In an inertial frame of reference, a series of experiments is conducted. in each experiment, two or three forces are applied to

seraphim [82]

The objects will remain at rest if net force acting on it is zero if their magnitude is same and they are acting in opposite direction then according to Newton's 2nd law the net force acting on the system is zero. Since the net force on the system is zero, the object will remain at rest.

8 0

2 years ago

Two runners ran side by side each holding one end of a horizontal pole. What would most likely happen if one of the runners bega

Anettt [7]

Answer:

See explanation.

Explanation:

If each runner was holding the pole, the runner in the water side of the pole would probably be behind the other runner. Since running in knee deep is hard and makes you slower, the pole would be slanted.

5 0

1 year ago