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2 years ago

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An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

, the tension in the cable was 6500 N. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N drag force. Part A

1 answer:

Assoli18 [71]2 years ago

7 0

Answer:

The tension in the cable when the craft was being lowered to the seafloor is 4700 N.

Explanation:

Given that,

When the craft was stationary, the tension in the cable was 6500 N.

When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N.

The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,

T + F = W

Here, T is tension

F = 1800 N

W = 6500 N

Tension becomes :

$T=W-F\\\\T=6500-1800\\\\T=4700\ N$

So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.

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Sketch the circuit labeling the meter and bulb as two separate resistors connected in parallel to the voltage source. Then show

Ksenya-84 [330]

Answer:

Show attached picture

Explanation:

Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call $R_M$ its internal resistance) and R indicates the resistance of the light bulb.

We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:

$R_M = 1000 R$ (1)

Both the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:

$V_M = V_R$

Using Ohm's law, $V=RI$ , we can rewrite the previous equation as:

$R_M I_M = R I_R$

where

$I_M$ is the current in the meter

$I_R$ is the current in the bulb

Using (1), this equation becomes

$(1000 R) I_M = R I_R \rightarrow I_M = \frac{I_R}{1000}$

so, the current in the meter is 1000 times less than through the bulb.

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Solnce55 [7]

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2 years ago

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2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de

kow [346]

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

u = 0

Therefore, the equation becomes

S = 1/2 gt²

Substituting the given values in the above equation

S = 0.5 x 9.8 x 1.5²

= 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

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2 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

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1 year ago

A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

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2 years ago