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Crazy boy [7]
3 years ago
10

arlene is to walk across a high wire strung horizontally between two buildings 10.0m apart. The sag in the rope when she is at t

he midpoint is 10.0 degrees, if her mass is 50.0 kg, what is the tension in the rope at this point?...can you help me...thanks!
Physics
2 answers:
Ipatiy [6.2K]3 years ago
4 0
We are given with a force in the middle of a rope measuring 10 m and an angle of sagging of 10 degrees from the horizontal. The tension of the rope is equal to the hypotenuse of the right triangle. 

sin 10 = 50 kg * 9.8 m/s2 / T
Tension = 2821. 80 N 
Greeley [361]3 years ago
3 0

Answer:

T = 1412.3 N

Explanation:

As we know that weight of the Arlene is counterbalanced by the tension in the string

So here it is given that string makes 10 degree angle at mid point

So we will have

T sin\theta + T sin\theta = mg

so we have

T = \frac{mg}{2sin\theta}

here we know that

m = 50 kg

\theta = 10 degree

now we have

T = \frac{50 \times 9.81}{2sin10}

T = 1412.3 N

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PSYCHO15rus [73]

The equivalent resistance of the two cylindrical conductors connected in parallel is 466 ohm.

<h3>Resistance</h3>

Resistance is a measure of the opposition to flow of electric current. It is measured in ohms.

It is given by the formula:

R=\rho\frac{l}{A} \\\\where\ l=length,A=area,\rho=resistivity

Given that R₂ = 469 ohm, hence:

R_2=\rho\frac{l_2}{A_2} \\\\469=\rho\frac{l_2}{\pi r_2^2}

But l₁ = 6l₂, r₁ = (1/5)r₂, hence:

R_1=\rho \frac{l_1}{A_1}=\rho *\frac{6l_2}{[\pi (1/5)r_2]^2} =150 * \rho \frac{l_2}{[\pi r_2]^2}=30*469=70350\ ohm

The equivalent resistance (R) is:

R=\frac{R_1R_2}{R_1+R_1}=\frac{469*70350}{469+70350}  =466\ ohm

The equivalent resistance of the two cylindrical conductors connected in parallel is 466 ohm.

Find out more on resistance at: brainly.com/question/17563681

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2 years ago
Which example provides the most complete description of an object's motion?
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Answer:

The hiker followed a road heading north for 2 miles in 30 minutes.

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The hiker followed a road heading north for 2 miles in 30 minutes. This describes the motion of hiker. The motion shows how fast the hiker is moving.  

Distance, d = 2 miles = 3218.6 m

times, t = 30 minutes = 1800 seconds

So, we can say that the hiker is moving with a speed of 1.78 m/s in north direction.

Hence, this is the required solution.

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Do you have a picture of the diagram?

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Which is the MOST accurate statement about the heating curve and energy?
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The answer is A). Moving from A to C the temperature and the kinetic energy increases.

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At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does
insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

V = V_R + V_L

The potential difference across the inductance is given by

V_L(t) = V e^{-\frac{t}{\tau}} (1)

where

\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s is the time constant of the circuit

At time t=0,

V_L(0) = V e^0 = V = 20 V

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

V_R = V-V_L = 20 V-20 V=0

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

t >> \tau

Since \tau is at the order of less than milliseconds.

Using eq.(1), we see that for t >> \tau, the exponential becomes zero, and therefore the potential difference across the coil is zero:

V_L = 0

Therefore, the potential difference across the resistor will be

V_R = V-V_L = 20 V- 0 = 20 V

(c) Yes

The two voltages will be equal when:

V_L = V_R (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

V=V_L +V_R

And rewriting this equation,

V_R = V-V_L

Substituting into (2) we find

V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V

So, the two voltages will be equal when they are both equal to 10 V.

(d) at t=5.77\cdot 10^{-4}s

We said that the two voltages will be equal when

V_L=\frac{V}{2}

Using eq.(1), and this last equation, this means

V e^{-\frac{t}{\tau}} = \frac{V}{2}

And solving the equation for t, we find the time t at which the two voltages are equal:

e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

I_0 = 3.20 A

The problem says this current is stable: this means that we are in a situation in which t>>\tau, so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

V_L(0)=.V

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

t>>\tau

If we use this approximation into the formula

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

We find that

V_L = 0

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0
3 years ago
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