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2 years ago

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The tension in a string from which a 4.0 kg object is suspended in an elevator is equal to 44 N. What is the acceleration of the

elevator, if upward is the positive direction. Write the numerical value only with 2 decimal places. Do not write the unit. If the acceleration is negative (downward) include the negative sign in your answer.

1 answer:

Ronch [10]2 years ago

8 0

Answer: 1

Explanation:

Given

Tension is the string $T=44\ N$

mass of object $m=4\ kg$

Tension is greater than the weight of the object i.e. elevator is moving upward

we can write

$\Rightarrow T-mg=ma\\\Rightarrow T=m(g+a)\\\Rightarrow 44=4(10+a)\\\Rightarrow 11=10+a\\\Rightarrow a=1\ m/s^2$

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: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

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3 years ago

A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea

Nostrana [21]

Hi there! :)

Reference the diagram below for clarification.

1.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
$R_T = R_1 + R_2 + ... + R_n$

In parallel:
$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

$\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega$

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
$R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}$

2.

We can use Ohm's law to solve for the current in the circuit.

$i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}$

3.

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

4.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

$V = iR\\\\V = (6)(15) = \boxed{90 V}$

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1 year ago

The massless spring of a spring gun has a force constant k=12~\text{N/cm}k=12 N/cm. When the gun is aimed vertically, a 15-g pro

ASHA 777 [7]

Answer:

0.011 m.

Explanation:

Energy stored in the spring = Energy of the projectile.

1/2ke² = mgh ................ Equation 1

Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.

make e the subject of the equation

e = √(2mgh/k)............................. Equation 2

Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×0.015×5/1200)

e = √(0.15/1200)

e = √(0.000125)

e = 0.011 m.

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2 years ago