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3 years ago

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a 5kg block on a rough horizontal surface is attached to a light spring (force constant=1.6kN/m). the block passes through its e

quilibrium position with a kinetic energy of 5J and is brought momentarily to rest after stretching the spring .06m. how much work is done by the frictional force on the block as it moves from its equilibrium position to the point of momentary rest?

1 answer:

natima [27]3 years ago

5 0

Answer:

2.12 J

Explanation:

Initial kinetic energy = final elastic energy + work by friction

KE = EE + W

KE = ½ kx² + W

5 J = ½ (1600 N/m) (0.06 m)² + W

W = 2.12 J

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A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

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3 years ago

A circuit consists of a battery connected to three resistors (65 ω, 25ω, and 170ω) in parallel. the total current through the re

White raven [17]

A. To find the total emf of the battery, just remember that in a parallel circuit, the voltage is the same throughout the circuit. So you can get the total voltage of the circuit by using Ohm's Law.

$I= \frac{V}{R}$

Where:
I = current (A)
V = Voltage (V) (emf)
R = Resitance (Ω)

Now you can derive the formula of Voltage by transposing the Resistance to the other side of the equation to isolate Voltage. The formula you will now use will be:
$V = IR$

However, you cannot solve this yet because the resistance you need is the total resistance in the circuit. To do this, you need to get the total resistance in this parallel circuit and the formula would be:

$\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}$

You have three resistors with the following resistance:
65Ω, 25Ω and 170Ω
$\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}$

$\frac{1}{R_{T}} = \frac{1}{R_{65}}+ \frac{1}{R_{25}}+ \frac{1}{R_{170}}$

$\frac{1}{R_{T}} =0.0153+0.04+0.006+0.0059$
$\frac{1}{R_{T}} =0.0613$

Get the reciprocal of both sides and divide:

$R_{T} = \frac{1}{0.0613} =16.32$

The total resistance then is 16.32Ω

Now that you have the total resistance, you can solve for the total voltage:
$V = IR$
$V = (1.8)(16.32)$
$V = 29.376V$

The emf of the battery is 29.376V

B. To find the resistance in each resistor, just apply Ohm's law again. In a parallel circuit, the voltage is the same, but the current that runs through it is different for each resistor. Now just solve for the current of each using the same voltage.

Resistor 1: 65Ω
$I= \frac{V}{R}$
$I= \frac{29.376}{65}$
$I= 0.45A$

The current flowing through resistor 1 with a resistance of 65Ω is 0.45A.

Resistor 2: 25Ω
$I= \frac{V}{R}$
$I= \frac{29.376}{25}$
$I= 1.18A$
The current flowing through resistor 2 with a resistance of 25Ω is 1.18A.

Resistor 3: 170Ω
$I= \frac{V}{R}$
$I= \frac{29.376}{170}$
$I= 0.17A$

The current flowing through resistor 3 with a resistance of 170Ω is 0.17A.

If you add up all their current it confirms the given that the total current running through all of them is 1.8A.

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3 years ago

The force generated by a single muscle fiber can be increased by increasing is called

mars1129 [50]

Answer:

The force generated by a single muscle fiber can be increased by increasing the frequency of action potentials

Explanation:

The force generated by a muscle fiber is the result of the shortening of the skeletal muscle, and this force is also know as muscle tension. The larger motor units shorten along with the smaller units to produce the muscle force. The time lapsed between the beginning of the action potential in the muscle and the beginning of the contraction is the latent period. Action potential is the result of the difference electrical potential as a result of passage of an impulse along the membrane of a muscle or nerve cell.

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3 years ago

A diver jumps off a diving platform that is 20 meters long. Describe the transfer of energy that occurs during the fall.

kobusy [5.1K]

Answer: gravitational potential energy is converted into kinetic energy

Explanation:

When the diver stands on the platform, at 20 m above the surface of the water, he has some gravitational potential energy, which is given by

$E=mgh$

where m is the man's mass, g is the gravitational acceleration and h is the height above the water. As he jumps, the gravitational potential energy starts decreasing, because its height h above the water decreases, and he acquires kinetic energy, which is given by

$K=\frac{1}{2}mv^2$

where v is the speed of the diver, which is increasing. When he touches the water, all the initial gravitational potential energy has been converted into kinetic energy.

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3 years ago

Complete the following:

masha68 [24]

When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

To find the answer, we have to know about the rules followed by drawing ray-diagram.

<h3>What are the rules obeyed by light rays?</h3>

If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.
The refracted ray becomes parallel to the major axis if the incident ray passes through the focus.
The refracted ray follows the same path if the incident light passes through the center of the curve.

Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

Learn more about refraction by a lens here:

brainly.com/question/13095658

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1 year ago