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3 years ago

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The time it takes for a dragster to cross the finish line is unknown. The dragster accelerates from rest at 26 m/s2 for a quarte

r-mile (402 m). What is the final velocity of the dragster?

1 answer:

exis [7]3 years ago

7 0

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${v}^{2} - {v(0)}^{2} = 2.a.∆x$

$v = final \: \: velocity$

$v(0) = the \: \: velocity \: \: in \: \: the \: \: rest \: \: which \: \: is \: \: zero \\$

$a = acceleration$

$∆x = covered \: \: distance$

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${v}^{2} - ({0})^{2} = 2 \times 26 \times 402$

$v = \sqrt{2 \times 26 \times 402}$

$v = \sqrt{20904}$

$v = 144.58 \: \: \frac{m}{s}$

Done...

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The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

$T_{1} + V_{1} = T_{2} + V_{2}$

$h_{ab} = h_{bc} = 0.18m$

potential energy at position 1,

$V1 = m_{ab} gh_{ab} + m_{bc} gh_{bc}$

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

$T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} + \frac{1}{2} m_{bc} v^{2}_{G} + \frac{1}{2} lw^{2}_{bc}$

$l_{ab} =\frac{m_{ab} l^{2}_{ab} }{3}$

= 1/3 *4 * (0.36)²

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$l =\frac{m_{bc} l^{2}_{bc} }{12}$

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on putting the values in above equation we get,

T₂ = 1.0667vb²

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vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

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1 year ago

A block of mass 10.0 kg is pulled to the right along a rough horizontal surface with a constant horizontal force of 20.0 N. The

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To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma

if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.

The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).

you will find that Fnet=15.0N

With that, plug in the values you know to calculate the acceleration of the block:

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3 years ago

Water drips from the nozzle of a shower onto the floor 193 cm below. The drops fall at regular (equal) intervals of time, the fi

Law Incorporation [45]

Answer: 108.81 cm and 48.66 cm

Explanation:

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What is know:

Height (h) = 193 cm

Gravity (g) = 981 $cm/s^{2}$

Initial Velocity = 0

First, we can know how long it take to the drop to travel to the floor. It can be done with the following equation:

$x = V_{0} t + \frac{1}{2} at^{2}$ (1)

Where:

x is the distance which is 193cm

Vo is the Initial Velocity which is zero

t is the time the time it takes the drop to travel from the shower to the floor

a is the aceleration, which in this case is the gravity.

With the Initial Velocity equals zero the equations simply:

$193 cm = \frac{1}{2}gt^{2}$

To search for the time:

$t =\sqrt{\frac{2*193cm}{981cm/s^{2} } }$

t = 0.627 s

This is the time it takes a drop to fall to the floor, with this time and knowing other 3 drops have driped from the shower by this time. We can calculate how much time it takes the shower to drip each drop.

Time for Drip = t/4

Time for Drip = 0.156

This time is the difference between each drop, using the same equation we can calculate where was each drop, because now it is know how much time had each drop after being drip from the shower.

Our first is already on the floor (193 cm) with 0.627 s, The second drops have been falling for (0.627s - 0.156) 0.471 s and our third drop for (0.627s - 0.156 - 0.156) 0.315 s

We can use (1) to know how far have each drop traveled on these times. We know the Initials Velocity are 0, know we need ot know the distances.

For the second drop:

$x = \frac{1}{2} (981cm/s^{2})(0.471s)^{2}$

$x =108.81 cm$

For the third drop:

$x = \frac{1}{2} (981cm/s^{2})(0.315s)^{2}$

$x = 48.66 cm$

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3 years ago