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2 years ago

6

A car is merging onto the expressway with an acceleration of 1.77 m/s^2. When it begins down the ramp the car is traveling at a

speed of 11.1 m/s. How long will it take for the car to reach a speed of 24.4 m/s

2 answers:

TiliK225 [7]2 years ago

7 0

Answer:

time taken = 7.514 seconds

$\sf{acceleration} = \dfrac{\Delta v}{\Delta t} = \dfrac{final \ speed - initial \ speed}{time \ taken}$

Here given:

acceleration: 1.77 m/s²

final speed: 24.4 m/s

initial speed: 11.1 m/s

using the given formula:

$\hookrightarrow \sf{1.77} = \dfrac{24.4 - 11.1}{time \ taken}$

$\hookrightarrow \sf{time \ taken} = \dfrac{24.4 - 11.1}{1.77}$

$\hookrightarrow \sf{time \ taken} = 7.514 \ seconds$

Jlenok [28]2 years ago

7 0

Formula :

$\frac{final \: speed - initial \: speed}{time \: taken}$

Given :

acceleration: 1.77 m/s²

final speed: 24.4 m/s

initial speed: 11.1 m/s

Solution :

$1.77 = \frac{24.4 - 11.1}{time \: taken}$

$time \: taken = \frac{24.4 - 11.1}{1.77}$

$taken \: time = 7.514 \: seconds$

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3 years ago

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$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

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$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

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The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

3 years ago

A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.

Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

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- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mb*g*H = 0.5*mb*vb^2

vb = √2*g*H

vb = ( 2*9.81*15 ) ^0.5

vb = 17.15517 m/s

- The angular momentum of system before collision is:

M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

M1 = M2

vo = 247.03448 / (5.10*3)

vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mo*g*h = 0.5*mo*vo^2

h = vo^2 / 2*g

h = 16.14604^2 / 2*(9.81)

h = 13.3 m

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3 years ago