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2 years ago

7

The speed of the current in a river is 6 mph. A ferry operator who works that part of the river is looking to buy a new boat for

his business. Every day, his route takes him 22.5 miles against the current and back to his dock, and he needs to make this trip in a total of 9 hours. He has a boat in mind, but he can only test it on a lake where there is no current. How fast must the boat go on the lake in order for it to serve the ferry operator’s needs?

1 answer:

user100 [1]2 years ago

7 0

The first step is to determine the rate using the given values. Distance formula 9(x-6)=45 is used where the rate is x-6. Value of x is equivalent to 11. The average speed that covers 45 miles in 9 hours is 5 miles per hour having no current.

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As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago

A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done

Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

$W = F*d$

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

$F = W_{x} = mgsin(\theta)$

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²

θ: is the degree angle to the horizontal = 30°

$F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N$

Now, we can find the work:

$W = F*d = 24.53 N*20 m = 490.6 J$

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0

2 years ago

A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee

Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

$s = ut +\frac{1}{2}at^{2}$

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

$s = \frac{1}{2}gt'^{2}$

$s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}$ (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

$s = v\times t''$

$s = 343\times t''$ (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t' (3)

Using eqn (2) and (3):

s = 343(3.5 - t') (4)

from eqn (1) and (4):

$4.9t'^{2} = 343(3.5 - t')$

$4.9t'^{2} + 343t' - 1200.5 = 0$

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

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3 years ago

A person driving her car at 48 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow

trasher [3.6K]

Answer:

She must stop the car before interception, distance traveled 12.66 m

Explanation:

We will take all units to the SI system

Vo = 48Km / h (1000m / 1Km) (1h / 3600s) = 13.33 m / s

V2 = 70 Km / h = 19.44 m / s

We calculate the distance traveled before stopping

X = Vo t + ½ to t²

Time is what it takes traffic light to turn red is t = 2.0 s

X = 13.33 2 + 1.2 (-7) 2²

X = 12.66 m

It stops car before reaching the traffic light turning to red

Let's analyze what happens if you accelerate, let's calculate the acceleration of the vehicle

V2 = Vo + a t2

a = (V2-Vo) / t2

a = (19.44-13.33) /6.6

a = 0.926 m / s2

This is the acceleration to try to pass the interception, now let's calculate the distance it travels in the time the traffic light changes from yellow to red (t = 2.0 s)

X = Vo t + ½ to t²

X = 13.33 2 + ½ 0.926 2²

X = 28.58 m

Since the vehicle was 30 m away, the interception does not happen

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3 years ago

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Phoenix [80]

Answer:

The temperature of the core raises by $2.8^{o}C$ every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

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energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

$\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s$

5 0

2 years ago