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3 years ago

12

A ball is thrown upward at time t=0 from the ground with an initial velocity of 4 m/s (~ 9 mph). Assume that g = 10 m/s^2. What

is the greatest height (in meters) reached by the ball?

1 answer:

aleksandrvk [35]3 years ago

4 0

Answer:

The greatest height reached by the ball is 0.8 m.

Explanation:

Given that,

Initial velocity = 4 m/s

We need to calculate the greatest height reached by the ball

Using equation of motion

$v^2=u^2+2gh$

Where, v = final velocity

u = initial velocity

g = acceleration due to gravity

Put the value in the equation

$0=4^2+2\times(-10)\times h$

$h =\dfrac{16}{20}$

$h =0.8\ m$

Hence, The greatest height reached by the ball is 0.8 m.

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Which point or points of view could be present in a business letter?

Studentka2010 [4]

Answer:

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Explanation:

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3 years ago

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

$d=\frac{1,280,000,000km}{37,000}=34,594.59km$

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

$v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}$

hence, the speed of the Hubble is approximately 384km/min

5 0

3 years ago

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is:

$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

4 0

2 years ago

Estimated speed of the vehicle

SSSSS [86.1K]

Really, Gundy ? ! ?

The formula for the car's speed is given and discussed in the box. The formula is

v = √(2·g·μ·d)

Then they <em>tell</em> you that μ is 0.750 , and then they <em>tell</em> you that d = 52.9 m . Also, everybody knows that 'g' is gravity = 9.8 m/s² .

They also tell us that the mass of the car is 1,000 kg, and they tell us that it took 3.8 seconds to skid to a stop. But we already <em>have</em> all the numbers in the formula <em>without</em> knowing the car's mass or how long it took to stop. The police don't need to weigh the car, and nobody was there to measure how long the car took to stop. All they need is the length of the skid mark, which they can measure, and they'll know how fast the guy was going when he hit the brakes !

Now, can you take the numbers and plug them into the formula ? ! ?

v = √(2·g·μ·d)

v = √( 2 · 9.8 m/s² · 0.75 · 52.9 m)

v = √( 777.63 m²/s²)

v = 27.886 m/s

Rounded to 3 digits, that's <em>27.9 m/s </em>.

That's about 62.4 mile/hour .

3 0

3 years ago

With what magnitude of force does a ball of mass 0.75 kilograms need to be hit so that it accelerates at the rate of 25 meters/s

SSSSS [86.1K]

Answer:

Assume that the ball undergoes motion along a straight line. ... F = m A Force = (mass) x (acceleration) The question tells you the mass and the acceleration. All YOU have to do is take the numbers and pluggum into Newton's 2nd law. F = m A = (0.75 kg) (25 m/s²) = (0.75 x 25) kg-m/s² = 18.75 Newtons .

Explanation:

i looked it up ok

8 0

3 years ago

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