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3 years ago

What is the final velocity of an object that is dropped if it falls a distance of 100 m?

Physics

1 answer:

Liula [17]3 years ago

4 0

Answer:

$44.27m/s$

Explanation:

The kinematic equation

$v_f^2=v_0^2+2ad$

gives the final velocity $v_f$ of the object given the initial velocity $v_0$ , the acceleration $a$ , and the distance traveled $d$ .

For our case, the object is dropped; therefore,

$v_0=0$

I.e. the initial velocity is zero. The acceleration due to gravity is

$a=9.8m/s^2$ ,

and the distance traveled is $d=100m$ .

Putting the values into the equation we get:

$v_f^2=0+2(9.8ms^{-2})(100m)\\\\v_f^2=1960\\\\v_f=\sqrt{1960} \\\\\boxed{v_f=44.27m/s}$

The final velocity of the object is 44.27 m/s.

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A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

fgiga [73]

The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

V = initial velocity = 36. 6 m/s

α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

Learn more about vertical velocity here:

brainly.com/question/24949996

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2 years ago

The magnetic field is 0.0 2M from the wire is 0.1 T what is the magnitude of the magnetic field 0.0 1M from the same wire

Kisachek [45]

Answer:

0.2 T

Explanation:

Magnetic field is inversely proportional to the distance from wire since the distance is halved therefore magnetic field will be doubled.

4 0

3 years ago

The magnitude of the weight of a 3.0 kg object on the surface of the earth is 29 N. True False

madreJ [45]

True

In fact, the weight of an object on the surface of the Earth is given by:
$F=mg$
where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration on Earth's surface. If we use the mass of the object, m=3.0 kg, we find
$F=mg=(3.0 kg)(9.81 m/s^2)=29 N$

8 0

3 years ago

Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres

inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )$

Here, $v_{1}$ is the velocity of water through wide ends of cylindrical pipe and $v_{2}$ is the velocity of water through narrow ends of cylindrical pipe.

Given, $v_{1} =1.4 m/s$

Now from equation continuity,

$v_{1} A_{1} = v_{2} A_{2}$ .

Here, $A_{1}$ and $A_{2}$ are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

$v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}$ .

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

$v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}$

$v_{2} = 4 \times 1.4 = 5.6 m/s$

Substituting these values with the density of water is $1000 \ kg/m^3$ in pressure difference formula we get.

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa$

3 0

3 years ago

Read 2 more answers

How does the atmosphere’s interaction with different wavelengths of light help to protect life on earth?

marishachu [46]

Answer:

Ozone layer in the upper atmosphere filters most of the harmful radiations of shorter wavelength. It actually absorbs the hazardous radiations like ultraviolet, gamma rays, x- rays and most of all those having shorter wavelength then the visible light. That's how the earth's atmosphere protects life on earth. But unfortunately, climate change and global warming is causing the depletion of ozone layer which is causing skin related diseases and harming not only the human life but also the plants and animals.

4 0

3 years ago