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1 year ago

4. A plane accelerates down a runway due to the constant resultant force of

Physics

1 answer:

Doss [256]1 year ago

4 0

Answer:

See below

Explanation:

Net force acting on the plane after overcoming frictional forces is

2.7 x 10^6 - 6.8 x 10^4 - 8.5 x 10^5 = 1 782 000 N

Work = F x d

= 1 782 000 x 1400 m = 2.5 x 10^9 J or 2.5 x 10^6 kJ

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What two measurements are needed to determine density?

g100num [7]

Density is defined as (mass) per unit (volume). So in order to calculate
the density of a glob of some substance, you pretty much have to measure
its mass and its volume.

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2 years ago

Read 2 more answers

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

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2 years ago

Wavelength is measured in hertz true or false?

g100num [7]

Wavelength is measured in various units of distance (mm, cm, m, etc.). Frequency uses hertz, but that is not what the question is asking. Therefore, the answer is false.

3 0

2 years ago

What is the speed of an object after falling (from rest) a distance of 9.80 meters near the surface of the Earth? Assume air

madam [21]

Hi there!

We can use the kinematic equation:

$v_f^2 = v_i^2 + 2ad$

vf = Final velocity (? m/s)

vi = initial velocity (0 m/s, dropped from rest)

a = acceleration (due to gravity, 9.8 m/s²)

d = distance (9.8 m)

Simplify the equation to solve for vf:

$v_f^2 = 0 + 2ad\\\\v_f = \sqrt{2ad}$

Substitute in the given values:

$v_f = \sqrt{2(9.8)(9.8)} = \boxed{13.86 m/s}$

8 0

2 years ago

An object, experiencing no friction, keeps moving at a constant speed. What can we say about the net force on the object?

sleet_krkn [62]

Answer:

Explanation:

Let's look at a mathematical representation of this. The equation for tis is just a souped up version of Newton's 2nd Law:

F - f = ma. It an object is moving at a constant speed, the acceleration of that object is 0. That changes this equation to

F = f which states that the applied Force equals the frictional force, choice a.

3 0

2 years ago