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liubo4ka [24]
2 years ago
5

The airport has a new plane that can go from 0 miles per hour to 100 miles per hour in 10 seconds. What is being described about

the new plane?
Physics
1 answer:
Nonamiya [84]2 years ago
6 0
<h3 />

<h2 /><h2 /><h2>Acceleration : </h2>

Acceleration is a vector quantity that is defined as ❝ rate of change in velocity ❞

And as per the question, the plane can go from 0 miles to 100 miles/hour in 10 seconds that shows rate of change in its velocity. that is :

from 0 miles/hour to 100 miles/hour. hence Acceleration is the quantity being described about the new plane.

  • \qquad \sf  \dashrightarrow \: a =  \dfrac{v - u}{t}

[ v = final velocity, u = initial velocity, t = time ]

  • \qquad \sf  \dashrightarrow \: a =  \dfrac{100 - 0}{10}

  • \qquad \sf  \dashrightarrow \: a =  \dfrac{100}{10}

  • \qquad \sf  \dashrightarrow \: a = 10 \:

So, the Acceleration of the new plane is 10 miles/hour²

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A ball is dropped from rest at point O . It passes a window with height 3.8 m in time interval tAB = 0.02 s.Identify the correct
Taya2010 [7]

Answer:

VB − VA = g tAB   &   (VA + VB)/2 = h / tAB

Explanation:

s = h = Displacement

tAB = t = Time taken

VA = u = Initial velocity

VB = v = Final velocity

a = g = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{3.8-\frac{1}{2}\times 9.8\times 0.02^2}{0.02}\\\Rightarrow u=189.902\ m/s

v=u+at\\\Rightarrow v=189.902+9.8\times 0.02\\\Rightarrow v=190.098\ m/s

\frac{v+u}{2}=\frac{190.098+189.902}{2}=190\ s

\frac{h}{t}=\frac{3.8}{0.02}=190\ s

Hence, the equations VB − VA = g tAB   &   (VA + VB)/2 = h / tAB will be used

3 0
3 years ago
A boat traveling across a river has a resultant velocity of 10 km/h and travels 34 degrees with respect to the shore. A) What is
vodka [1.7K]

Answer:

a) 1.55 m/s

b) 2.3 m/s

Explanation:

We know that the boat travels across the river, if we define the river as the x-axis, then the velocity of the boat is only on the y-axis.

Then we can write the velocity of the boat in still water as:

S = (0, B)

Now, when the boat is on the river, the velocity of the boat will be equal to the velocity of the boat in still water plus the velocity of the river.

The velocity of the river is:

v = (R, 0).

Then the velocity of the boat in that river is:

V' = (0, B) + (R, 0) = (R, B)

Now, we know that the velocity of the boat is 10km/h, and it travels at an angle of 34° with respect to the shore.

We can use the Pythagoreans theorem to write the components of this velocity as:

x-axis component = 10km/h*cos(34°) = 8.29 km/h

y-axis component = 10km/h*sin(34°) =  5.59 km/h

Then the velocity of the boat can be written in components as:

velocity = ( 8.29 km/h,  5.59 km/h)

And we knew that the velocity of the boat was written as  (R, B)

Then we must have:

R = 8.29 km/h

B = 5.59 km/h

a) The speed of the boat in m/s:

We know that the speed of the boat is 5.59 km/h.

First, we know that:

1km = 1000m, then:

5.59 km/h = 5.59*(1000m)h = 5,590 m/h

And we know that:

1h = 3600s

Then we can write:

5,590 m/h = 5,590 m/(3600s) = 1.55 m/s

b) The speed of the river in m/s:

We know that the speed of the river is 8.29 km/h

Using the same reasoning as above, we can do the change of units as follows:

8.29 km/h = 8.29 (1000m)/(3600s) = 2.3 m/s

6 0
2 years ago
The charge per unit length on a long, straight filament is -92.0 μC/m. Find the electric field 10.0 cm above the filament.
Pepsi [2]

Answer:

E = 1.655 x 10⁷ N/C towards the filament

Explanation:

Electric field due to a line charge is given by the expression

E = [tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}[/tex]

where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is

8.85 x 10⁻¹².

Putting the given values in the equation given above

E = \frac{92\times10^{-6}}{2\times3.14\times8.85\times10^{-12}\times10^{-1}}

E = 1.655 x 10⁷ N/C

4 0
3 years ago
g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th
kati45 [8]

The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

<em>y</em> = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²

-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

<em>v</em>₀ = 3.12 m/s

3 0
2 years ago
7. Describe about the difficultites that would arise due to lack of non-uniformity in measurement. ​
r-ruslan [8.4K]

Answer:

Many difficulties would arise if there was a lack of uniformity in the measurement of various weights and measures between business, industry, individuals and countries. The biggest implications for a lack of uniformity are in health and safety, equity and sustainability.

Explanation:

palike nlng po

6 0
3 years ago
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