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2 years ago

8

The air speed of a plane is defined as its velocity with respect to the surrounding air, or in other words how fast the plane wo

uld move in still air. The ground speed is the speed of the plane as measured from the ground. A Nimbus-4 glider is traveling at an air speed of 144 km/h. a) What is the glider’s ground speed if it is experiencing a 48 km/h headwind (wind coming from the front)?

1 answer:

amid [387]2 years ago

7 0

Answer:

3

Explanation:

because

You might be interested in

Developed countries typically have _______.

Anna35 [415]

Answer:

Good government

Explanation:

Many people will think its natural resources but its not. Africa is a proof. They have resources but leadership is what makes most of their countries 3rd world. #BADGOVERNANCE

8 0

2 years ago

A stone is thrown vertically into the air at an initial velocity of 96 ft/s. On Mars, the height s (in feet) of the stone above

vladimir1956 [14]

Answer:

240 ft

Explanation:

t = Time taken

u = Initial velocity = 96 ft/s

v = Final velocity

s = Displacement

a = Acceleration = 12 m/s² on Mars 32 ft/s² on Earth negative due to upward direction

Mars

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -12\times t^2\\\Rightarrow s=96t-6t^2\ ft$

Earth

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -32\times t^2\\\Rightarrow s=96t-16t^2\ ft$

Differentiating the first equation with respect to time we get

$\frac{ds}{dt}=96-12t$

Equating with zero

$0=96-12t\\\Rightarrow t=\frac{96}{12}=8\ s$

Differentiating the second equation with respect to time we get

$\frac{ds}{dt}=96-32t$

Equating with zero

$0=96-32t\\\Rightarrow t=\frac{96}{32}=3\ s$

Applying the time taken to the above equations, we get

$s=96t-6t^2\ ft\\\Rightarrow s=96\times 8-6\times 8^2\\\Rightarrow s=384$

$s=96t-16t^2\\\Rightarrow s=96\times 3-16\times 3^2\\\Rightarrow s=144$

Difference in height = 384-144 = 240 ft

The stone will travel 240 ft higher on Mars

6 0

3 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$

$\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}$

Using the formulas for $V_{ox}, V_{oy}:$

$\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}$

Simplifying

$4cos\theta sin\theta=3sin^2\theta$

Dividing by $sin\theta$

$4cos\theta=3sin\theta$

Rearranging

$tan\theta=\frac{4}{3}$

$\theta=arctan\frac{4}{3}$

$\theta=53.13^o$

4 0

3 years ago

Question 1 of 10

Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field $E$ at a distance $R$ from a charge $Q$ is given by

$E = k\dfrac{Q}{R^2}$

where $k = 9*10^9Nm/C$ is the coulomb's constant.

Now, in our case

$R = 0.0075m$

$Q = 0.0035C$ ;

therefore,

$E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}$

$\boxed{E = 5.6*10^{11}N/C.}$

which is choice C from the options given<em> (at least it resembles it).</em>

6 0

3 years ago

a distant galaxy is studied with a radio telescope x ray telescope and optical light telescope. the images form which set of mus

aivan3 [116]

Answer:

The modern instruments or we can say the different levels of telescopes are used to explore and study the distant galaxies. i.e the Hubble telescope is out there providing the data regarding the different properties of the celestial entities which in other case is not visible to the human naked eye.

Explanation:

Scientists and research workers are in constant search for more answers as they explore the universe and implement the laws of physics on the celestial entities. But, most of the objects inside the universe are not visible to human naked eye, as they are far from sight and thus more advanced form of instruments like the x-ray, optical, and light telescopes are used to determine the different properties of the celestial entities inside the universe.

As, these telescopes includes the most recent "Hubble telescope", which is out there inside the space to explore the universe and more over the galaxies by subjecting them with x-rays and then provide us with a very rough but valid results to study the distant galaxies.

6 0

3 years ago