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Keith_Richards [23]

3 years ago

12

A string attached to an airborne kite was maintained at an angle of 40.0 with the ground. if 120m of string was reeled in to ret

urn the kite back to the ground. what was the horizontal displacement of the kite

1 answer:

AleksAgata [21]3 years ago

5 0

The hypotenuse is measured at 120 meters of string, and you need to solve for the leg of the triangle that is horizontal. The degree is 40, so use trigonometry to figure it out.
Cosin (40) is equal to around .766
Adjacent/Hypotenuse
x/120 = cos40
Answer: 91.92533.
If you use 3 significant figures it should be 91.9 meters.

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A proton is confined to a space 1 fm wide (about the size of an atomic nucleus). What's the minimum uncertainty in its velocity?

Daniel [21]

Answer:

Minimum uncertainty in velocity of a proton, $\Delta v\ge 3.15\times 10^7\ m/s$

Explanation:

It is given that,

A proton is confined to a space 1 fm wide, $\Delta x=10^{-15}\ m$

We need to find the minimum uncertainty in its velocity. We know that the Heisenberg Uncertainty principle gives the uncertainty between position and the momentum such that,

$\Delta p.\Delta x\ge \dfrac{h}{4\pi}$

Since, p = mv

$\Delta (mv).\Delta x\ge \dfrac{h}{4\pi}$

$m \Delta v.\Delta x\ge \dfrac{h}{4\pi}$

$\Delta v\ge \dfrac{h}{4\pi m\Delta x}$

$\Delta v\ge \dfrac{6.63\times 10^{-34}}{4\pi \times 1.67\times 10^{-27}\times 10^{-15}}$

$\Delta v\ge 3.15\times 10^7\ m/s$

So, the minimum uncertainty in its velocity is greater than $3.15\times 10^7\ m/s$ . Hence, this is the required solution.

3 0

3 years ago

Is baking soda less dense or more dense than water?

boyakko [2]

It’s more dense than air and less dense than liquid!

6 0

2 years ago

Which of these objects must have the greatest force acting on it? pls help fast

8_murik_8 [283]

Let's look at Newton's second law

F=ma

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If mass is more force will be more.

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8 0

1 year ago

Suppose the coefficient of static friction between the road and the tires on a car is 0.638 and the car has no negative lift. Wh

larisa [96]

Answer:

12.6332454263 m/s

Explanation:

m = Mass of car

v = Velocity of the car

$\mu$ = Coefficient of static friction = 0.638

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of turn = 25.5 m

When the car is on the verge of sliding we have the force equation

$\dfrac{mv^2}{r}=\mu mg\\\Rightarrow v=\sqrt{\mu gr}\\\Rightarrow v=\sqrt{0.638\times 9.81\times 25.5}\\\Rightarrow v=12.6332454263\ m/s$

The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s

4 0

3 years ago

Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.

pogonyaev

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is $D = 0.59 \ m$

Explanation:

From the question we are told that

The best resolution is $\theta = 0.3 \ arcsecond$

The wavelength is $\lambda = 700 \ nm = 700 *10^{-9 } \ m$

Generally the

1 arcminute = > 60 arcseconds

=> x arcminute => 0.3 arcsecond

So

$x = \frac{0.3}{60 }$

=> $x = 0.005 \ arcminutes$

Now

60 arcminutes => 1 degree

0.005 arcminutes = > z degrees

=> $z = \frac{0.005}{60 }$

=> $z = 8.333 *10^{-5} \ degree$

Converting to radian

$\theta = z = 8.333 *10^{-5} * 0.01745 = 1.454 *10^{-6} \ radian$

Generally the resolution is mathematically represented as

$\theta = \frac{1.22 * \lambda }{ D}$

=> $D = \frac{1.22 * \lambda }{\theta }$

=> $D = \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }$

=> $D = 0.59 \ m$

4 0

2 years ago