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Basile [38]
3 years ago
6

A tire swing hanging from a branch reaches nearly to the ground. How could you estimate the height of the branch using only a st

opwatch?
Physics
1 answer:
AlexFokin [52]3 years ago
4 0

Answer:

Explanation:

With the help of expression of  time period of pendulum we can calculate the height of the branch . The swinging tire can be considered equivalent to swinging bob of a pendulum . Here length of pendulum will be equal to height of branch .

Let it be h . Let the time period of swing of tire be T then from the formula of time period of pendulum

T = 2\pi\sqrt{\frac{l}{g} }  where l is length of pendulum .

here l = h so

T = 2\pi\sqrt{\frac{h}{g} }  

h = \frac{T^2g}{4\pi^2}

If we calculate the time period of swing of tire , we can calculate the height of branch .

The time period of swing of tire can be estimated with the help of a stop watch . Time period is time that the tire will take in going from one extreme point to the other end and then coming back . We can easily estimate it with the help of stop watch .

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Two 0.50 g spheres are charged equally and placed 2.5 cm apart. When released, they begin to accelerate at 170 m/s^2 .What is th
vitfil [10]

Answer:

q=7.65*10^{-8}C

Explanation:

Using Newton's second law, we calculate the magnitude of the electric force between the spheres:

F=ma\\F=0.5*10^{-3}kg(170\frac{m}{s^2})\\F=0.085N

The magnitude of the charge in both spheres is the same. So, we calculate the charge, using Coulomb's law:

F=\frac{kq^2}{d^2}\\q=\sqrt\frac{Fd^2}{k}\\q=\sqrt\frac{(0.085N)(2.5*10^{-2}m)^2}{8.99*10^9\frac{N\cdot m^2}{C^2}}\\q=7.65*10^{-8}C

8 0
3 years ago
The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr
Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by  the following equation:

dsin\theta_{n}=n\lambda   (1)

Where:

d is the width of the slit

\lambda  is the wavelength of the light

n is an integer different from zero.

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda   (2)

Now we have to find the value of \theta_{1}:

sin\theta_{1}=\frac{\lambda}{d}  

\theta_{1}=arcsin(\frac{\lambda}{d})   (3)

We know:

\lambda=104nm=104(10)^{-9}m

In addition we are told the diffraction grating has 5000 slits per mm, this means:

d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}

Substituting the known values in (3):

\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})

\theta_{1}=arcsin(0.52)

<u>Finally:</u>

\theta_{1}=31.33\º >>>This is the first-order diffraction angle

4 0
2 years ago
When a physical change in a sample occurs what does not change
Lelu [443]

Answer:

When a physical change in a sample occurs, composition of the sample does not change. It stays the same. It stays the same. Also, the properties of the sample will still be the same

5 0
2 years ago
Hi, I am having some difficulties in solving this question. Could someone please explain this question to me in detail. The thin
aniked [119]

Answer:

a) 3.0×10⁸ m

b) 0 m

Explanation:

Displacement is the distance from the starting position to the final position.

a) In half a year, the Earth travels from one point on the circle to the point on the exact opposite side of the circle (from 0° to 180°).  The distance between the points is the diameter of the circle.

x = 2r

x = 2 (1.5×10⁸ m)

x = 3.0×10⁸ m

b) In a full year, the Earth travels one full revolution, so it ends up back where it started.  The displacement is therefore 0 m.

8 0
3 years ago
Jak możemy służyć innym ludziom. Pilne
Troyanec [42]
Could u put it in english u can use goggle translation if u do not know how to
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