All categories

3 years ago

An ant crawling on a sphere is which type of motion - 1d 2d or 3d

Physics

2 answers:

Wittaler [7]3 years ago

8 0

The surface of a sphere is a 3D shape. So an ant crawling on it is moving in 3D.

Phoenix [80]3 years ago

4 0

The ant crawling on a sphere is 3D motion.

You might be interested in

How would the amount of carbon dioxide in the atmosphere change if they were no plants?

Flauer [41]

There would be significantly more CO2 in the atmosphere because plants take in CO2 during photosynthesis and fix the carbon into glucose.

8 0

3 years ago

A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla

kotegsom [21]

Answer:

The power output of the oscillator is 0.350 watt.

Explanation:

Given that,

Diameter = 1.0 mm

Tension = 5.7 N

Frequency = 57.0 Hz

Amplitude = 0.54 cm

We need to calculate the power output of the oscillator

Using formula of the power

$P=\dfrac{1}{2}\times\mu\times\omega^2\times a^2\times v$

Put the value into the formula

$P=\dfrac{1}{2}\times A\times\rho\times\omega^2\times a^2\times\dfrac{\sqrt{T}}{\mu}$

$P=\dfrac{1}{2}\times3.14\times(0.0005)^2\times7850\times(2\times\pi\times57.0)^2\times(0.54\times10^{-2})^2\times\sqrt{\dfrac{5.7}{3.14\times(0.0005)^2\times7850}}$

$P=0.350\ Watt$

Hence, The power output of the oscillator is 0.350 watt.

3 0

3 years ago

Which of the following is an example of acceleration? I. A car speeds up. II. A car slows down. III. A car travels in a straight

svetlana [45]

I., II., and IV. are examples of acceleration. III. isn't.

6 0

4 years ago

A basketball player standing up with the hoop launches the ball straight up with an initial velocity of v_o = 3.75 m/s from 2.5

denis23 [38]

Answer:

a) The maximum height the ball will achieve above the launch point is 0.2 m.

b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.

Explanation:

For the height reached, we use 3rd equation of motion:

2gh = Vf² - Vo²

Here,

Vo = 3.75 m/s

Vf = 0m/s, since ball stops at the highest point

g = -9.8 m/s² (negative sign for upward motion)

h = maximum height reached by ball

therefore, eqn becomes:

2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²

h = 0.2 m

To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:

2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²

(Vo)² = 19.6 m²/s²

Vo = √19.6 m²/s²

Vo = 4.43 m/s

Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)

Vo = 0.174 in/ms

6 0

3 years ago

A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp

Anna71 [15]

$\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}$

Explanation:

Natural length of a spring is $0.5\text{ }m$ . The spring is streched by $0.02\text{ }m$ . The resultant energy of the spring is $0.5\text{ }J$ .

The potential energy of an ideal spring with spring constant $k$ and elongation $x$ is given by $\dfrac{1}{2}kx^{2}$ .

So, in the current problem, the natural length of the spring is not required to find the spring constant $k$ .

$\text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}$

∴ The spring constant of the spring = $2500\text{ }\frac{kg}{s^{2}}$

4 0

4 years ago