What happens when a boxer doesn't make weight?

On Mars, the acceleration due to gravity is 3.77 m / s 2 . 3.77 m/s2. How far would a 15 g 15 g rock fall from rest in 2.5 s 2.5

Zinaida [17]

Answer:

s = 11.78 m

Explanation:

given,

acceleration due to gravity, g = 3.77 m/s²

mass of the rock = 15 g

time = 2.5 s

distance traveled = ?

using equation of motion

$s = ut +\dfrac{1}{2}at^2$

initial speed = 0 m/s

$s = \dfrac{1}{2}at^2$

$s = \dfrac{1}{2}\times 3.77 \times 2.5^2$

s = 11.78 m

distance traveled by the rock is equal to 11.78 m.

7 0

3 years ago

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3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of

Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

$n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555$

Again using Snell's law for blue light as :

$n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283$

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0

3 years ago

A certain wave motion has a frequency of 10 cycles / s and a wavelength of 3 m. so its propagation speed is

kodGreya [7K]

3.5m is ur answer ask for more questions anytime

5 0

3 years ago

A gas, behaving ideally, has a pressure P1 and at a volume V1. The pressure of the gas is changed to P2. Using Avogadro’s, Charl

Bond [772]

Answer:

Boyle's Law

$\therefore P_1.V_1=P_2.V_2$

Explanation:

Given that:

initially:

pressure of gas, $= P_1$

volume of gas, $= V_1$

finally:

pressure of gas, $= P_2$

volume of gas, $= V_2$

To solve for final volume $V_2$

According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.

According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.

But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.

Here nothing is said about the temperature, so we consider the Boyle's Law which states that at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.

Mathematically:

$P_1\propto \frac{1}{V_1}$