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3 years ago

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A student throws a rock horizontally off a 5.0 m tall building. The rock's initial speed is 6.0 m/s. How long will it take the r

ock to reach the ground? Air resistance is negligible.'

2 answers:

Archy [21]3 years ago

5 0

Answer:

The time taken by the rock to reach the ground is 0.569 seconds.

Explanation:

Given that,

A student throws a rock horizontally off a 5.0 m tall building, s = 5 m

The initial speed of the rock, u = 6 m/s

We need to find the time taken by the rock to reach the ground. Using second equation of motion to find it. We get :

$s=ut+\dfrac{1}{2}gt^2\\\\5=6t+\dfrac{1}{2}\times 9.8t^2\\\\t=0.569\ seconds$

So, the time taken by the rock to reach the ground is 0.569 seconds. Hence, this is the required solution.

trapecia [35]3 years ago

5 0

Answer:

The rock to reach the ground in 0.569 sec.

Explanation:

Given that,

Height = 5.0 m

Speed = 6.0 m/s

A student throws a rock horizontally off a 0.5 m tall building.

This is the distance.

So, s = 0.5 m

We need to calculate the time

Using equation of motion

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

Where, s = distance

t = time

g = acceleration due to gravity

Put the value into the formula

$5.0=6.0\times t+0.5\times9.8t^2$

$4.9t^2+6.0 t-5.0 =0$

$t = 0.569\ sec$

Neglect the negative value.

Hence, The rock to reach the ground in 0.569 sec.

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Suppose light from a 632.8 nm helium-neon laser shines through a diffraction grating ruled at 520 lines/mm. How many bright line

Leya [2.2K]

Answer:

1 bright fringe every 33 cm.

Explanation:

The formula to calculate the position of the m-th order brigh line (constructive interference) produced by diffraction of light through a diffraction grating is:

$y=\frac{m\lambda D}{d}$

where

m is the order of the maximum

$\lambda$ is the wavelength of the light

D is the distance of the screen

d is the separation between two adjacent slit

Here we have:

$\lambda=632.8 nm = 632.8\cdot 10^{-9} m$ is the wavelength of the light

D = 1 m is the distance of the screen (not given in the problem, so we assume it to be 1 meter)

$n=520 lines/mm$ is the number of lines per mm, so the spacing between two lines is

$d=\frac{1}{n}=\frac{1}{520}=1.92\cdot 10^{-3} mm = 1.92\cdot 10^{-6} m$

Therefore, substituting m = 1, we find:

$y=\frac{(632.8\cdot 10^{-9})(1)}{1.92\cdot 10^{-6}}=0.330 m$

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6 0

3 years ago

A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanba

Phantasy [73]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

$m_1$ = Mass of bean bag = 0.354 kg

$m_2$ = Mass of empty crate = 3.77 kg

$v_1$ = Speed of the bean bag

$v_2$ = Speed of the crate

Acceleration

$a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g$

$a=--9.81\times 0.48=4.7088\ m/s^2$

From equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s$

In this system the momentum is conserved

$m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s$

The speed of the bean bag is 31.42383 m/s

8 0

3 years ago

Find the momentum of a particl with a mass of one gram moving with half the speed of light.

joja [24]

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

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To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

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p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

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2 years ago