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3 years ago

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A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

that the acceleration due to gravity is -32 feet per second squared, how high is the stone 6 seconds later?

1 answer:

Black_prince [1.1K]3 years ago

4 0

<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y$ is the height of the stone at 6s (the value we want to find)

$y_{o}=925ft$ is the initial height of the stone

$V_{o}=20ft/s$ is the initial velocity of the stone

$t=6s$ is the time at which we need to find the height

$g=32ft/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $y$ from (1):

$y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2}$ (2)

Finally:

$y=469ft$ This is the height of the stone at t=6s

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During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient

ElenaW [278]

a) $1.58\cdot 10^6 J$

b) $1.15\cdot 10^6 J$

c) $0.43\cdot 10^6 J$

d) 35.8 m/s

Explanation:

a)

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

$U=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

$g=9.8 m/s^2$

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

$U=(670)(9.8)(240)=1.58\cdot 10^6 J$

b)

The energy transferred to thermal energy during the slide is equal to the work done by friction, which is:

$W=F_f d$

where

$F_f$ is the force of friction

d = 740 m is the displacement of the rock along the ramp

The force of friction is given by:

$F_f=-\mu mg cos \theta$

where

$\mu=0.25$ is the coefficient of friction

m = 670 kg is the mass of the rock

$\theta$ is the angle of the ramp

Since we know the lenght of the ramp (d = 740 m) and the height (h = 240 m), we can find the angle:

$\theta=sin^{-1}(\frac{h}{d})=sin^{-1}(\frac{240}{740})=18.9^{\circ}$

Therefore, the work done by friction is:

$W=-\mu m g cos \theta d =-(0.25)(670)(9.8)(cos 18.9^{\circ})(740)=-1.15\cdot 10^6 J$

So, the energy transferred to thermal energy is $1.15\cdot 10^6 J$ .

c)

According to the law of conservation of energy, the kinetic energy of the rock as it reaches the bottom of the hill will be equal to the initial potential energy (at the top) minus the energy transformed into thermal energy.

Therefore, we have:

$K_f = U_i -E_t$

where here we have:

$U_i=1.58\cdot 10^6 J$ is the potential energy of the rock at the top of the hill

$E_t=1.15\cdot 10^6 J$ is the energy converted into thermal energy

Substituting, we find

$K_f=1.58\cdot 10^6-1.15\cdot 10^6=0.43\cdot 10^6 J$

So, this is the kinetic energy of the rock at the bottom of the hill.

d)

The kinetic energy of the rock at the bottom of the hill can be rewritten as

$K_f=\frac{1}{2}mv^2$

where

m is the mass of the rock

v is its final speed

In this problem, we have:

$K_f=0.43\cdot 10^6 J$ is the final kinetic energy of the hill

m = 670 kg is the mass of the rock

Therefore, the final speed of the rock is:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.43\cdot 10^6)}{670}}=35.8 m/s$

7 0

4 years ago

Which is an example of radiation?

oksano4ka [1.4K]

I believe The snowman should be the answer

5 0

3 years ago

Read 2 more answers

A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

elena55 [62]

Answer:

$3.2\times 10^{-7}\ m$ or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = $9.38\times 10^{14}\ Hz$

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = $3\times 10^8\ m/s$

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

$v=f\lambda$

Now, expressing the above equation in terms of wavelength 'λ', we have:

$\lambda=\frac{v}{f}$

Now, plug in the given values and solve for 'λ'. This gives,

$\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m$

Therefore, the wavelength of the radiations absorbed by the ozone is nearly $3.2\times 10^{-7}\ m$ or 0.32 μm.

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3 years ago

A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a

olga nikolaevna [1]

Answer:

4 hoop, disk, sphere

Explanation:

Because

We are given data that

Hoop, disk, sphere have Same mass and radius

So let

And Initial angular velocity, = 0

The Force on each be F

And Time = t

Also let

Radius of each = r

So let's find the inertia shall we!!

I1 = m r² /2

= 0.5 mr² the his is for dis

I2 = m r² for hoop

And

Moment of inertia of sphere wiil be

I3 = (2/5) mr²

= 0.4 mr²

So

ωf = ωi + α t

= 0 + ( τ / I ) t

= ( F r / I ) t

So we can see that

ωf is inversely proportional to moment of inertia.

And so we take the

Order of I ( least to greatest ) :

I3 (sphere) , I1 (disk) , I2 (hoop) , ,

Order of ωf: ( least to greatest)

That of omega xf is the reverse of inertial so

hoop, disk, sphere

Option - 4

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4 years ago

About ________ percent of wais scores fall between 70 and 130.

stiv31 [10]

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