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3 years ago

a hammer drops from a height of 8 meters. calculate the speed with which it hits the ground. show work

Physics

1 answer:

ioda3 years ago

3 0

Answer:

12.5 m/s

Explanation:

The motion of the hammer is a free fall motion, so a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2-u^2=2as$

Where, taking downward as positive direction, we have:

s = 8 m is the displacement of the hammer

u = 0 is the initial velocity (it is dropped from rest)

v is the final velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving the equation for v, we find the final velocity:

$v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(8)}=12.5 m/s$

So, the final speed is 12.5 m/s.

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Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person

sertanlavr [38]

Answer:

v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

f ’= f (v + v₀) / (v- $v_{s}$ )

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

v₀ = v_{s} = v ’

we substitute

f ’= f (v + v’) / (v - v ’)

f ’/ f (v-v’) = v + v ’

v (f ’/ f -1) = v’ (1 + f ’/ f)

v ’= (f’ / f-1) / (1 + f ’/ f) v

v ’= (f’-f) / (f + f’) v

let's calculate

v ’= (3400 -3000) / (3000 +3400) 343

v ’= 400/6400 343

v ’= 21.44 m / s

3 0

2 years ago

Find an expression for the square of the orbital period.

jenyasd209 [6]

Answer:

T²= 4π²R³/GM

Explanation:

First we know that

Fg= Fc

Because centripetal force must equal gravitational force

GMm/R² = Mv²/R

But velocity is 2πR/T

So by substitution we have

GMm/R²= M (2πR/T)/T

We have

T²= 4π²R³/GM as period

8 0

2 years ago

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unch

Alenkinab [10]

Answer:

Potential difference and charge will also increase.

Explanation:

Asking that :

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unchanged?

The charge is directly proportional to area of the plate. That is, increase in area of the plate of a capacitor will lead to the increase in the charges between the plates.

And since charge is also proportional to the magnitude of potential difference between the plates from the definition of capacitance of a capacitor which says that:

Q = CV

Therefore, increase in the area of the plate will also lead to increase in potential difference between the plates.

Therefore, if the plate area were increased while the plate separation remains unchanged, the charge and potential difference between them will also increase.

5 0

3 years ago

A sound from a source has an intensity of 270 dB when it is 1 m from the source.

Rufina [12.5K]

Remember that sound intensity decreases in inverse proportion to the distance squared. So, to solve this we are going to use the inverse square formula: $\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
where
$I_{2}$ is the intensity at distance 2
$I_{1}$ is the intensity at distance 1
$d_{2}$ is distance 2
$d_{1}$ is distance 1

We can infer for our problem that $I_{1}=270$ , $d_{1}=1$ , and $d_{2}=3$ . Lets replace those values in our formula to find $I_{2}$ :
$\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
$\frac{I_{2} }{270} =( \frac{1}{3} )^2$
$\frac{I_{2} }{270} = \frac{1}{9}$
$I_{2}= \frac{270}{9}$
$I_{2}=30$ dB

We can conclude that the intensity of the sound when is <span>3 m from the source is 30 dB.</span>

8 0

3 years ago

A rope pulls on a metal box at an angle of 60.0° with a force of 255 N. The box moves horizontally for 10.0 m. How much work was

N76 [4]

w = F•s•cos θ

= 255•10•cos 60°

= 255•10•½

= 1275 joule

7 0

3 years ago