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3 years ago

3 A 100 g steel ball falls from a height of 1.8 m on to a metal plate and rebounds to a height of 1.25 m.

Physics

2 answers:

BigorU [14]3 years ago

6 0

Given values:

Mass of the steel ball, m = 100 g = 0.1 kg

Height of the steel ball, h1 = 1.8 m

Rebound height, h2 = 1.25 m

a. PE= mgh

0.1 x 9.8 x 1.8 =

1.764 Joules

b. KE = PE ->

1.764 Joules

c. KE= 1/2 mv square

so v = square root 2ke/m

square root 2 x 1.764/ 0.1

= 5.93 m/s

d. KE=PE=mgh square

0.1 x 9.8 x 1.21 =

1.186 joules

velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s

svetlana [45]3 years ago

3 0

It makes a distinct "clang", even if there is nobody there to hear it.

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Answer:

v = 15.65 m/s

Explanation:

We use conservation of mechanical energy between initial (i) and final (f) states:

Pi + KEi = Pf + KEf

At the top of the cave at the instant the bat starts to fall, there is only potential energy since the bat's velocity is zero.

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and the KEi = 0 J (no velocity)

Knowing the height of the cave's roof (12.8 m) , we can find the mass of the bat:

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Using conservation of mechanical energy, the final state is:

Pf + KEf = 600 J

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3 years ago

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Answer:

False.

Explanation:

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4 years ago

Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha

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The density of seawater at a depth where the pressure is 500 atm is $1124kg/m^3$

Explanation:

The relationship between bulk modulus and pressure is the following:

$B=\rho_0 \frac{\Delta p}{\Delta \rho}$

where

B is the bulk modulus

$\rho_0$ is the density at surface

$\Delta p$ is the variation of pressure

$\Delta \rho$ is the variation of density

In this problem, we have:

$B=2.3\cdot 10^9 N/m^2$ is the bulk modulus

$\rho_0 =1100 kg/m^3$

$\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa$ is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

$\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3$

Learn more about pressure in a fluid:

brainly.com/question/9805263

#LearnwithBrainly

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3 years ago

3 A 100 g steel ball falls from a height of 1.8 m on to a metal plate and rebounds to a height of 1.25 m.​

3 A 100 g steel ball falls from a height of 1.8 m on to a metal plate and rebounds to a height of 1.25 m.