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Schach [20]
2 years ago
8

21

Physics
1 answer:
sergeinik [125]2 years ago
5 0
Distance between the centers of the two objects are the second object
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What are the formulas to calculate acceleration
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In middle school, the formula you'll use most often when you're
working with acceleration is . . .

Acceleration = (change in speed during some time) / (time for the change)


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3 years ago
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What is the magnitude of the linear momentum of a 7.30 kg bowling ball going down the
padilas [110]
Momentum = mass x velocity
m=7.3kg
v=20 m/s

momentum= 7.3(20) = 146 kg m/s
3 0
3 years ago
2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the
frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

 

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum =  2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = \frac{1}{2} mv²

v² = 2gh₁  

v = √(2gh₁)

from the second equation; s = ut +  \frac{1}{2} at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )    

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

 

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{3}mR²) ω²

v = √( \frac{6}{5}gh₁ )

so we substitute √( \frac{6}{5}gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( \frac{6}{5}gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0
3 years ago
______ is the percent ionization of a weak acid in water increases as the concentration of acid decreases.
Nuetrik [128]

The ratio of concentration of ionized acid to the initial concentration of acid multiplied by 100 will give the percent ionization of a weak acid in water increases as the concentration of acid decreases.

Explanation:

Percent ionization is used for quantifying the number of ions present in the weak acid when dissolved in a solution. So it is similar to the pKa value. The percent ionization value can be determined as negative log of dissociation constant. Also the as the number of ions increases in weak acid, the concentration of acid will be decreasing . It can be calculated using the formula for percent ionization as follows:

Percent ionization = \frac{Ionized acid concentration}{Initial concentration of acid} * 100

As the water volume or concentration increases, the acid will get diluted much more thus leading to decrease in the concentration of acid.

So the ratio of concentration of ionized acid to the initial concentration of acid multiplied by 100 will give the percent ionization of a weak acid in water increases as the concentration of acid decreases.

7 0
2 years ago
For a sound coming from a point source, the amplitude of the sound is inversely
gayaneshka [121]

Answer ) Sound level equation

The intensity of a sound wave is related to its amplitude squared by the following relationship: I=(Δp)22ρvw I = ( Δ p ) 2 2 ρ v w . Here Δp is the pressure variation or pressure amplitude (half the difference between the maximum and minimum pressure in the sound wave) in units of pascals (Pa) or N/m2.

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2 years ago
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