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3 years ago

According to Galileo, this quantity is not needed to keep a body in motion under ideal conditions.

2 answers:

8 0

This condition is called Galileo's Law of Inertia which states that all bodies accelerate at the smart rate , no matter what are their masses or size. Inertia is that tendency of matter to resist changes in its velocity. <span>Isaac Newton's first law of motion captures the concept of inertia. </span>

nordsb [41]3 years ago

3 0

Answer:

According to Galileo, quantity does not matter under ideal conditions. The ideal conditions to experiment under is when there is no air resistance only then it can be seen that the statement is true i.e. quantity is not a factor

If you drop a ball and a feather under the ideal condition i.e. no air resistance you will see that both the ball and feather which weigh different will fall at the same time

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Consider this statement: Air is matter. Which facts best support the statement?

lyudmila [28]

A. Balloons can be filled with air.
C. Air has mass.

7 0

3 years ago

Read 2 more answers

Does sound travel much faster than light ?

Kitty [74]

Answer:

nothing travels faster than light

Example:

You’ll always see lightning before you hear it, because typically lightning will be a mile away, two miles away.

4 0

3 years ago

The velocities of light in air and glass are 3.0 x 10^8ms and 2.0×10^8ms respectively. If the angle of refraction is 30°, the si

JulijaS [17]

Answer:

0.75

Explanation:

$refractive \: index \: = \frac{3.0 \times {10}^{2} }{2.0 \times {10}^{2} }$

= 1.5

$refractive \: index = \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) }$

$1.5 = \ \frac{ \sin(i) }{ \sin(30) }$

1.5 × ½ = sin(i)

$\sin(i) = 0.75$

5 0

3 years ago

One object (m1 = 0.220 kg) is moving to the right with a speed of 2.10 m/s when it is struck from behind by another object (m2 =

blagie [28]

Answer:

vf₁ = 6.86 m/s , to the right

vf₂ = 2.96 m/s, to the right

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

p=m*v

where

p:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

P₀ :Initial linear momentum quantity

Pf : Final linear momentum quantity

Data

m₁= 0.220 kg : mass of object₁

m₂= 0.345 kg : mass of object₂

v₀₁ = 2.1 m/s ₁ , to the right : initial velocity of m₁

v₀₂= 6 m/s, to the right i :initial velocity of m₂

Problem development

We appy the formula (1):

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂

We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:

(0.22)*(2.1) + (0.345)*(6) = (0.22)*vf₁ +(0.345)*vf₂

2.532 = (0.22)*vf₁ +(0.345)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

$e= \frac{v_{f2}-v_{f1} }{v_{o1} -v_{o2} }$

1*(v₀₁ - v₀₂ ) = (vf₂ -vf₁)

(2.1 - 6 ) = (vf₂ -vf₁)

-3.9 = (vf₂ -vf₁)

vf₂ = vf₁ - 3.9

vf₂ = vf₁ - 3.9 Equation (2)

We replace Equation (2) in the Equation (1)

2.532 = (0.22)*vf₁ +(0.345)*( vf₁ - 3.9)

2.532 = (0.22)*vf₁ +(0.345)* (vf₁) -(0.345)( 3.9)

2.532 + 1.3455 = (0.565)*vf₁

3.8775 = (0.565)*vf₁

vf₁ = (3.8775) / (0.565)

vf₁ = 6.86 m/s, to the right

We replace vf₁ = 6.86 m/s in the Equation (2)

vf₂ = 6.86 - 3.9

vf₂ = 2.96 m/s, to the right

8 0

3 years ago

The three components of velocity in a velocity field are given by u = Ax + By + Cz, v = Dx + Ey + Fz, and w = Gx + Hy + Jz. Dete

Alexxandr [17]

Answer:

The relationship is only between the coefficients A, E and J which is:

$A + E + J = 0$ . The remaining coefficients can be anything without any constraints.

Explanation:

Given:

The three components of velocity is a velocity field are given as:

$u = Ax + By + Cz\\\\v = Dx + Ey + Fz\\\\w = Gx + Hy + Jz$

The fluid is incompressible.

We know that, for an incompressible fluid flow, the sum of the partial derivatives of each component relative to its direction is always 0. Therefore,

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0$

Now, let us find the partial derivative of each component.

$\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(Ax+By+Cz)\\\\\frac{\partial u}{\partial x}=A+0+0=A\\\\\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(Dx+Ey+Fz)\\\\\frac{\partial v}{\partial y}=0+E+0=E\\\\\frac{\partial w}{\partial z}=\frac{\partial }{\partial z}(Gx+Hy+Jz)\\\\\frac{\partial w}{\partial z}=0+0+J=J$

Hence, the relationship between the coefficients is:

$A+E+J=0$

There is no such constraints on other coefficients. So, we can choose any value for the remaining coefficients B, C, D, F, G and H.

6 0

3 years ago