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2 years ago

DIylpst any three properties of 9.

1 answer:

7 0

Answer:Commutative property of multiplication: Changing the order of factors does not change the product. For example, 4 \times 3 = 3 \times 44×3=3×44, times, 3, equals, 3, times, 4.

Associative property of multiplication: Changing the grouping of factors does not change the product. For example, (2 \times 3) \times 4 = 2 \times (3 \times 4)(2×3)×4=2×(3×4)left parenthesis, 2, times, 3, right parenthesis, times, 4, equals, 2, times, left parenthesis, 3, times, 4, right parenthesis.

Identity property o

Explanation:

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The properties of elements in compounds are usually _______ the properties of the elements alone.

KATRIN_1 [288]

are usually different from the property. only mixture takes the similar property of its constituent element

7 0

3 years ago

If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?

Lena [83]

Answer:2 volts

Explanation:

Power=current x voltage

6=3 x voltage

Divide both sides by 3

6/3=(3 x voltage)/3

2=voltage

Voltage=2volts

5 0

3 years ago

A fisherman casts his lure at an angle of 33 degrees above the horizontal. The lure reaches a maximum height of 2.3 m. Assuming

omeli [17]

Answer:

12.32 m/s

Explanation:

Using the formula of maximum height of a projectile,

H = u²sin²Ф/2g................... Equation 1

Where H = maximum height, u = initial velocity, Ф = angle of projection, g = acceleration due to gravity

make u the subject of the equation

u = √(2Hg/sin²Ф)............ Equation 2

Given: H = 2.3 m, Ф = 33°, g = 9.8 m/s²

Substitute into equation 2

u = √[(2×2.3×9.8)/sin²33°]

u =√ [45.08/(0.545)²]

u = 45.08/0.297

u = √(151.785)

u = 12.32 m/s

3 0

3 years ago

A glass ball of radius 3.74 cm sits at the bottom of a container of milk that has a density of 1.04 g/cm3. The normal force on t

Gelneren [198K]

Answer:

The mass of the ball is 0.23 kg

Explanation:

Given that

radius ,r= 3.74 cm

Density of the milk ,ρ = 1.04 g/cm³ = 1.04 x 10⁻³ kg/cm³

Normal force ,N= 9.03 x 10⁻² N

The volume of the ball V

$V=\dfrac{4}{3}\pi r^3$

$V=\dfrac{4}{3}\times \pi \times 3.74^3\ cm^3$

V= 219.13 cm³

The bouncy force on the ball = Fb

Fb = ρ V g

Fb + N = m g

m=Mass of the ball = Density x volume

m = γ V , γ =Density of the Ball

ρ V g + N = γ V g ( take g= 10 m/s²)

$\gamma =\dfrac{N+\rho V g}{V g}$

$\gamma =\dfrac{9.03\times 10^{-2}+1.04\times 10^{-3}\times 219.13\times 10}{219.13\times 10}$

γ = 0.00108 kg/cm³

m = γ V

m = 0.00108 x 219.13

m= 0.23 kg

The mass of the ball is 0.23 kg

5 0

3 years ago

Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

$h = \frac{v^{2}}{2g}$

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

$\frac{mv^{2}}{2} = mgh$

$h = \frac{v^{2}}{2g}$

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

7 0

2 years ago

DIylpst any three properties of 9.​

DIylpst any three properties of 9.