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3 years ago

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Which of the following is not a potential sign of chemical change? a. release of gas c. change of color b. evaporation of water

d. production of gas Please select the best answer from the choices provided A B C D

1 answer:

vitfil [10]3 years ago

5 0

B is the answer. because it is a physical change

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Physics double pivot question

andriy [413]

Explanation:

Assuming the wall is frictionless, there are four forces acting on the ladder.

Weight pulling down at the center of the ladder (mg).

Reaction force pushing to the left at the wall (Rw).

Reaction force pushing up at the foot of the ladder (Rf).

Friction force pushing to the right at the foot of the ladder (Ff).

(a) Calculate the reaction force at the wall.

Take the sum of the moments about the foot of the ladder.

∑τ = Iα

Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0

Rw (3.0 sin 60°) = mg (1.5 cos 60°)

Rw = mg / (2 tan 60°)

Rw = (10 kg) (9.8 m/s²) / (2√3)

Rw = 28 N

(b) State the friction at the foot of the ladder.

Take the sum of the forces in the x direction.

∑F = ma

Ff − Rw = 0

Ff = Rw

Ff = 28 N

(c) State the reaction at the foot of the ladder.

Take the sum of the forces in the y direction.

∑F = ma

Rf − mg = 0

Rf = mg

Rf = 98 N

3 0

3 years ago

Calculat the acceleration of a person at latitude<br> 40degreesowing to the rotation of the earth.

beks73 [17]

Answer:

acceleration of person = 9.77 m/s²

Explanation:

given data

latitude = 40 degree

to find out

Calculate the acceleration of a person

solution

we know that here 40 degree = 0.698 rad

so

acceleration of person = g - ω²R ...............1

and 1 rotation complete in 24 hours = 360 degree

here g is 9.81

so we know Earth angular speed ω = 7.27 × $10^{-5}$ rad/s and R is earth radius that is 6.37 × $10^{6}$ m

so

put here value in equation 1 we get

acceleration of person = g - ω²R

acceleration of person = 9.81 - (7.27 × $10^{-5}$ )² × 6.37 × $10^{6}$

acceleration of person = 9.77 m/s²

6 0

3 years ago

Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. It has a centripetal force. F acting on it. If bod

cluponka [151]

Answer:

The centripetal force on body 2 is 8 times of the centripetal force in body 1.

Explanation:

Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. The centripetal force acting on it is given by :

$F=\dfrac{mv^2}{r}$

Body 2 has a mass 2m and its moving in a circle of radius 4r at a speed 4v. The centripetal force on body 2 is :

$F'=\dfrac{2m\times (4v)^2}{4r}\\\\F'=\dfrac{2m\times 16v^2}{4r}\\\\F'=8\dfrac{mv^2}{r}\\\\F'=8F$

So, the centripetal force on body 2 is 8 times of the centripetal force in body 1.

8 0

3 years ago

Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time (when they are not slee

user100 [1]

Answer:

I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s , I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s , I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s and I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

Explanation:

The impulse is equal to the variation of the moment, to apply this relationship to our case, we will assume that initially the mouse was at rest

I = Δp = m $v_{f}$ -m v₀

I = m ( $v_{f}$ -v₀)

Bold indicates vector quantities, let's calculate the momentum of each mouse in for the x and y axes

We recommend bringing all units to the SI system

Mouse 1.

It has a mass of 22.3 g = 22.3 10⁻³ kg, a final velocity of (v = 0.349 i ^ - 0.301 j ^) m / s with an initial velocity of zero

Iₓ = m ( $v_{fx}$ - v₀ₓ)

Iₓ = 22.3 10⁻³ (0.349 -0)

Iₓ = 7.78 10⁻³ J s

$I_{y}$ = m ( $v_{fy}$ - $v_{oy}$ )

$I_{y}$ = 22.3 10⁻³ (-0.301)

$I_{y}$ = -6.71 10⁻³ J s

I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s

Mouse 2

Mass 17.9 g = 17.9 10⁻³ kg

Speed (-0.699 i ^ - 0.815 j ^) m / s

Iₓ = m ( $v_{fx}$ - v₀ₓ)

Iₓ = 17.9 10⁻³ (-0.699 -0)

Iₓ = -12.5 10⁻³ J s

$I_{y}$ = 17.9 10⁻³ (-0.815 - 0)

$I_{y}$ = -14.6 10⁻³ J s

I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s

Mouse 3

Mass 19.1 g = 19.1 10⁻³ kg

Speed (0.745i ^ + 0.975 j ^) m / s

Iₓ = 19.1 10⁻³ (0.745 -0)

Iₓ = 14.2 10⁻³ J s

$I_{y}$ = 19.1 10⁻³(0.975 -0)

$I_{y}$ = 18.6 10⁻³ J s

I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s

Mouse 4

Mass 10.1 g = 10.1 10⁻³ kg

Speed (-0.905i ^ + 0.717j ^) m / s

Iₓ = 10.1 10⁻³ (-0.905 -0)

Iₓ = -9.14 10⁻³ J s

$I_{y}$ = 10.1 10⁻³ (0.717 -0)

$I_{y}$ = 7.24 10⁻³ J s

I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

8 0

3 years ago

What is the buoyant force of a dog displacing 10 pounds of water?

scZoUnD [109]

Answer:

find V using d = M/V

Explanation:

F = d*V*g

d = density of fluid (in this case, 1000)

V = volume of object

g = gravity

3 0

2 years ago