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3 years ago

When a piece of wood burns, it makes heat, light, and ash. Therefore, it is an example of what type of change

Physics

2 answers:

Nastasia [14]3 years ago

8 0

It is a chemical change

Ivahew [28]3 years ago

3 0

This is a chemical change because you cannot change the ash back to wood.

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“Why can I sometimes see the moon during the day?“Why can I sometimes see the moon during the day?

asambeis [7]

Answer:

its bc of the way earth spins

Explanation:

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3 years ago

How should we take good care of your hygiene when playing basketball and volleyball.

kolezko [41]

Answer taking care of the body, skin, hair, eyes, teeth, legs, as well as our clothes. Physical exercises and supporting activities, sleep and rest and nutritious food come under personal hygiene. drink enough of water of your bodyweight and stretch.

Explanation:

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3 years ago

A doorframe is twice as tall as it is wide. There is a positive charge on the top left corner and an equal but negative charge i

Andrei [34K]

Answer:

α = 141.5° (counterclockwise)

Explanation:

q₁ = +q

q₂ = -q

q₃ < 0

b = 2*a

We apply Coulomb's Law as follows

F₁₃ = K*q₁*q₃ / d₁₃² = + K*q*q₃ / (2*a)² = + K*q*q₃ / (4*a²)

F₂₃ = K*q₂*q₃ / d₂₃² = - K*q*q₃ / (5*a²)

(d₂₃² = a² + (2a)² = 5*a²)

Then

∅ = tan⁻¹(2a/a) = tan⁻¹(2) = 63.435°

we apply

F₃x = - F₂₃*Cos ∅ = - (K*q*q₃ / (5*a²))* Cos 63.435°

⇒ F₃x = - 0.0894*K*q*q₃ / a²

F₃y = - F₂₃*Sin ∅ + F₁₃

⇒ F₃y = - (K*q*q₃ / (5*a²))* Sin 63.435° + (K*q*q₃ / (4*a²))

⇒ F₃y = 0.0711*K*q*q₃ / a²

Now, we use the formula

α = tan⁻¹(F₃y / F₃x)

⇒ α = tan⁻¹((0.0711*K*q*q₃ / a²) / (- 0.0894*K*q*q₃ / a²)) = - 38.5°

The real angle is

α = 180° - 38.5° = 141.5° (counterclockwise)

4 0

4 years ago

A pool is to be filled with 60 m3 water from a garden hose of 2.5 cm diameter flowing water at 2 m/s. Find the mass flow rate of

stealth61 [152]

Answer:

Time= 6.12*10^4s

mass flow rate m=0.98kg/s

Explanation:

Given

Volume= 60m^3

diamter= 2.5cm= 0.025m

radius= 0.0125m

area A= πr^2

area A= 3.142*0.0125^2

area A= 4.9*10^-4m^2

the velocity of the flow 2m/s

volume flow rate

V=vA

V=2* 4.9*10^-4

V=9.82*10^-4 m^3/s

Time taken to fill the pool

time= volume/volume flow rate

time= 60/9.82*10^-4

time= 6.12*10^4s

Mass flow rate

m= density *volume flow rate

Assuming the density of water to be 997kg/m^3

m= 997*9.82*10^-4

m=0.98kg/s

6 0

3 years ago

Consider two planets of mass m and 2m,

Rzqust [24]

Answer:

Part a)

$\frac{F_1}{F_2} = 10.125$

Part b)

$\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12$

Part c)

$\frac{T_1}{T_2} = 9.54$

Explanation:

Part a)

As we know that the gravitational force is given as

$F = \frac{GMm}{r^2}$

so we will have to find the ratio of force on two planets due to star

so here we have

$\frac{F_1}{F_2} = \frac{m_1r_2^2}{m_2r_1^2}$

$\frac{F_1}{F_2} = \frac{m (4.5r)^2}{(2m) r}$

$\frac{F_1}{F_2} = 10.125$

Part b)

Orbital speed is given as

$v = \sqrt{\frac{GM}{r}}$

so the ratio of two orbital speed is given as

$\frac{v_1}{v_2} = \frac{r_2}{r_1}$

$\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12$

Part c)

Time period is given as

$T = 2\pi\sqrt{\frac{r^3}{GM}}$

so the ratio of two time period is given as

$\frac{T_1}{T_2} = \sqrt{\frac{r_1^3}{r_2^3}}$

$\frac{T_1}{T_2} = \sqrt{\frac{4.5r^3}{r^3}}$

$\frac{T_1}{T_2} = 9.54$

8 0

3 years ago

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