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BlackZzzverrR [31]
3 years ago
15

Which of these is an example of a physical change? Question options: wood decaying iron melting paper burning steel rusting

Physics
1 answer:
marin [14]3 years ago
4 0
Im pretty sure burning paper

its either that or melting iron but im 90 percent sure its paper burning
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A sprinter starts from rest and reaches a speed of 15 m/s in 4.25 s. Find his acceleration
kakasveta [241]

Answer:

a=3.53 m/s^2

Explanation:

Vo=0 m/s (because he is not moving at the start)

V1=15 m/s

t= 4.25 s

a = (V1-Vo) / t = 15/4.25 = 3.53 m/s^2

5 0
3 years ago
What is the difference between real and apparent weightlessness?
MrRissso [65]

Answer:

In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.

Explanation:

Hope this helps:)

4 0
2 years ago
Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains
OLEGan [10]

Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

C = 252000 J/326.7 kg°C

<u>C = 771.35 J/kg°C</u>

5 0
3 years ago
A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg
yulyashka [42]

Answer:

T=245.76N

Explanation:

We know that the frequency of the nth harmonic is given by f_n=nf, where f is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:

f_{n+1}-f_n=(n+1)f-nf=nf+f-nf=f

Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

f=f_{n+1}-f_n=480Hz-400Hz=80Hz

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

f=\frac{1}{2L} \sqrt{\frac{T}{\mu}}

So the tension is:

T=\mu(2Lf)^2

Which for our values is:

T=(0.0006kg/m)(2(4m)(80Hz))^2=245.76N

6 0
3 years ago
A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa
Jlenok [28]

Answer:

The answer is 18 N.

Explanation:

A force can be divided into components x and y components. The component along the x-axis is called the horizontal component and along the y-axis is called the vertical component. In this case, as the force is in a horizontal direction and is also known as x-component of force. The x- component of force is  

Fx = Fcosθ

Fx = 22(cos 35°)

Fx = 22 x 0.819

Fx =  18 N

Child's horizontal pull forces are equal to that of frictional resistance force on the wagon.

5 0
3 years ago
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