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4 years ago

Which of Newton’s laws example why your hands get red when you press them against a wall

Physics

1 answer:

snow_tiger [21]4 years ago

3 0

A obviously because hand against walls equals gravity to me

I hope this helped you on whatever you're doing.

Have a lovely day <3

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If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the m

Xelga [282]

Answer:

The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is $\frac{135.9}{A}N$

Explanation:

Here we have

Initial temperature of air T₁ = 20 °C = ‪293.15 K

Final temperature of air T₁ = 120 °C = 393.15 K

Initial pressure P₁ = 1 atm = ‪101325 Pa

Final pressure P₂ = Required

Area = A

Therefore we have for the pressure cooker, the volume is constant that is does not change

By Chales law

P₁/T₁ = P₂/T₂

P₂ = T₂×P₁/T₁ = 393.15 K× (‪101325 Pa/‪293.15 K) = ‭135,889.22 Pa

∴ P₂ = 135.88922 KPa = 135.9 kPa

Where Force = $\frac{Pressure}{Area}$ we have

Force = $F_{120}=\frac{135.9}{A}N$ .

4 0

4 years ago

A hanging nickel wire with diameter 0.12 cm is initially 2.7 m long. when a 117 kg mass is hung from it,

dedylja [7]

Yoooooo You want to mbe asd asdas

5 0

3 years ago

A machine that uses 200 w of power moves an object a distance of 15 m in 25 sec. Find

Sergeeva-Olga [200]

Answer:dummy

Explanation:

Dum dum

6 0

3 years ago

Someone please help me

Trava [24]

Answer:

The answer to your question is: D) Ф₂ = 49.71°

Explanation:

Data

n₁ = 1.33

Ф₁ = 35°C

n₂ = 1

Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)

Process

Substitution

Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)

Ф₂ = sin⁻¹ (1.33 sin 35/1)

Ф₂ = sin⁻¹ (1.33 x 0.574/ 1)

Ф₂ = sin⁻¹ ( 0.7628 / 1)

Ф₂ = sin⁻¹ (0.7628)

Ф₂ = 49.71°

8 0

4 years ago

Sarah is making bread. She measures 0.6 kg of hot water into a bowl. The water needs to be at a temperature of

Mrrafil [7]

Answer:

$\Delta m = 0.171\,kg$

Explanation:

The process of water adding is described by the First Law of Thermodynamics:

$m_{w,o} \cdot h_{w,o} + \Delta m \cdot h_{w} = (m_{w,o} + \Delta m)\cdot h_{w,f}$

The amount of additional mass is:

$m_{w,o}\cdot h_{w,o} - m_{w,o}\cdot h_{w,f} = \Delta m\cdot (h_{w,f}-h_{w})$

$\Delta m = \frac{m_{w,o}\cdot(h_{w,o}-h_{w,f})}{h_{w,f}-h_{w}}$

Given that water is incompressible, the equation can be further simplified:

$\Delta m = m_{w,o}\cdot \frac{c_{p,w}\cdot (T_{w,o}-T_{w,f})}{c_{p,w}\cdot (T_{w,f}-T_{w})}$

$\Delta m = m_{w,o}\cdot \left(\frac{T_{w,o}-T_{w,f}}{T_{w,f}-T_{w}} \right)$

$\Delta m = (0.6\,kg)\cdot \left(\frac{40 ^{\circ}C - 42^{\circ}C}{42^{\circ}C-49^{\circ}C} \right)$

$\Delta m = 0.171\,kg$

6 0

3 years ago