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dezoksy [38]
3 years ago
5

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

an is only 0.350 N. Knowing the coefficient of kinetic friction between the two materials (0.04), he quickly calculates the normal force. What is it (in N)?
Physics
1 answer:
wel3 years ago
8 0

Answer:

Normal force = 8.75 N

Explanation:

given,

frictional force between the steel spatula and the Teflon frying pan=0.350 N

coefficient of friction between material =0.04

normal force = ?

using formula,

Frictional force = coefficient of friction × normal force

normal\ force = \dfrac{Frictional\ force}{coefficient\ of\ friction}

normal\ force = \dfrac{0.350}{0.04}

Normal force = 8.75 N

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A 2.0 kg mass weighs 10 Newtons on planet X. what is the acceleration due to gravity on planet X? Show the work.
KengaRu [80]

Answer: The gravitational acceleration on planet X is 5 N/kg

On Earth (with the gravitational accelartion g_E) the mass of 2kg will correspond to

F_E = m\cdot g_E = 2.0 \mbox{kg}\cdot 9.8 \frac{N}{kg} = 19.6 N

On planet X we are told the same measure is only 10N. Since there is a proportional relationship between g and F, we can calculate g_X:

\frac{F_E}{g_E}=\frac{F_X}{g_X} \implies\\g_x = \frac{g_E\cdot F_X}{F_E}=\frac{9.8 N/kg \cdot 10 N }{19.6N}=5 \frac{N}{kg}

8 0
3 years ago
Consider an electron that is 10-10 m from an alpha particle (9 = 3.2 x 10-19 C). (Enter the magnitudes.) (a) What is the electri
Studentka2010 [4]

Answer:

a)E=2.88*10^{11}N/C

b)E=1.44*10^{11}N/C

c)F=4.61*10^{-8}N

Explanation:

We use the definition of a electric field produced by a point charge:

E=k*q/r^2

<u>a)Electric Field  due to the alpha particle:</u>

E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(10^{-10})^2=2.88*10^{11}N/C

<u>b)Electric Field  due to electron:</u>

E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(10^{-10})^2=1.44*10^{11}N/C

<u>c)Electric Force on the alpha particle, on the electron:</u>

The alpha particle and electron feel the same force but with opposite direction:

F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(10^{-10})^2=4.61*10^{-8}N

4 0
3 years ago
Where do we hear loud sound, at antinode or node?​
tester [92]

Answer:

node

Explanation:

on the graph node is higher than antinode

so it can get or hear loud sounds faster

4 0
3 years ago
In the desert enviroment the chemical weathering of rocks is generally reduced because
faust18 [17]
<span>In the desert environment the chemical weathering of rocks is generally reduced because there is a lack of rain, in which most chemical weathering is caused by. No rain, no chemicals, no chemical weathering.</span>
8 0
3 years ago
What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each
suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

  • ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
  • A = area of plates and
  • d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

  • σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
  • a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

7 0
2 years ago
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