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quester [9]
2 years ago
12

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 0.400 N. Knowing the coefficient of kinetic friction between the two materials (0.04), he quickly calculates the normal force. What is it (in N)? N
Physics
1 answer:
IRINA_888 [86]2 years ago
5 0

Answer:

normal force = 10 N

Explanation:

Given data

frictional force = 0.400 N

coefficient of kinetic friction = 0.04

Solution

we get here normal force that is express as

normal force = \frac{Frictional\ force}{coefficient\ of\ friction}        ............1

put here value and we will get value

normal force = \frac{0.400}{0.04}  

solve it we get

normal force = 10 N

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   Mass= 15 million kg        Mass= 5 million kg

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A circuit has a voltage of 10 V and a current of 5 A. What must the resistance be?
azamat

Answer:

R=V/I

R= 2

Explanation:

R = 10V/5A

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3 0
2 years ago
Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th
8090 [49]

Answer:

r_{cm} = 0.074 m from the position of the center of the Sun

Explanation:

As we know that mass of Sun and Jupiter is given as

M_s = 1.98 \times 10^{30} kg

M_j = 1.89 \times 10^{27} kg

distance between Sun and Jupiter is given as

r = 7.78 \times 10^{11} m

now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them

so we will have

r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}

r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}

r_{cm} = 0.074 m from the position of the center of the Sun

3 0
3 years ago
Calculate the mass 9f the earth, assuring that uts is sphere with radius 6.67×10^6m.​
MA_775_DIABLO [31]

Answer:

6.86 × 10²⁴ kg

Explanation:

The mass of the earth m = density of earth, ρ × volume of earth, V

m = ρV

The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m

Since m = ρV

m = ρ4πr³/3

So, substituting the values of the variables into the equation for the mass of the earth, m, we have

m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3

m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3

m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3

m = 6546105.64378π × 10¹⁸ kg/3

m = 20565197.400122 × 10¹⁸ kg/3

m = 6855065.8 × 10¹⁸ kg

m = 6.8550658 × 10²⁴ kg

m ≅ 6.86 × 10²⁴ kg

8 0
2 years ago
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