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creativ13 [48]
3 years ago
6

A pillow with mass of 0.3 kg sits on a bed with a coefficient of static friction of 0.6. What is the maximum force of static fri

ction?
Physics
2 answers:
lys-0071 [83]3 years ago
5 0
The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body. 
 
                        P = W = mass x acceleration due to gravity
    
                    P = (0.3 kg) x (9.8 m/s²) = 2.94 kg m/s² = 2.94 N

Solving for the static friction force (F), 
                                              F = P x c 
 
                                      F = (2.94 N) x 0.6 = 1.794 N

Therefore, the maximum force of static friction is 1.794 N. 



bazaltina [42]3 years ago
5 0

Answer:

1.8 N

Explanation:

A P E X verified

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A car accelerates uniformly from rest to 23 m/s over a distance of 30 meters. What is the acceleration of the car?
kolezko [41]

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

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u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

velocity= distance/time

23= 30/t

t= 30/23

t= 1.30 seconds

hence

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A solid sphere of uniform density has a mass of 3.0 × 104 kg and a radius of 1.0 m. What is the magnitude of the gravitational f
elena55 [62]

Answer:

a) Fg = 9.495x10⁻⁶N

b) Fg = 3.908x10⁻⁶N

c)

F_{g} =\frac{Gm_{1}m_{2}R  }{r^{3} }

Explanation:

Given:

m₁ = mass = 3x10⁴kg

r = radius = 1 m

m₂ = 9.3 kg

Questions:

a) What is the magnitude of the gravitational force due to the sphere located at R = 1.4 m, Fg = ?

b) What is the magnitude of the gravitational force due to the sphere located at R= 0.21 m, Fg = ?

c) Write a general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 m from the center of the sphere.

a) Since R > r, the equation for the gravitational force is:

F_{g} =\frac{Gm_{1}m_{2}  }{R^{2} }

Here,

G = gravitational constant = 6.67x10⁻¹¹m³/s² kg

Substituting values:

F_{g} =\frac{6.67x10^{-11}*3x10^{4}*9.3  }{1.4^{2} } =9.495x10^{-6} N

b) Since R < r, the equation for the gravitational force is:

F_{g} =\frac{Gm_{1}m_{2}R  }{r^{3} } =\frac{6.67x10^{-11}*3x10^{4}*9.3*0.21  }{1^{3} } =3.908x10^{-6} N

c) The general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 is the same to b)

F_{g} =\frac{Gm_{1}m_{2}  R}{r^{3} }

8 0
3 years ago
What are the si units for average speed?
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Consider a telescope with a small circular aperture of diameter 2.0 centimeters. Calculate the angular separation θ1 at which tw
kogti [31]

To solve this problem it is necessary to apply the concepts related to diffraction through a circular opening.

By definition the angular resolution is given by

\theta = 1.22\frac{\lambda}{D}

Where,

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\lambda = 600*10^{-9}m

d = 1.5*10^{-2}m

Therefore replacing,

\theta = 1.22\frac{600*10^{-9}}{1.5*10^{-2}}

\theta = 4.88*10^{-5} rad

Therefore the angular separation is 4.88*10^{-5} rad

7 0
3 years ago
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