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iogann1982 [59]
3 years ago
7

The weakest type of friction that occurs between solid surfaces is _____ friction

Physics
2 answers:
irga5000 [103]3 years ago
8 0
Rolling friction, hope this helps
Evgen [1.6K]3 years ago
4 0

Answer:

Static, sliding, and rolling friction occur between solid surfaces. Static friction is strongest, followed by sliding friction, and then rolling friction, which is weakest. Fluid friction occurs in fluids, which are liquids or gases.

Explanation:

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How can speed be defined
Anna11 [10]

Explanation:

C) distance / time

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6 0
2 years ago
Read 2 more answers
A student tries to measure the period of a pendulum that is already swinging
Alex17521 [72]

Answer:

a. The student's mistake was that the student did not swing the pendulum and start the watch at the same time.

b. 1.2 s per swing.

c. The likely effect of her reaction time is that they will should subtract two seconds off the time.

Explanation:

5 0
2 years ago
The specific weight of sea water is 10.1 kN/m^3. Convert to lbs/in^3.
Viktor [21]

Answer:

0.03719 lbs/in³

Explanation:

Specific weight is given by multiplying the density of an object to the acceleration due to gravity.

\gamma =\rho g\\\Rightarrow \rho=\frac{gamma}{g}\\\Rightarrow \rho=\frac{10.1\times 10^3}{9.81}\\\Rightarrow \rho=1029.562\ kg/m^3

1\ kg=2.20462\ lb

1\ m=39.3701\ in

\\\Rightarrow 1029.562\ kg/m^3=\frac{1029.562\times 2.20462}{39.3701^3}=0.03719\ lbs/in^3

So,

10.1\ kN/m^3=0.03719\ lbs/in^3

8 0
3 years ago
A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

Therefore, the extension of the wire is 0.362 mm.

8 0
3 years ago
A small package is attached to a helium-filled balloon rising at 2 m/s. The package drops from the balloon when it is 14 meters
scZoUnD [109]
I would assume air resistance is negligible and so the acceleration of the package would be approximately 9.81 m/s².

Taking downwards as positive, use v²=u²+2as.
v²=(-2)²+2(9.81)(14)
v=16.7 m/s
4 0
3 years ago
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