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2 years ago

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Friction depends on the types of surfaces involved and how hard the surfaces push together. Please select the best answer from t

he choices provided T F

1 answer:

LenKa [72]2 years ago

5 0

True: Friction depends on the types of surfaces involved and how hard the surfaces push together.

You might be interested in

food product with 10 kg mass is being transported to thesurface of the moon,where the acceleration due to gravity is1.624 m/s2;

larisa [96]

Answer:

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

Explanation:

g = Acceleration due to gravity

m = Mass = 10 kg

Weight on Earth

$W=mg\\\Rightarrow W=10\times 9.81\\\Rightarrow W=98.1\ N$

Converting to lbf

$98.1\times 0.22481=22.053861\ lbf$

On Moon

$W=10\times 1.624\\\Rightarrow W=16.24\ N$

Converting to lbf

$16.24\times 0.22481=3.6509144\ lbf$

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

5 0

2 years ago

A demented scientist creates a new temperature scale, the "Z scale." He decides to call the boiling point of nitrogen 0°Z and th

slavikrds [6]

Answer:

A) $Z = 0.577 C +112.931$

$Z = 0.577*(100) +112.931=170.631 Z$

B) $C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C$

C) $K = C +273.15$

$K = -22.41 +273.15 =250.739 K$

Explanation:

For this case we want to create a function like this:

$Z = a C + b$

Where Z represent the degrees for the Z scale C the Celsius grades and tha valus a and b parameters for the model.

The boiling point of nitrogen is -195,8 °C

The melting point of iron is 1538 °C

We know the following equivalences:

-195.8 °C = 0 °Z

1538 °C = 1000 °Z

Let's say that one point its (1538C, 1000 Z) and other one is (-195.8 C, 0Z)

So then we can calculate the slope for the linear model like this:

$a = \frac{z_2 -z_1}{c_2 -c_1}= \frac{1000 Z- 0Z}{1538C -(-195.8 C)}=\frac{1000 Z}{1733.8 C}=0.577 \frac{Z}{C}$

And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:

$0 = 0.577 (-195.8) + b$

And if we solve for b we got:

$b = 0.577*195.8 =112.931 Z$

So then our lineal model would be:

$Z = 0.577 C +112.931$

Part A

The boiling point of water is 100C so we just need to replace in the model and see what we got:

$Z = 0.577*(100) +112.931=170.631 Z$

Part B

For this case we have Z =100 and we want to solve for C, so we can do this:

$Z-112.931 = 0.577 C$

$C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C$

Part C

For this case we know that $K = C +273.15$

And we can use the result from part B to solve for K like this:

$K = -22.41 +273.15 =250.739 K$

6 0

3 years ago

19. What do vibrations create?<br><br> Wavelengths<br> sound waves<br> energy<br> electricity

OverLord2011 [107]

Answer:

Energy

Explanation:

When an object vibrates, it creates kinetic energy that is transmitted by molecules in the medium. As the vibrating sound wave comes in contact with air particles passes its kinetic energy to nearby molecules. As these energized molecules begin to move, they energize other molecules that repeat the process.

5 0

2 years ago

Read 2 more answers

Master of physics needed

Delicious77 [7]

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0

3 years ago

A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The

VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

$\dfrac{1}{2}mu^2 = m g d sin \theta$

$u^2 = 2 g d sin30^0$

$u= \sqrt{2\times 9.8 \times 10 sin30^0}$

u = 9.89 m/s

Using second law of motion

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

Using third equation of motion ,

v² - u² = 2 a s

0² - 9.89² = 2 x 2.45 x s

s = 20 m

the distance moved before stopping is 20 m

3 0

3 years ago