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mestny [16]
3 years ago
8

3. Write any two types of force.​

Physics
1 answer:
Lerok [7]3 years ago
8 0

Answer:

An <u>applied force</u> is a force that is applied to an object by a person or another object. If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person.

A <u>friction force</u> is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object. For example, if a book slides across the surface of a desk, then the desk exerts a friction force in the opposite direction of its motion. Friction results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The maximum amount of friction force that a surface can exert upon an object can be calculated using the formula below:

F_{frict} = µ • F_{norm}

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marissa [1.9K]
Deforestation, the chopping off of the trees that can take thousands or even millions of years to grow again<span />
6 0
3 years ago
Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others com
Bas_tet [7]

Answer:

Failure rate   = 20%

MTBF = 880 hours

Explanation:

given data

batteries = 10

tested = 200 hours

one failed = 20 hours

another fail at =  140 hours

solution

we know that Mean Time between Failures is express as = (Total up time) ÷  (number of breakdowns)    ....................1

so here Total up time will be

Total up time = 200 × 10

Total up time = 2000

and here

Number of breakdown = 1 at 20 hour and another at 140 hour = 2

so it will be  = (Total up time) ÷ (number of breakdowns)      .......2

=  \frac{2000}{2}   =  1000  

so here gap between occurrences is

gap between occurrences=  140 - 20

gap between occurrences = 120 hour

and

MTBF  will be

MTBF = 1000 - 120

MTBF = 880 hours  

and

Failure rate (FR)  will be

Failure rate (FR) =  1 ÷ MTBF    ................3

Failure rate (FR) = R÷T     ......................4

as here R is the number of failures and T is total time

so Failure rate (FR)  = 20%

4 0
3 years ago
A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.
GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is 8.75 m/s^2 in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, F_f = 15 N, pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, W=mg, where m is the mass of the book and g=9.8 m/s^2 is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

W=(4)(9.8)=39.2 N

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

N-W=ma

where a is the acceleration; however, since N = W, this becomes

a=0

And since the book is initially at rest on the desk, this means that there is no motion.

c)

We said there are two forces acting in the horizontal direction:

- The applied force, F = 50 N, forward

- The frictional force, F_f = 15 N, backward

Since they act along the same line, we can calculate their resultant as

\sum F = F - F_f = 50 - 15 = 35 N

and therefore the net force is 35 N in the forward direction.

d)

The net force is obtained as the resultant  of the net forces in the horizontal and vertical direction. However, we have:

- The net force in the horizontal direction is 35 N

- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction

Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.

e)

The acceleration of the book can be calculated by using Newton's second law:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

Here we have:

\sum F = 35 N (in the forward direction)

m = 4 kg

Therefore, the acceleration is

a=\frac{\sum F}{m}=\frac{35}{4}=8.75 m/s^2 (forward)

Learn more about forces, weight and Newton's second law:

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8 0
3 years ago
The sound produced by the loudspeaker in the drawing has a frequency of 11999 Hz and arrives at the microphone via two different
const2013 [10]

The speed at which sound travels through the gas in the tube is 719.94m/s

<u>Explanation:</u>

Given:

Frequency, f = 11999Hz

Wavelength, λ = 0.03m

Velocity, v = ?

Sound speed in the tube is calculated by multiplying the frequency v by the wavelength λ.

As the sound loudness changed from a maximum to a minimum, then we know the sound interference in the case changed from constructive interference (the two sound waves are in phase, i.e. peaks are in a line with peaks and so the troughs), to a destructive interference (peaks coinciding with troughs). The least distance change required to cause such a change is a half wavelength distance, so:

λ/2 = 0.03/2

 λ  = 0.06m

We know,

v = λf

v = 0.06 X 11999Hz

v = 719.94m/s

Therefore, the speed at which sound travels through the gas in the tube is 719.94m/s

3 0
3 years ago
A balloon filled with helium gas at 20°C occupies 4.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196°C, whil
mrs_skeptik [129]

Answer:

0.25 L

Explanation:

P_1 = Initial pressure = 1 atm

T_1 = Initial Temperature = 20 °C

V_1 = Initial volume = 4.91 L

P_2 = Final pressure = 5.2 atm

T_2 = Final Temperature = -196 °C

V_2= Final volume

From ideal gas law we have

\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\\Rightarrow P_2=\dfrac{1\times 4.91(273.15-196)}{(20+273.15)\times 5.2}\\\Rightarrow V_2=0.24849\ L\approx 0.25\ L

The pressure experienced by the balloon is 0.25 L

7 0
3 years ago
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