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2 years ago

How do computer get input

2 answers:

6 0

Computers get input through INPUT DEVICES. These devices are designed to insert data into a computer. Examples are:
Keyboard
Mouse
Joystick
Hope that helped. Have a nice day

Vitek1552 [10]2 years ago

5 0

an input device is (a piece of computer hardware equipment) used to provide data and control signals to an information processing system such as acomputer or information appliance.

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_____ of rough surfaces reduces friction

Radda [10]

Answer:

Hey shaikaadil700 !

Lubricating  of rough surfaces reduces friction.

Explanation:

• Lubricating is the smoothening or polishing of the surfaces

$.$

3 0

2 years ago

On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4

anygoal [31]

Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$

b) $\Delta s=31868131.8681\ J.K^{-1}$

Explanation:

Given:

amount of heat transfer occurred, $dQ=3.5\times 10^6\ J$

initial temperature of car, $T_i=36.5+273=309.5\ K$

final temperature of car, $T_f=44.4+273=317.4\ K$

We know that the change in entropy is given by:

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)}$ (heat is transferred into the system of car)

$\Delta s=443037.9747\ J.K^{-1}$

amount of heat transfer form the system of house, $dQ=5.8\times 10^8\ J$

initial temperature of house, $T_i=23.5+273=296.5\ K$

final temperature of house, $T_f=5.3+273=278.3\ K$

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{5.8\times 10^8}{278.3-296.5}$

$\Delta s=31868131.8681\ J.K^{-1}$

6 0

2 years ago

What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (see section 17.6;

antiseptic1488 [7]

<h3>Answer;</h3>

volume = 6.3 × 10^-2 L

<h3>Explanation;</h3>

Volume = mass/density

Mass = 0.0565 Kg,

Density = 900 kg/m³

= 0.0565 kg/ 900 kg /m³

= 6.3 × 10^-5 M³

but; 1000 L = 1 m³

Hence, volume = 6.3 × 10^-2 L

8 0

2 years ago

A puck of mass 0.5100.510kg is attached to the end of a cord 0.8270.827m long. The puck moves in a horizontal circle without fri

yanalaym [24]

Answer: 2.75 1/sec

Explanation:

The only external force (neglecting gravity) acting on the puck, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r (1)

Our unknown, is the frequency at which the puck can go around the circle, which is the inverse of the period Tp.

By definition, a period is the time needed by the puck to complete one entire circle.

By definition also , angular velocity is the rate of change of the angle advanced, so we can express this way:

ω = ∆θ / ∆t

The angle advanced during one period, is exactly (by angle definition) 2 π radians.

So, we can always write the angular velocity, ω, as follows:

ω = 2π / Tp = 2πf

Now, there is a relationship between linear and angular velocity, that can be found applying simply the definition of velocity and of an angle too, as follows:

v = ∆s / ∆t = r ∆θ/∆t = ω r

Replacing in (1), we have:

T = mω2 r2 / r = m ω2r (2)

We have just found that ω= 2πf, so, replacing in (2) :

T = m (2π)2 f2 r

Solving for f:

f = 1/2π√(T/mr) = 1/2π 17.28 1/sec = 2.75 1/sec

6 0

2 years ago

Determine the value of the resultant and its location from O. see attach image.

xxTIMURxx [149]

Answer:

Explanation:

In the x direction the force will be

½(-w₀)L/2 = -¼w₀L

acting ⅔(L/2) = L/3 below the x axis.

In the y direction the force will be

½(-w₀)L + ½w₀L/2 = -¼w₀L

the magnitude of the resultant will be

F = w₀L √((-¼)² + (-¼)²) = w₀L√⅛

in the direction

θ = arctan(-¼w₀L / -¼w₀L) = 225°

to find the distance, we balance moments

(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]

(√⅛)[d] = ½ [⅔L] + ¼ [⅔L/2] - ¼ [L - ⅓L/2]

(√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]

(√⅛)[d] = ⅓L + ⅟₁₂L - ¼L + ⅟₂₄L

(√⅛)[d] = 5L/24

d = 5L/24 / (√⅛)

d = 5√⅛L/3

8 0

2 years ago