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3 years ago

What kind of mouse is a battery-powered mouse that uses radio waves or bluetooth technology to communicate with a device?

1 answer:

8 0

Wireless/cordless mouse. The same goes for a wireless keyboard/cordless keyboard. They are devices that are battery powered and transmit data to the system unit using wireless technology such as radio waves (Bluetooth) or infrared light waves (IrDA).

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Imagine that you are going to visit your friend. Before you get there, you decide to stop at the variety store. If you walk 200

SashulF [63]

Answer:

400m

Explanation:

Brainliest? :))

Let your initial displacement from your home to the store be

1 and your displacement from the store to your friend’s house

be Dd

Given: Dd

1 = 200 m [N]; Dd

2 = 600 m [S]

Required: Dd

Analysis: Dd

T 5 Dd

1 1 Dd

Solution: Figure 6 shows the given vectors, with the tip of Dd

joined to the tail of Dd

2. The resultant vector Dd

T is drawn in red,

from the tail of Dd

1 to the tip of Dd

2. The direction of Dd

T is [S].

T measures 4 cm in length in Figure 6, so using the scale of

1 cm : 100 m, the actual magnitude of Dd

T is 400 m.

Statement: Relative to your starting point at your home, your

total displacement is 400 m [S].

6 0

2 years ago

A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55

Anit [1.1K]

Explanation:

We have,

Mass of a baseball is 0.147 kg

Initial velocity of the baseball is 44.5 m/s

The ball is moved in the opposite direction with a velocity of 55.5 m/s

It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Change in momentum,

$\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s$

Impulse = 14.7 kg-m/s

Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s

4 0

3 years ago

You place a solid cylinder of mass M on a ramp that is inclined at an angle β to the horizontal. The coefficient of static frict

zlopas [31]

Answer:

Explanation:

Let the frictional force required be f .

frictional force is responsible in creating rotational motion in the cylinder.

torque created by frictional force = f R

if angular acceleration be α

α = f R / I , I is moment of inertia of cylinder .

α = a / R , a is linear acceleration.

f R / I = a / R

a = f R² / I

linear acceleration a of cylinder down the slope

ma = mgsinθ - f , ( f force is acting upwards and mgsinθ is acting downwards )

mf R² / I = mgsinθ -f

f ( m R² / I + 1) = mgsinθ

f = mgsinθ / ( m R² / I + 1)

= mgsinθ / ( m R² / mk² + 1) , k is radius of gyration of cylinder.

= mgsinθ / ( R² / k² + 1)

Putting the given values

f = Mgsinβ /( R² / k² + 1)

for cylinder , R² / k² = 2

f = Mgsinβ /3

6 0

3 years ago

Suppose Galileo dropped a lead ball (100 kilograms) and a glass ball (1 kilogram) from the Leaning Tower of Pisa. Which one hit

pogonyaev

If he dropped them both at the same time, then as close as anyone could tell, they both hit the ground at the same time.

8 0

3 years ago

1. A 14-cm tall object is placed 26 cm from a converging lens that has a focal length of 13 cm.

AURORKA [14]

Answer:

a) Please find attached the required drawing of light passing through the lens

By the use of similar triangles;

The image distance from the lens = 26 cm

The height of the image = 14 cm

c) The image distance from the lens = 26 cm

The height of the image = 14 cm

Explanation:

Question;

a) Determine the image distance and the height of the image

b) Calculate the image position and height

The given parameters are;

The height of the object, h = 14 cm

The distance of the object from the mirror, u = 26 cm

The focal length of the mirror, f = 13 cm

The location of the object = 2 × The focal length

Therefore, given that the center of curvature ≈ 2 × The focal length, we have;

The location of the object ≈ The center of curvature of the lens

The diagram of the object, lens and image created with MS Visio is attached

From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m

The height of the image = The height of the object - 14 cm

b) The lens equation is used for finding the image distance from the lens as follows;

$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$

Where;

v = The image distance from the lens

We get;

$v = \dfrac{u \times f}{u - f}$

Therefore;

$v = \dfrac{26 \times 13}{26 - 13} = \dfrac{26 \times 13}{13} = 26$

The distance of the image from the lens, v = 26 cm

The magnification, M =v/u

∴ M = 26/26 = 1, therefore, the object and the image are the same size

Therefore;

The height of the image = The height of the object = 14 cm.

5 0

3 years ago