Different between these term<br> A)scalar and vector and give these example<br> B)speed and velocity

A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5m/s2 for 4s. After t

Pavlova-9 [17]

Taking right movement to be positive means leftward movement is negative.
Hence we have a deceleration of
$a = - 2.5 {ms}^{ - 2}$

$t = 4s$
$v = 3.0 {ms}^{ - 1}$
Using this 'suvat' equation
$v = u + at$
we can determine the initial velocity

$3.0= u + -2.5(4)$

$3.0+2.5(4)=u$

$13.0 = u$

Hence the initial velocity is 13.0 meters per seconds

3 0

3 years ago

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The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

3 0

3 years ago

What committee decides which bills move to the House floor for debate and sets the terms for the debate?

spayn [35]

i think the answer is B-House Rules Comittee

6 0

3 years ago

A television remote control uses infrared light with a wavelength of 940 nm. What is the frequency of the light?

Ronch [10]

Answer:

Frequency = 3.19 * 10^14 Hz or 1/s

Explanation:

Relationship b/w frequency and wavelength can be expressed as:

C = wavelength * frequency, where c is speed of light in vacuum which is 3.0*10^8 m/s.

Now simply input value (but before that convert wavelength into meters to match the units, you do this by multiply it by 10^-9 so it will be 940*10^-9)

3.0 * 10^8 = Frequency * 940 x 10^-9

Frequency = 3.19 * 10^14 Hz or 1/s

5 0

3 years ago

Bettina spoke into a microphone during the school play to increase the sound of her voice so the audience could hear her speak.

grandymaker [24]

The loudness was increased by the amplifier which converted electrical energy into sound energy.

7 0

3 years ago

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Different between these term A)scalar and vector and give these example B)speed and velocity