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3 years ago

8

What is the acceleration of a Ford Mustang GT that can go from 0.00 to 27.8 m/s in 5.15 seconds?

1 answer:

bixtya [17]3 years ago

4 0

Finding acceleration= final speed-initial speed/time taken (or A=V-U\T)

Finial speed= 27.8s
Initial speed= 0s
Time taken= 5.15

So..

27.8-0/5.15= 5.40m/s (rounded to two decimal places)

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During which segments are two states of matter present?

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Answer:

the correct answer is D.) 2,4

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4 years ago

In the photoelectric effect, it is found that incident photons with energy 5.00ev will produce electrons with a maximum kinetic

Yakvenalex [24]

From Literature:

The amount of energy in the photons is given by this equation:

E = hf

where E = energy
h = Planck's constant = 6.63 * 10^-34 Joule seconds
f = frequency of the light, Hz

Given:

E= 3.00 eV and Planck's constant

To solve for the frequency, E = 3.00 eV

1 electronvolt = 1.60218 x 10^-19 Joules

3 * 1.60218 x 10^-19 Joules = 6.63 * 10^-34 Joule seconds * f
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3 years ago

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If the Moon's orbit were perpendicular to the ecliptic, would eclipses be possible? What would be the most common phase of the M

tangare [24]

Answer:no

Explanation:

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4 years ago

A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

<u>Time taken to descent the total height:</u>

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

<u>Now the total time taken by the ball to hit the ground:</u>

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago

A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be

lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × $10^{-3}$ m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ) ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v / 2 ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 ) / 2

0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 = 108

ω² = 108 / 7.160 × $10^{-5}$

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0

4 years ago