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brilliants [131]

3 years ago

12

Question 11 of 29

1 answer:

lara [203]3 years ago

3 0

Answer:

the answer is a

Explanation:

wawawawawawawawawawawawawa

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Find the initial velocity

Ivenika [448]

a = dv/t = (330-x)/1.5 =50

x= 330-75 = 255 m/s

Intial velocity 255 m/s

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3 years ago

I need help in Physiology!!!!!!

Alisiya [41]

Perception is the sensory experience of the world. It involves both recognizing environmental stimuli and actions in response to these stimuli. Through the perceptual process, we gain information about the properties and elements of the environment that are critical to our survival.

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2 years ago

A clock clicks 50 times in 12 seconds. Calculate the frequency and period of the clock.

Otrada [13]

This would be an example on what to do, just replace the fraction numbers with your own! Good luck!

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3 years ago

Read 2 more answers

An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha

Lynna [10]

Answer: 6.45 s

Explanation:

We have the following equation:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height when the rock hits the ground

$y_{o}=75 m$ the height at the edge of the cilff

$V_{o}=20 m/s$ the initial velocity

$g=9.8 m/s^{2}$ acceleration due gravity

$t$ time

$0=75 m+(20 m/s)t-(4.9 m/s^{2})t^{2}$ (2)

Rearranging the equation:

$-(4.9 m/s^{2})t^{2} + (20 m/s)t + 75 m=0$ (3)

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , and we have to use the quadratic formula if we want to find $t$ :

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (4)

Where $a=-4.9$ , $b=20$ , $c=75$

Substituting the known values and choosing the positive result of the equation:

$t=\frac{-20\pm\sqrt{20^{2}-4(-4.9)(75)}}{2(-4.9)}$ (5)

$t=6.453 s$ This is the time it takes to the rock to hit the ground

8 0

3 years ago

A 10 m long steel beam is accidentally dropped by a construction crane from a height of 4.89 m. The horizontal component of the

Otrada [13]

Answer:

e = 1.21 mV

Explanation:

given,

length of rod = 10 m

height of drop = 4.89 m

Earth’s magnetic field = 12.4 µT

acceleration of gravity = 9.8 m/s²

velocity of the beam

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8 \times 4.89}$

v = 9.79 m/s

emf of the beam

e = B l v

e = 12.4 x 10⁻⁶ x 9.79 x 10

e = 1.21 x 10⁻³ V

e = 1.21 mV

4 0

3 years ago