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3 years ago

A) make U subject of the formula E= m/2g(v^2-u^2)

Mathematics

1 answer:

ollegr [7]3 years ago

7 0

Answer:

u = v2−2Egm−−−−−−−√

Step-by-step explanation:

2Eg = 2g (M2g(V2−U2)

2Eg = m(V2 - U2)

2Eg - mV2 - mU2

mU2 = mV2 - 2Eg

mU2m=mV2−2Egm

U2 = mV2−2Egm

= mV2m−2Egm

U2 = V2 - 2Egm

U = V2−2Egm−−−−−−−−√

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3 years ago

Which ordered pair is a solution of the equation y = x - 2?

Liula [17]

Answer:

One order pair is

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Step-by-step explanation:

What ordered pairs are the options?

Pick a value for

and solve for

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8 0

3 years ago

Using appropriate properties, find the value of 249² - 248²

Lady_Fox [76]

Answer:

$\boxed{\bold{\huge{\boxed{497}}}}$

Step-by-step explanation:

=> $\sf 249^2 - 248^2$

Using Formula $\sf a^2 - b^2 = (a+b)(a-b)$

Where

a = 249

b = 248

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3 0

3 years ago

33. Suppose you were on a planet where the

tatyana61 [14]

Answer:

a) 3.675 m

b) 3.67m

Explanation:

We are given acceleration due to gravity on earth = $9.8ms^-2$

And on planet given = $2.0ms^-2$

A) Since the maximum jump height is given by the formula

$\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}$

Where H = max jump height,

v0 = velocity of jump,

Ø = angle of jump and

g = acceleration due to gravity

Considering velocity and angle in both cases

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}$

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)

g1 = 2.0 $ms^-2$ and

g2 = 9.8 $ms^-2$

Substituting these values we get H1 = 3.675m which is the required answer

B) Formula to find height of ball thrown is given by

$\mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}$

which is due to projectile motion of ball

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,

a = acceleration = g = acceleration due to gravity

Considering t = same on both places we can write

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}$

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

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We get h1 = 3.67m which is the required height

6 0

4 years ago

What is the area of rectangle with length 18cm and width 4cm?

miv72 [106K]

I hope this helps you

Area=width×length

Area=18×4

Area=72

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3 years ago