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3 years ago

Identify the two pieces of information you need to know the velocity of an object

Physics

1 answer:

Georgia [21]3 years ago

8 0

Manuel with and height?

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PLEASE SOLVE QUICKLY!!! Solve for A, B, and C from graph

hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

Explanation:

We know,

$x = \frac{L}{\frac{W}{F} +1}$

and L = x+y

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

$L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm$

Therefore, position weight placed from fulcrum is 59.35cm

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

$47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8 = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g$

Therefore, weight of the beam is 196.56g

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

$50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8 = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g$

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

4 0

3 years ago

A physics student with too much free time drops a watermelon from a roof of a building, hears the sound of the watermelon going

tatiyna

Answer:

28.6260196842 m

Explanation:

Let h be the height of the building

t = Time taken by the watermelon to fall to the ground

Time taken to hear the sound is 2.5 seconds

Time taken by the sound to travel the height of the cliff = 2.5-t

Speed of sound in air = 340 m/s

For the watermelon falling

$s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow h=\frac{1}{2}\times 9.81\times t^2$

For the sound

Distance = Speed × Time

$\text{Distance}=340\times (2.5-t)$

Here, distance traveled by the stone and sound is equal

$\frac{1}{2}\times 9.81\times t^2=340\times (2.5-t)\\\Rightarrow 4.905t^2=340\times (2.5-t)\\\Rightarrow t^2=\frac{340}{4.905}(2.5-t)\\\Rightarrow t^2+69.3170234455t-173.292558614=0$

$t=\frac{-69.31702\dots +\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1},\:t=\frac{-69.31702\dots -\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1}\\\Rightarrow t=2.4158\ s\ or\ -71\ seconds$

The time taken to fall down is 2.4158 seconds

$h=\frac{1}{2}\times 9.81\times 2.4158^2=28.6260196842\ m$

Height of the buidling is 28.6260196842 m

7 0

3 years ago

Please select the word from the list that best fits the definition I study French so I can talk to my cousin who lives in Paris

salantis [7]

Answer:

There is no list...

Explanation:

Post the list and words so I can help you.

8 0

3 years ago

Read 2 more answers

Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of t

Lelu [443]

Answer:

c) At a distance greater than r

Explanation:

If G= Gravitational constant

M= Mass of earth

r= distance from earth center

then orbital speed is ;

v = $\sqrt{\frac{GM}{r} }$

==> v²=GM/r

If speed of first satellite = V₁

==> V₁² = GM/r

==> r = GM/V₁²

If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.

So our answer is c

5 0

3 years ago

Two girls,(masses m1 and m2) are on roller skates and stand at rest, close to each other and face-to-face. Girl 1 pushes squarel

Arturiano [62]

Answer:

$\vec{v}_1 = -\frac{\vec{v}_2m_2}{m_1}$

Explanation:

The center of mass of the system (two girls) is constant, as the velocity of the center of mass of the system is also constant.

$\vec{v}_{cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}$

The initial velocity of the system is zero, since both girls are at rest. So the velocity of the total system at any point should be zero as well.

$0 = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}\\\vec{v}_1 = -\frac{\vec{v}_2m_2}{m_1}$

This is true, because there is no friction between the girls and the ground. Otherwise, the velocity of the center of mass wouldn't be constant.

5 0

3 years ago