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3 years ago

9

Suppose you want to move a big rock in your yard. It is about waist high and 4 feet long, has a volume of 1.2 m3 and a density o

f 2.5 g/cm3. How much does it weigh, in units of lbs?

1 answer:

beks73 [17]3 years ago

3 0

Answer:

6613.87 lbs

Explanation:

1.2 m³ = 1200000 cm³

Mass = Density * Volume

M = (2.5 g/cm³) * 1200000 cm³ = 3000000 g

1 lb = 453.592 g

3000000 g * (1 lbs / 453.592 g) = 6613.87 lbs

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2. In any energy transformation, energy is _____. A created B conserved C destroyed

algol13

Answer:

B energy can't be created or destroyed

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3 years ago

siniylev [52]

Answer:

a:it speed up

b:it should be positive since final

velocity is larger than initial velocity

c:acceleration is approximately 4.5

m/s^2

Explanation:

initial velocity=u=4.47m/s

Final velocity=v=17.9m/s

Time=t=3 seconds

a:the car speed up since the velocity

increased

b:change in velocity is positive

because final velocity is larger than

initial velocity

17.9-4.47=13.43 m/s

c: acceleration=(v-u)/t

acceleration=(17.9-4.47)/3

acceleration=13.43/3

acceleration=4.5 m/s^2

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3 years ago

When water vapor is cooled it three forms into water droplets why does this represent a conservation in mass

AVprozaik [17]

In the conservation of mass, mass is never created or destroyed in chemical reactions in the same way water is not created or destroyed it is only transferred from one form to another and its mass is always conserved.

8 0

3 years ago

a pitcher throws a 0.35 kg ball, giving it an acceleration of 10.0 m/s. what would the acceleration be, if the pitcher threw the

german

Answer:

New acceleration is 20.0 m/s².

Explanation:

F = m*a

F = 0.35 kg * 10.0m/s²

If Force will be doubled , and mass will be the same

we can write

2F = 2*0.35 kg*10.0 m/s²=0.35kg*20.0m/s²

New acceleration is 20.0 m/s².

8 0

4 years ago

A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi

bogdanovich [222]

Answer:

1) $W_{object} = 400 N$

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

$\sum F_{Tot} = 0$

In this case we just have y-direction forces.

$\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0$

Now, let's solve the equation for W(object).

$W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N$

2 ) To find the position of the heaviest weight we need to use the torque definition.

$\sum \tau = 0$

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

$L = 1.50 m$ ; total length of the bar.

$D_{AP} = x$ ; distance between Tension A and P point.

$D_{BP} = L-x$ ; distance between Tension B and P point.

$D_{W_{bar}P} = \frac{L}{2}-x$ ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

$\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0$

Finally we just need to solve it for x.

$x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}$

$x = 0.28 m$

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0

4 years ago