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4 years ago

13

Anna pushes a box with a force of 8.00 newtons. She generates a power of 3.00 watts. How much time does it take for Anna to move

the box the distance of 9.72 meters?

2 answers:

QveST [7]4 years ago

8 0

Power is the energy in a system per time. It will have units of Watts which is equal to joules per second. It can be expressed as:

P = E / t

where E = Force x distance

P = Fd / t
t = Fd / P
t = 8 (9.72) / 3.0
t = 25.92 s

Ratling [72]4 years ago

8 0

Answer:

time taken, t = 25.92 seconds

Explanation:

It is given that,

Force acting on the box, F = 8 N

Power generated on the box, P = 3 watts

Distance traveled by the box, d = 9.72 m

We know that the power generated by an object is equal to the total work done divided by total time taken. Its formula is given by :

$P=\dfrac{W}{t}$

Since, W = F d

$P=\dfrac{Fd}{t}$

$t=\dfrac{Fd}{P}$

$t=\dfrac{8\ N\times 9.72\ m}{3\ W}$

t = 25.92 seconds

So, the time taken by Anna to move the box to the distance of 9.72 meters is 25.92 seconds. Hence, this is the required solution.

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Answer:

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<u>Instant Acceleration</u>

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Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

$\displaystyle v(t)=\frac{df}{dt}$

And the acceleration is

$\displaystyle a(t)=\frac{dv}{dt}$

Or equivalently

$\displaystyle a(t)=\frac{d^2f}{d^2t}$

The given height of a projectile is

$f(t)=-16t^2 +238t+3$

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$\displaystyle v(t)=-32t+238$

And the acceleration

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$\boxed{\displaystyle a(5)=-32}$

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3 years ago

What is the formula of the compound between calcium and sulfur that has the percent composition 55.6?

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CaS is the empirical formula of the compound between calcium and sulphur that has the percent composition 55.6.

When percentages are given, we take a total mass of 100 grams.

Therefore the mass of each element is equal to the percentage given.

Mass of Ca = 55.6 g (given) of

S Mass = 44.4 g (100 - 55.6 = 44.4)

Step 1: Convert the given mass to moles.

moles Ca = given mass Ca / molar mass Ca

moles = 55.6 / 40 = 1.39 moles

mol S = specific mass S / molar mass S

mol = 44.4 / 32 = 1.39 mol

Step 2: Divide the molar ratio of each molar value by the smallest number of moles calculated.

For Ca = 1.39 / 1.39 = 1

For S = 1.39 / 1.39 = 1

The ratio of Ca : S = 1:1

Hence the empirical formula of the given compound will be CaS.

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Answer:

Conservation of angular momentum

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3 years ago

A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of

kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

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The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

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3 years ago

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The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $\text{heat}_{absorbed}=\text{heat}_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 39 g

$m_2$ = mass of coffee = 166 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $83^oC$

$c_1$ = specific heat of aluminium = $0.904J/g^oC$

$c_2$ = specific heat of coffee= $4.1801J/g^oC$

Putting all the values in equation 1, we get:

$39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]$

$T_{final}=80.14^oC$

Hence, the final temperature of the solution is $80.14^oC$

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4 years ago