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3 years ago

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A man of mass 50kg ascends a flight of stairs 5m high in 5 seconds. If acceleration due to gravity is 10ms-2, the power expended

is

1 answer:

Setler79 [48]3 years ago

7 0

The power expended is 500 W

Explanation:

First of all, we start by calculating the work done by the man in order to ascend: this is equal to the gravitational potential energy gained by the man, which is

$W=mg\Delta h$

where

m = 50 kg is the mass of the man

$g=10 m/s^2$ is the acceleration of gravity

$\Delta h = 5 m$ is the change in height

Substituting,

$W=(50)(10)(5)=2500 J$

Now we can calculate the power expended, which is given by

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

where

W = 2500 J is the work done

t = 5 s is the time elapsed

Substituting, we find

$P=\frac{2500}{5}=500 W$

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4. State in words how acceleration is calculated.

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You multiply force times friction

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3 years ago

Find the quantity of heat needed

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Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

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3 years ago

If the force applied to an object is not greater than the starting friction, what will happen to the object? Which answer is cor

AleksandrR [38]

Answer:

The object will move in the opposite direction of the force applied. - 2.

8 0

3 years ago

A cannon is fired straight up into the air. If the cannon ball comes back down to the launch point in 5 seconds, what was the ma

Nana76 [90]

Answer:

30.63 m

Explanation:

From the question given above, the following data were obtained:

Total time (T) spent by the ball in air = 5 s

Maximum height (h) =.?

Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:

Total time (T) spent by the ball in air = 5 s

Time (t) taken to reach the maximum height =.?

T = 2t

5 = 2t

Divide both side by 2

t = 5/2

t = 2.5 s

Thus, the time (t) taken to reach the maximum height is 2.5 s

Finally, we shall determine the maximum height reached by the ball as follow:

Time (t) taken to reach the maximum height = 2.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =.?

h = ½gt²

h = ½ × 9.8 × 2.5²

h = 4.9 × 6.25

h = 30.625 ≈ 30.63 m

Therefore, the maximum height reached by the cannon ball is 30.63 m

3 0

3 years ago

A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m = 0.900 fg (where a femtogra

just olya [345]

Answer:

9.73 x 10⁻¹⁰ m

Explanation:

According to Heisenberg uncertainty principle

Uncertainty in position x uncertainty in momentum ≥ h / 4π

Δ X x Δp ≥ h / 4π

Δp = mΔV

ΔV = Uncertainty in velocity

= 2 x 10⁻⁶ x 3 / 100

= 6 x 10⁻⁸

mass m = 0.9 x 10⁻¹⁵ x 10⁻³ kg

m = 9 x 10⁻¹⁹

Δp = mΔV

= 9 x 10⁻¹⁹ x 6 x 10⁻⁸

= 54 x 10⁻²⁷

Δ X x Δp ≥ h / 4π

Δ X x 54 x 10⁻²⁷ ≥ h / 4π

Δ X = h / 4π x 1 / 54 x 10⁻²⁷

= $\frac{6.6\times10^{-34}}{4\times3.14\times54\times10^{-27}}$

= 9.73 x 10⁻¹⁰ m

7 0

3 years ago