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julsineya [31]
2 years ago
11

Campaign against crime​

Physics
1 answer:
erica [24]2 years ago
8 0

Answer:

ok what do you need to know about ''campaign against crime​''?

Explanation:

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Charlie runs at an average speed of 6.5 km/hr. If he runs for 1.5 hours, how far has he traveled?
VikaD [51]

Answer:

9.75 km

Explanation:

Charlie runs 6.5 km/hr

-> Charlie wants to run for 1.5 hours

6.5km + 6.5km/2

= 6.5 km + 3.25km

= 9.75 km

8 0
2 years ago
A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume calorimeter. Consequently,
White raven [17]
Mass of methanol = 1.922g; Change in temperature = 5.14° C; Heat capacity of the bomb calorimeter + water = 8.69kJ/°C. Number of moles.
3 0
2 years ago
A 60​-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the
Mariulka [41]

Answer:

part (a). 176580 J

part (b). 197381 J

Explanation:

Given,

  • Density of the chain = \rho\ =\ 10\ kg/m.
  • Length of the chain = L = 60 m
  • Acceleration due to gravity = g = 9.81 m/s^2

part (a)

Let dy be the small element of the chain at a distance of 'y' from the ground.

mass of the small element of the chain = \rho dy

Work done due to the small element,

dw\ =\ \rho g (60\ -\ y)dy\\

Total work done to wind the entire chain = w

w\ =\ \displaystyle\int_{0}^{L} \rho g(60\ -\ y)dy\\\Rightarrow  w\ =\ \rho g\left |(60y\ -\ \dfrac{y^2}{2})\ \right |_{0}^{60}\\\Rightarrow w\ =\ 10\times 9.81\times (60\times 60\ -\ \dfrac{60^2}{2})\\\Rightarrow w\ =\ 176580\ J

part (b)

  • mass of the block connected to the chain = m = 35 kg

Total work done to wind the chain = work done due to the chain + work done due to the mass

\therefore W\ =\ w\ +\ mgL\\\Rightarrow W\ =\ 176580\ +\ 35\times 9.81\times 60\\\Rightarrow W\ =\ 176580\ +\ 20601\\\Rightarrow W\ =\ 197381\ J

4 0
2 years ago
A gas‑forming reaction produces 1.90 m 3 1.90 m3 of gas against a constant pressure of 179.0 kPa. 179.0 kPa. Calculate the work
goldfiish [28.3K]

Answer:

The work done is 3.4 × 10⁵ J.

Explanation:

Given:

Pressure of the gas produced (P) = 179 kPa

Volume of the gas produced (ΔV) = 1.90 m³

We need to find the work done in joules. For that, we don't need any conversion as the units are already in SI units which will give the result in Joules only.

Now, let us verify our results by using conversion factors and without using them.

Using conversion factors:

1 m³ = 1000 L

So, 1.90 m³ = 1.90 m³ × 1000 \frac{L}{m^3} = 1900 L

Also, 1 atm = 101.325 kPa

So, 179 kPa = 179 kPa × \frac{1\ atm}{101.325\ kPa} = 1.767 atm

Now, work done in a constant pressure process is given as:

Work = Pressure × Volume change

Work = P × ΔV

Work = 1.767 atm × 1900 L

Work = 3.36 × 10³ atm-L

Now, again using the energy conversion for work.

1 atm-L = 101.325 J

So, 3.36 × 10³ atm-L = 3.36 × 10³ atm-L × \frac{101.325\ J}{1\ atm-L} = 3.4 × 10⁵ J

Therefore, the work done is 3.4 × 10⁵ J.

Now, let us verify the above result without any conversion.

Work = P × ΔV = 179 × 1000 × 1.90 = 3.4 × 10⁵ J.

Therefore, the work done is same by both ways.

Hence the work done is 3.4 × 10⁵ J.

8 0
3 years ago
A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simp
masha68 [24]

To solve this problem we must rely on the equations of the simple harmonic movement that define the period as a function of length and gravity as

T = 2\pi \sqrt{\frac{l}{g}}

Where

l = Length

g = Gravity

Re-arrange to find L,

L = g (\frac{T}{2\pi})^2

Our values are given as

g = 9.81m/s

T = 10.1s

Replacing,

L = g (\frac{T}{2\pi})^2

L = (9.81) (\frac{10.1}{2\pi})^2

L = 25.348m

Therefore the height would be 25.348m

5 0
3 years ago
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