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3 years ago

What is a pickel in a nasty way do?

Physics

2 answers:

katen-ka-za [31]3 years ago

8 0

Answer:

pickle rick

Explanation:

zvonat [6]3 years ago

6 0

Answer:

um not sure but sounds nasty

Explanation:

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Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

kramer

Answer:

The height reached by the material on Earth is 91 km.

Explanation:

Given that,

Mass $M_{Io}=8.93\times10^{22}\ kg$

Radius = 1821 km

Height $h_{Io}=500\ km$

Suppose we need to find that how high would this material go on earth if it were ejected with the same speed as on Io?

We need to calculate the acceleration due to gravity on Io

Using formula of gravity

$g =\dfrac{GM_{Io}}{(R_{Io})^2}$

Put the value into the formula

$g=\dfrac{6.67\times10^{-11}\times8.93\times10^{22}}{(1821\times10^{3})^2}$

$g=1.79\ m/s^2$

Let v be the speed at which the material is ejected.

We need to calculate the height

Using the formula of height

$H=\dfrac{v^2}{2g}$

Using ratio of height of earth and height of Io

$\dfrac{H_{e}}{H_{Io}}=\dfrac{\dfrac{v^2}{2g_{e}}}{\dfrac{v^2}{2g_{Io}}}$

$\dfrac{H_{e}}{H_{Io}}=\dfrac{g_{Io}}{g_{e}}$

Put the value into the formula

$\dfrac{H_{e}}{H_{Io}}=\dfrac{1.79}{9.8}$

$\dfrac{H_{e}}{H_{Io}}=0.182$

$H_{e}=0.182\times H_{Io}$

$H_{e}=0.182\times500$

$H_{e}=91\ km$

Hence, The height reached by the material on Earth is 91 km.

3 0

3 years ago

Which statement is true about vectors?

lora16 [44]

Vectors need direction and magnitude to exist, otherwise, it becomes scalars that do not require direction to be expressed.

eg - Velocity is a vector as it needs direction, but speed is scalar and does not need direction

hope this helps

5 0

3 years ago

Read 2 more answers

A roller coaster at an amusement park girls from the top of a tall hill to the bottom during this time of the roller coaster

Mrac [35]

D), because the roller coaster it's falling, therefore, losing height, but gaining velocity

8 0

4 years ago

Steam at 400C has a specific volume of 0.02m3/kg. Determine the pressure of the steam based on a) the ideal gas equation b) the

nikklg [1K]

Answer:

by ideal gas pressure = 15529.475 kPa

by compressibility chart pressure = 12576 kPa

by steam tables Pressure = 12517 kPa

Explanation:

given data

temperature T = 400°C = 673 K

volume v = 0.02 m³/kg

to find out

pressure by ideal gas, compressibility chart and steam tables

solution

we know here by table

gas constant R is 0.4615 kJ/ kg-K

and critical temp Tc = 647.1 K

and critical pressure Pc = 22064 kPa

so by ideal gas pressure is

pressure = R×T / v

pressure = 0.4615 × 673 / 0.02

pressure = 15529.475 kPa

and

by compressibility chart

temperature reduce is = T/ Tc

temperature reduce Tr = 673 / 647.1

Tr = 1.040 K

so pseudo reduce volume is here

reduce volume Vr = v / ( RTc/Pc)

reduce volume Vr = $\frac{0.02}{\frac{461.5(647.1)}{22064*10^{3} } }$

0.02 / ( 461.5(647.1) / 22064×10³)

reduce volume = 1.48

and we know by compressibility chart

reduce pressure Pr is 0.57

pressure = Pr × Pc

pressure = 0.57 × 22064 × 10³

pressure = 12576 kPa

and

from steam table

pressure is 12.5 MPa at 673 K and 0.020030 m³/kg

pressure is 15 MPa at 673 K and 0.015671 m³/kg

pressure P is

$\frac{0.02 - 0.020030}{0.015671 - 0.020030}$ = $\frac{ P - 12.5}{15 - 12.5}$

Pressure = 12517 kPa

4 0

3 years ago

The latent heat of fusion of water at 0 °C is 6.025 kJ mol'' and the molar heat

Scilla [17]

Answer:

$\Delta H_{tot} = 2258.025\,kJ$

Explanation:

The amount of heat released from water is equal to the sum of latent and sensible heats. Let suppose that water is initially at a temperature of $25^{\circ}C$ . Then:

$\Delta H_{tot} = \Delta H_{s, w} + \Delta H_{f,w} + \Delta H_{s,i}$

$\Delta H_{tot} = n\cdot (c_{w}\cdot \Delta T_{w} + L_{f} + c_{i}\cdot \Delta T_{i})$

Finally, the amount of heat released from water is now computed by replacing variables:

$\Delta H_{tot} = (1\,mol)\cdot \left[\left(75.3\,\frac{kJ}{mol\cdot K} \right)\cdot (25^{\circ}C-0^{\circ}C)+ 6.025\,\frac{kJ}{mol} + \left(37.7\,\frac{kJ}{mol\cdot K} \right)\cdot (0 + 10^{\circ}C)\right]$ $\Delta H_{tot} = 2258.025\,kJ$

4 0

3 years ago