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3 years ago

8

a horizontal force of 100N is required to push a crate across a factory floor at a constant speed. What is the net force acting

on the crate/ What is the force of friction acting on the crate?

1 answer:

Bas_tet [7]3 years ago

8 0

If the crate is moving along the floor in the same direction with a constant speed, it is in dynamic equilibrium. Equilibrium means there is no net force acting on the crate.

Since there is no net force on the crate, there must be a friction force on the crate equal in magnitude and opposite in direction to the applied horizontal force. Therefore the force of friction acting on the crate is 100N.

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Phosphorus-32, a radioactive isotope of phosphorus-31 (atomic number 15), undergoes a form of radioactive decay whereby a neutro

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Answer:

The product of the decay its Sulfur-32

Explanation:

Phosphorus-32 ( lets write it $_{15}^{32}P$ , where the number above its the atomic mass and the number below the atomic number) decays turning a neutron into a proton and emitting radiation on the form of a electron. This is the beta minus decay, and, actually, an electronic antineutrino its also produced. We can write this decay for an X isotope with a Y isotope produced as:

$_{Z}^{A}X \to _{Z+1}^{A}Y + e^- + \bar{\nu_e}$

where $e^-$ its the electron, and $\bar{\nu_e}$ the electronic antineutrino . We can see that the atomic number increases by one (cause a proton it produced and retained into the nucleus), and the atomic mass is approximately the same (there is a small difference between the neutron and proton mass, but its very small).

So, Phosphorus-32 (atomic number 15) will turn to an element with atomic number 16, and atomic mass 32, as:

$_{15}^{32}P \to _{15+1}^{32}Y + e^- + \bar{\nu_e}$ .

$_{15}^{32}P \to _{16}^{32}Y + e^- + \bar{\nu_e}$ .

The Y isotope must have an atomic number of 16 and an atomic mass of 32. The element with atomic number 16 its Sulfur (S), so, our decay its

$_{15}^{32}P \to _{16}^{32}S + e^- + \bar{\nu_e}$ .

and the product of such decay its Sulfur-32

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4 years ago

How many micrometers are there in 4cm

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40000

Explanation:

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3 years ago

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On May 20, 1999, 37-year old Robbie Knievel, son of the famed daredevil Evel

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3 years ago

Rank the angular speed of the following objects, from highest to lowest:- A bowling ball of radius 12.3 cm rotating at 8.21 radi

dlinn [17]

Answer:

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

Explanation:

<u>To rank the angular speed (ω) of the objects, we need first calculate its value for every object:</u>

A bowling ball of radius 12.3cm rotating at 8.21 radians per second:

ω = 8.21 rad/s

A tire of radius 0.321m rotating at 75.8 rpm:

$\omega = 75.8 \frac{rev}{min}\cdot \frac{2\pi rad}{1rev}\cdot \frac{1min}{60s} = 7.94rad/s$

A 6.84cm diameter top spinning at 375 degrees per second:

$\omega = 375 \frac{^\circ}{s}\cdot \frac{2\pi rad}{360^ \circ} = 6.54rad/s$

A square with sides (b) 0.458m long, whose corners are moving with tangential speed (v) 2.51 m/s as it rotates about its center:

$\omega = \frac{v}{r} = \frac{v}{\frac{b}{2}\sqrt 2} = \frac{2.51 m/s}{\frac{0.458 m}{2} \sqrt 2} = 7.75rad/s$

A rock on a string, being swung in a circle of radius 0.521 m with a centripetal acceleration (a) of 25.4 m/s²:

$\omega = \sqrt \frac{a}{r} = \sqrt \frac{25.4 m/s^{2}}{0.521m} = 6.98rad/s$

<u>Now, the rank of the angular speed of the objects, from highest to lowest is: </u>

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

I hope it helps you!

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3 years ago

A two-liter bottle of your favorite beverage has just been removed from the trunk of your car. The temperature of the beverage i

Ksivusya [100]

Answer:

a) 209.3 kilojoules must be removed from two liter of beverage, b) A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles, c) Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

Explanation:

a) <em>How much heat energy must be removed from your two liters of beverage?</em>

At first we suppose that the beverage has the mass and specific heat of water and that there are no energy interactions between the bottle and its surroundings.

From the First Law of Thermodynamics and definition of sensible heat, we get that amount of removed heat ( $Q$ ), measured in kilojoules, is represented by the following formula:

$Q = \rho \cdot V\cdot c\cdot (T_{o}-T_{f})$ (Eq. 1)

Where:

$\rho$ - Density of the beverage, measured in kilograms per cubic meter.

$V$ - Volume of the bottle, measured in cubic meters.

$c$ - Specific heat of water, measured in kilojoules per kilogram-Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures, measured in Celsius.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $V = 2\times 10^{-3}\,m^{3}$ , $c = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}$ , $T_{o} = 35\,^{\circ}C$ and $T_{f} = 10\,^{\circ}C$ , then:

$Q = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (2\times 10^{-3}\,m^{3})\cdot \left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right) \cdot (35\,^{\circ}C-10\,^{\circ}C)$

$Q = 209.3\,kJ$

209.3 kilojoules must be removed from two liter of beverage.

b) <em>You are having a party and need to cool 10 of these two-liter bottles in one-half hour. What rate of heat removal, in kW, is required?</em>

The total amount of heat that must be removed from 10 2-L bottles is:

$Q_{T} = 10\cdot (209.3\,kJ)$

$Q_{T} = 2093\,kJ$

If we suppose that bottles are cooled at constant rate, then, rate of heat removal is determined by this formula:

$\dot Q = \frac{Q_{T}}{\Delta t}$ (Eq. 2)

Where:

$Q_{T}$ - Total heat, measured in kilojoules.

$\Delta t$ - Time, measured in seconds.

$\dot Q$ - Rate of heat removal, measured in kilowatts.

If we know that $Q_{T} = 2093\,kJ$ and $\Delta t = 1800\,s$ , we find that rate of heat removal is:

$\dot Q = \frac{2093\,kJ}{1800\,s}$

$\dot Q = 1.163\,kW$

A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles.

c) <em>Assuming that your refrigerator can accomplish this and that electricity costs 8.5 cents per kW-hr, how much will it cost to cool these 10 bottles (in $)?</em>

A kilowatt-hour equals 3600 kilojoules. The electricity cost is equal to the removal heat of 10 bottles ( $Q_{T}$ ), measured in kilojoules, and unit electricity cost ( $c$ ), measured in US dollars per kilowatt-hour. That is:

$C = c\cdot Q_{T}$

If we know that $c = 0.085\,\frac{USD}{kWh}$ and $Q_{T} = 2093\,kJ$ , the total cost of cooling 10 bottles is:

$C = \left(0.085\,\frac{USD}{kWh}\right)\cdot \left(2093\,kJ\right)\cdot \left(\frac{1}{3600}\,\frac{kWh}{kJ} \right)$

$C = 0.049\,USD$

Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

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4 years ago