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DanielleElmas
4 days ago
7

While watching the crane in operation, an observer mentions to you that for a given load there is a maximum angle θmax between 0

∘ and 90∘ that the crane arm can make with the horizontal without tipping the crane over. is this correct?
Physics
1 answer:
ValentinkaMS [3.3K]4 days ago
7 0
<span>I think this could be right, however, it is not safe. Personally, I would not approve the scenario as it poses risks to both the crane operators and the company. In my view, the answer is NO because although the crane might technically perform the movement, it shouldn't happen</span>.
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A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
ValentinkaMS [3343]
No one is going to handle that for a mere 5 points lol.
8 0
1 month ago
When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (
inna [2989]
The answer is (C) 4 beats per second. The number of beats is computed as the difference between the frequencies of the two tuning forks. Plugging in the frequency values yields a result. Thus, the number of observable beats per second will be 4.
3 0
21 day ago
Consider the uniform electric field \vec{E} =(4000~\hat{j}+3000~\hat{k})~\text{N/C} ​E ​⃗ ​​ =(4000 ​j ​^ ​​ +3000 ​k ​^ ​​ ) N/
kicyunya [3154]

Answer:

Electric flux is calculated as \phi=31562.63\ Nm^2/C

Explanation:

We start with the given parameters:

The electric field impacting the circular surface is E=(4000j+3000k)\ N/C

Our objective is to ascertain the electric flux passing through a circular region with a radius of 1.83 m situated in the xy-plane. The area vector is oriented in the z direction. The formula for electric flux is expressed as:

\phi=E{\cdot}A

\phi=(4000j+3000k){\cdot}Ak

Applying properties of the dot product, we calculate the electric flux as:

\phi=3000\times Ak

\phi=3000\times \pi (1.83)^2

\phi=31562.63\ Nm^2/C

Consequently, the electric flux for the circular area is \phi=31562.63\ Nm^2/C. Thus, this represents the required answer.

4 0
1 month ago
Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.
inna [2989]
The heat required to raise the temperature of a substance by \Delta T is represented by
Q=m C_p \Delta T
where m stands for the mass of the substance and C_p indicates the specific heat of the substance. In this situation, we possess m=100~g=0.1~Kg and C_p=4.19~KJ/(Kg K), the specific heat of water.
Consequently, we can ascertain the temperature rise \Delta T:
\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C
Initially, the water's temperature was 25^{\circ}C, so the end temperature should be
T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C
Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ
where C_L denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of 100^{\circ}C (the temperature at which the phase change begins)

4 0
1 month ago
A 1.00 l sample of a gas at 25.0◦c and 1.00 atm contains 50.0 % helium and 50.0 % neon by mass. what is the partial pressure of
inna [2989]
V = Volume of gas sample = 1.00 L = 0.001 m³T = temperature of gas = 25.0 °C = 25 + 273 = 298 K P = pressure = 1.00 atm = 101325 Pa n = number of moles of gas using ideal gas law:PV = n RT101325 (0.001) = n (8.314) (298)n = 0.041 n₁ = moles of heliumn₂ = moles of neonm₁ = mass of helium = n₁ (4) = 4 n₁m₂ = mass of neon = n₂ (20.2) = 20.2 n₂given that:m₁ = m₂4 n₁ = 20.2 n₂n₁ = 5.05 n₂also n₁ + n₂ = n5.05 n₂ + n₂ = 0.041n₂ = 0.0068mole fraction of neon is mole fraction = n₂ /n = 0.0068/0.041 = 0.166P₂ = partial pressure of neon =(mole fraction) P P₂ = (0.166) (1)P₂ = 0.166 atm
5 0
9 days ago
Read 2 more answers
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