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3 years ago

I hate life do u know why

Physics

2 answers:

Katena32 [7]3 years ago

6 0

Answer:

Most of us have experienced that pivotal peak of pain, anger or frustration in which we want to scream “I hate my life.” Yet, the feeling that a dark cloud has specifically settled over us and our experiences can feel pretty isolating. The truth is, no matter how singled out or overwhelmed we feel, and no matter what area we are struggling in, we are not alone. More than half of U.S. workers are unhappy with their job. One in 10 Americans struggles with depression. All of us have moments of utter despair. Escaping from this hopeless-seeming state may feel impossible. Yet, in reality, we are not doomed, and we are not powerless. No matter what our circumstances, we can all learn tools to help us emerge from the darkest moments in our lives.

viktelen [127]3 years ago

5 0

Answer:ummmm no why

Explanation:??

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A 100-kg astronaut is floating in outer space. If the astronaut throws a 2-kilogram wrench at a speed of 10 meters per second, w

sergey [27]

Since Astronaut and wrench system is isolated in the space and there is no external force on it

So here momentum of the system will remain conserved

so here we can say

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

initially both are at rest

so here plug in all values

$0 = 100 v_{1f} + 2\times 10$

$v_{1f} = -0.20 m/s$

so here the astronaut will move in opposite direction and its speed will be equal to 0.20 m/s

3 0

3 years ago

How is the q of an rlc parallel resonant circuit calculated?

notka56 [123]

Answer:

It is calculated by dividing Resistance, R, by Inductive reactance, XL.

Explanation:

Q is called the Q factor of a resonance circuit. In a parallel resonance circuit, it is calculated by finding the ratio of the power stored in the circuit to the power distributed in the circuit. It is a way of measuring the quality of a circuit or how effective the circuit is.

Q factor is the inverse in the resonance series circuit.

Q factor of a resonance parallel circuit,

R = Resistance

XL = Inductive reactance

3 0

3 years ago

A car drives around a curve with radius 400 m at a speed of 32 m/s. The road is banked at 7.0 degree. The mass of the car is 150

Ronch [10]

Answer:

The magnitude of the centripetal force to make the turn is 3,840 N.

Explanation:

Given;

radius of the cured road, r = 400 m

speed of the car, v = 32 m/s

mass of the car, m = 1500 kg

The magnitude of the centripetal force to make the turn is given as;

$F_c = \frac{mv^2}{r}$

where;

Fc is the centripetal force

$F_c = \frac{mv^2}{r} \\\\F_c = \frac{(1500)(32)^2}{400}\\\\F_c = 3,840 \ N$

Therefore, the magnitude of the centripetal force to make the turn is 3,840 N.

3 0

3 years ago

Starburst galaxies:

alexgriva [62]

Answer:

are often associated with a galaxy that is colliding with another galaxy.

Explanation:

A starburst galaxy is a galaxy that undergoes very fast formation of stars. The rate at which stars are born is 100 times more than 3 solar masses per year of the Milky Way. The starburst is stage of the formation of a galaxy. After this stage is complete the stars will have used almost all the gas in it. As the star formation rate is very fast the difference between the age of the stars and the galaxy itself is very less. The star formation is triggered by mergers and tidal interactions between gas-rich galaxies.

6 0

3 years ago

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

$p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}$

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

$F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N$

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

$F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}$

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0

3 years ago