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qwelly [4]
3 years ago
12

Hey idk what to do anymore. I am just about to give up and I feel like just running off to the woods and live alone at least I c

an feel like I am not failing anyone. I hate that I am so nice to people and I let people stomp on me like I am nothing and act like it doesn't hurt me. I have been hurt for so long that I don't know what to do anymore I have no real friends in my real life I sometimes feel like I am worth nothing and I am trapped like this but I don't let go I still try and people don't understand that is hurts so bad to the point that I have gone numb. I stay strong just bc I am hoping that it will get better but it doesn't. I cry every night just to let go of everything from the night sometimes I just wish that I never woke up bc of how amazing it is to get away from everything and worry about nothing but it doesn't last forever. I hate this feeling and it is worse bc I can't tell anybody.
Physics
1 answer:
Alona [7]3 years ago
3 0

Answer:

Don't feel that way ok , every thing is going to be ok in the end an if it's not ok its the end. you can fight through it don't let anyone or anything get to you , just know that we all are here for you an, every one on this is unique in their own way ! just love your self an others will do the same love you all

You might be interested in
A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)
Alekssandra [29.7K]

Answer:

The electric field strength is 6.6\times10^{5}\ V/m

Explanation:

Given that,

Magnetic field = 0.150 T

Speed v= 4.40\times10^{6}\ m/s

We need to calculate the electric field strength

Using formula of velocity

v=\dfrac{E}{B}

E=v\times B

Where, v = speed

B = magnetic field

Put the value into the formula

E=4.40\times10^{6}\times0.150

E=660000\ V/m

E=6.6\times10^{5}\ V/m

Hence, The electric field strength is 6.6\times10^{5}\ V/m

4 0
3 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
2) Why do you think fossil fueled cars are still out numbering alternative and replacement fueled cars even though fossil fuels
Misha Larkins [42]

Answer:

Fossil-fueled cars are still outnumbering the alternative and replacement-fueled cars because they are generally cost-effective and are efficient.

Explanation:

<em>Fossil fuels</em> are non-renewable energy that cannot be replenished. Although many people know how harmful using them can be to the environment, still, people continue to buy cars that are powered by it. This is because, <em>using fossil-fueled cars can help you save money</em> compared to using<em> replacement-fueled cars.</em><em> Its engines are also more powerful,</em> which means they can arrive at their destination in just a few minutes. <u>This makes it efficient.</u> The cost of maintaining a fossil-fueled car is less expensive compared to<em> replacement-fueled cars.</em> People are naturally keen when it comes to budgeting, thus, many people still buy cars powered by fossil fuels.

3 0
3 years ago
Sue and jenny kick a soccer ball at exactly the same time. sue's foot exerts a force of 57.6 n to the north. jenny's foot exerts
Alika [10]
It is 17.58 my friennnd
8 0
4 years ago
Solve for work when
BlackZzzverrR [31]

So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

  • If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
  • If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.
<h3>Formula Used</h3>

The work done by a moving object can be expressed in the equation:

If the Angle Is Ignored

\boxed{\sf{\bold{W = F \times s}}}

If the Angle Effect on Work

\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}

With the following condition:

  • W = work that done by object (J)
  • F = force that applied (N)
  • s = shift or distance (m)
  • \sf{\theta} = angle of elevation (°)

<h3>Solution</h3>

We know that :

  • F = force that applied = \sf{1.41 \times 10^4} N
  • s = shift or distance = 84.9 m
  • \sf{\theta} = angle of elevation = 45°

What was asked ?

  • W = work that done by object = ... J

Step by step :

\sf{W = F \times s \times \cos(\theta)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}

\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}

\sf{W = 59.8545 \sqrt{2}}

\boxed{\sf{W \approx 84.65 \: J}}

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

3 0
1 year ago
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