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Salsk061 [2.6K]

3 years ago

12

i just want a reason to live if my mom is always on me about everything and screams in my face to move out i have zero will to l

ive

2 answers:

Nookie1986 [14]3 years ago

8 0

Answer:

All i can say is, same lol i don't understand why parents complain about spending money and doing stuff for us when we didn't ask to be born and as our parents, its their job.

Explanation:

prisoha [69]3 years ago

5 0

Answer:

I understand how u feel if u need someone to talk to I will be here for u my stepmom is like that i will be here for u if u wana talk

Explanation:

You might be interested in

Paula has walked in a straight line, 30.5° north of west, for 1650 meters. How far south and east should she walks to return to

olchik [2.2K]

The answer is A.
Sy = 1650 x sin30.5 = 837.4 m toward south
Sx = 1650 x cos30.5 = 1421.7 m toward east

8 0

3 years ago

Your cousin Jannik skis down a blue square ski slope, with an initial speed of 3.6 m/s. He travels 15 m down the mountain side b

fenix001 [56]

Answer: The loss of energy due to friction is equal to 1,253 J.

Explanation:

The problem tells us that the skier has an initial speed of 3.6 m/s, which means that his initial kinetic energy is as follows:

K₁ = 1/2 m v₁² = 1/2 . 58.0 Kg. (3.6)² (m/s)² = 376 J

After coming to a flat landing, his final speed is 7.8 m/s, so the final kinetic energy is as follows:

K₂ = 1/2 m v₂² = 1/2. 58.0 Kg. (7.8)² (m/s)² = 1,764 J

Now, when skying down the slope the increase in kinetic energy only can come from another type of energy, in this case, gravitational potential energy.

If we take the ground flat level as a Zero reference, the initial gravitational potential energy, can be written as follows, by definition:

U₁ = m.g. h (1)

Now, we don't know the value of the height h, but we know that the incline has a 18º angle above the horizontal, and that the distance travelled along the incline is 15 m.

By definition, the sinus of an angle, is equal to the proportion between the height and the hypotenuse , so we can write the following equation:

sin 18º = h / 15 m ⇒ h = 15 m. sin 18º = 4.6 m

Replacing in (1), we get:

U₁ = 58.0 Kg. 9.8 m/s². 4.6 m = 2,641 J

So, we can get the total initial mechanical energy, as follows:

E₁ = K₁ + U₁ = 376 J + 2,641 J = 3,017 J

After arriving to the flat zone, all potential energy has become in kinetic energy, even though not completely, due to the effect of friction.

This remaining kinetic energy can be written as follows:

E₂ = K₂ = 1,764 J

The difference E₂-E₁, is the loss of energy due to friction forces acting during the travel along the 15 m path, and is as follows:

ΔE= E₂ - E₁ = 1,764 J - 3,017 J = -1,253 J

8 0

3 years ago

Which one of the following is not equivalent to 2.50 miles?

lisabon 2012 [21]

Answer:

Choice C is not equivalent to 2.50 miles.

Explanation:

The given data is now converted into feet, inches, kilometers, yards and centimeters:

mi - ft

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)$

$x = 13200\,ft$

$x = 1.320\times 10^{4}\,ft$ (Choice A)

mi - in

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)\cdot \left(12\,\frac{in}{ft} \right)$

$x = 158400\,in$

$x = 1.584\times 10^{5}$ (Choice B)

mi - km

$x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)$

$x = 4.025\,km$ (Different from Choice C)

mi - yd

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi}\right) \cdot \left(\frac{1}{3}\,\frac{yd}{ft} \right)$

$x = 4400\,yd$

$x = 4.40\times 10^{3}\,yd$ (Choice D)

mi - cm

$x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)\cdot \left(100000\,\frac{cm}{km}\right)$

$x = 402500\,cm$

$x = 4.025\times 10^{5}\,cm$ (Choice E)

Choice C is not equivalent to 2.50 miles.

6 0

4 years ago

Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T

Oksanka [162]

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

$T-f-T'=m_Ba$

Where $m_B$ =Mass of block B

Substitute the values

$50-10-T'=m_B\times 0=0$

$T'==40 N$

For block A

$T'-f'=m_Aa$

Where $m_A=$ Mass of block A

Substitute the values

$40-f'=m_A\times 0=0$

$f'=40 N$

Hence, the friction force acting on block A=40 N

3 0

3 years ago

How does an inclined plane change distance and how does it change direction

ElenaW [278]

Objects are known to go down because of a unbalanced force

7 0

3 years ago