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Maurinko [17]
2 years ago
15

How do u say i like u to ur crush but not make it akward

Physics
2 answers:
Sever21 [200]2 years ago
4 0

Answer:

bark at then

Explanation:

Tcecarenko [31]2 years ago
3 0

Answer:

first, become bsf with him.

then, laugh at his jokes even is it wasn't funny and make eye contact.

NEVER tell him u like him until u r 100% sure he has the slightest bit of interest towards u other than just being friends

Explanation:

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A man can throw a ball a maximum horizontal distance of 56.1523 m. The acceleration of gravity is 9.8 m/s 2 . How far can he thr
IceJOKER [234]

Answer:

hmax= 28.2 m

Explanation:

Horizontal movement:

  • For the horizontal movement, as once released, nothing exerts any influence on the ball in the horizontal direction, it will keep moving at a constant speed, so, the equation for the horizontal displacement is as follows:

        x = v₀*t = 56.1523 m

Vertical movement:

  • For the vertical movement, once thrown upward, the ball is under the sole influence of gravity, g, which causes the ball to accelerate downward, and which magnitude is 9.8 m/s².
  • At any time, applying the definition of acceleration, we can find the value of the velocity, as follows:

        vf = vo-g*t

  • When the ball reaches to the maximum height, the ball will come momentarily at rest, so vf =0.
  • At this point, we can find the time at which  the ball reached to its highest point, as follows:

        t = \frac{vo}{g}

  • In the same way, we can find the maximum height reached by the ball, just replacing this value of time in the equation for the displacement, in the vertical direction, as follows:

        h = vo*t -\frac{1}{2} *g*t^{2}

        hmax = vo*(\frac{vo}{g}) -\frac{1}{2} *g*(\frac{vo}{g} )^{2} =\frac{vo^{2}}{2*g}

  • Now, returning to the horizontal movement, as the time must be the same for both movements, we can replace the value for time we have just found, in the equation for the horizontal displacement:

       x = v0x * t = v0x* \frac{voy}{g}

  • But as we know that vox = voy, we can rewrite the equation above as follows:

       x = v0* t = v0* \frac{vo}{g} =\frac{vo^{2} }{g} = 56.1523 m

  • We can solve for v₀, as follows:

        vo = \sqrt{56.1523 m *9.8 m/s2} =23.5 m/s

  • Now, we can replace this value in the expression for hmax, as follows:
  • hmax = \frac{vo^{2}}{2*g} =\frac{(23.5m/s)^{2}}{2*9.8 m/s2} =  28.2 m
4 0
3 years ago
A 2578-kg van runs into the back of a 825-kg compact car at rest. They move off together at 8.5 m/s. Assuming the friction with
allochka39001 [22]

Answer:

<em>The initial speed of the car = 11.22 m/s</em>

Explanation:

Law of conservation of energy: It states that when two bodied collide in a closed system, the sum of momentum before collision is equal to the sum of momentum after collision.

<em>Note:</em> A close system is one that is free from from external forces. E.g Frictional force.

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

m₁u₁ + m₂u₂ = (m₁ + m₂)V..................... Equation 1

Where m₁ = mass of the van, m₂ = mass of the car, u₁ = initial velocity of the van, u₂ = initial velocity of the car, V = Common velocity.

<em>Given: m₁ = 2578 kg, m₂ = 825 kg, u₂ = 0 ( the car was at rest), V= 8.5 m/s</em>

<em>Substituting these values into equation 1, and solving for u₁</em>

<em>2578(u₁) + (825 × 0) = (2578 + 825)8.5</em>

<em>2578u₁ = 28925.5</em>

<em>Dividing both side of the equation by the coefficient of u₁</em>

<em>2578u₁/2578 = 28925.5/2578</em>

<em>u₁ = 11.22 m/s</em>

<em>The initial speed of the car = 11.22 m/s</em>

3 0
3 years ago
A cannon shoots an artillery shell with an initial velocity of 400 meters/second at an
OLga [1]

It will land at 14139.19 m away.

Explanation:

The expression for range d on level ground is given by;

d=v² sin (2Ф) /g where Ф is the fire angle and g is acceleration due to gravity

Given v=400m/s ,Ф= 60° and g=9.8 so,

d= 400² sin(120°) /9.8

d=(400²×0.86602540378) / 9.8

d=14139.19 m

Motion for falling object : brainly.com/question/11799308

Keyword : initial velocity, angle, range

#LearnwithBrainly

6 0
3 years ago
Free tingsssssssssss
il63 [147K]

Explanation:

yayyy thx u soo muchihuhuhu

5 0
2 years ago
Read 2 more answers
The strength of the electric field 0.5 m from a 6 µc charge is n/c. (use k = 8.99 × 109 n•meters squared per coulomb squared and
diamong [38]

53....................................

Explanation:

6 0
2 years ago
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