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3 years ago

What is the length of AR?

Physics

1 answer:

AlladinOne [14]3 years ago

6 0

Answer:Assault rifle?

Explanation:

20 m16 size and a 14.5 barrel and get a pump shot and get the blue version and go to tilted and one pump even tho forkknife is garbage.:)

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A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of

kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

The mass of the pole vaulter, m = 57 kg

The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

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3 years ago

1. Approximately how many people watch the March Madness tournament?

Stolb23 [73]

40 million
1. Approximately how many people watch the March Madness tournament? There are more than 140 million that watch March Madness.

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3 years ago

Do humans depend on photosynthesis? Explain how, why, or why not.

Elanso [62]

Humans rely on photosynthesis to generate food as well as provide energy for heating, lighting, and electricity.

4 0

2 years ago

Read 2 more answers

A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d

andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x² - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0

3 years ago

bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc

USPshnik [31]

Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

$m_A$ = mass of car A;

$v_{A1}$ = the initial velocity of car A;

$v_{A2}$ = the final velocity of car A;

and

$m_B$ = mass of car B;

$v_{B1}$ = the initial velocity of car B;

$v_{B2}$ = the final velocity of car B.

Then, the law of conservation of momentum demands that

$m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}$

And the conservation of kinetic energy says that

$\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2$

These two equations are solved for final velocities $v_{A2}$ and $v_{B2}$ to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

$\boxed{v_{A2} = -1.05m/s}$

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

$\boxed{v_{B2} = 3.49m/s}$

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.

4 0

3 years ago

What is the length of AR?​

What is the length of AR?