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3 years ago

In the diagram, q1, q2, and q3 are in a straight line.

Physics

1 answer:

schepotkina [342]3 years ago

3 0

Answer:

did anyone figure it out

Explanation:

please help

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The mass of an object is 45 kilograms. its weight on earth is what?

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Why is velocity most important in space launch?

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3 years ago

Read 2 more answers

I will gib brainlyest or whatever.

astraxan [27]

Answer:

Range of the projectile: approximately $1.06 \times 10^{3}\; {\rm m}$ .

Maximum height of the projectile: approximately $80\; {\rm m}$ (approximately $45.0\; {\rm m}$ above the top of the cliff.)

The projectile was in the air for approximately $7.07\; {\rm s}$ .

The speed of the projectile would be approximately $155\; {\rm m \cdot s^{-1}}$ right before landing.

(Assumptions: drag is negligible, and that $g = 9.81\; {\rm m\cdot s^{-1}}$ .)

Explanation:

If drag is negligible, the vertical acceleration of this projectile will be constantly $a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}$ . The SUVAT equations will apply.

Let $\theta$ denote the initial angle of elevation of this projectile.

Initial velocity of the projectile:

vertical component: $u_{y} = u\, \sin(\theta) = 153\, \sin(11.2^{\circ}) \approx 29.71786\; {\rm m\cdot s^{-1}}$
horizontal component: $u_{x} = u\, \cos(\theta) = 153\, \cos(11.2^{\circ}) \approx 150.086\; {\rm m\cdot s^{-1}}$ .

Final vertical displacement of the projectile: $x_{y} = (-35)\; {\rm m}$ (the projectile landed $35\: {\rm m}$ below the top of the cliff.)

Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the final vertical velocity $v_{y}$ of this projectile:

${v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}$ .

$\begin{aligned} v_{y} &= -\sqrt{{u_{y}}^{2} + 2\, a_{y} \, x_{y}} \\ &= -\sqrt{(29.71786)^{2} + 2\, (-9.81)\, (-35)} \\ &\approx (-39.621)\; {\rm m\cdot s^{-1}}\end{aligned}$ .

(Negative since the projectile will be travelling downward towards the ground.)

Since drag is negligible, the horizontal velocity of this projectile will be a constant value. Thus, the final horizontal velocity of this projectile will be equal to the initial horizontal velocity: $v_{x} = u_{x}$ .

The overall final velocity of this projectile will be:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \sqrt{(150.086)^{2} + (-39.621)^{2}} \\ &\approx 155\; {\rm m\cdot s^{-1}} \end{aligned}$ .

Change in the vertical component of the velocity of this projectile:

$\begin{aligned} \Delta v_{y} &= v_{y} - u_{y} \\ &\approx (-39.621) - 29.71786 \\ &\approx 69.3386 \end{aligned}$ .

Divide the change in velocity by acceleration (rate of change in velocity) to find the time required to achieve such change:

$\begin{aligned}t &= \frac{\Delta v_{y}}{a_{y}} \\ &\approx \frac{69.3386}{(-9.81)} \\ &\approx 7.0682\; {\rm s}\end{aligned}$ .

Hence, the projectile would be in the air for approximately $7.07\; {\rm s}$ .

Also the horizontal velocity of this projectile is $u_{x} \approx 150.086\; {\rm m\cdot s^{-1}}$ throughout the flight, the range of this projectile will be:

$\begin{aligned}x_{x} &= u_{x}\, t \\ &\approx (150.086)\, (7.0682) \\ &\approx 1.06 \times 10^{3}\; {\rm m} \end{aligned}$ .

When this projectile is at maximum height, its vertical velocity will be $0$ . Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the maximum height of the projectile (relative to the top of the $35\; {\rm m}$ cliff.)

$\begin{aligned}x &= \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a} \\ &\approx \frac{0^{2} - 29.71786^{2}}{2\, (-9.81)} \\ &\approx 45.0\; {\rm m}\end{aligned}$ .

Thus, the maximum height of the projectile relative to the ground will be approximately $45.0\; {\rm m} + 35\; {\rm m} = 80\; {\rm m}$ .

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1 year ago

What is the momentum of an object with 9.74 kg and 15 m/s?

dolphi86 [110]

Answer: 146.1 kg/m/s

Explanation: Momentum is equal to mass times velocity, so you do 9.74 kg times 15 m/s and get 146.1 kg/m/s.

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3 years ago