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3 years ago

If you increase the mass but keep force constant what will happen to the acceleration of the object

Physics

1 answer:

drek231 [11]3 years ago

3 0

Answer:accelaration will decrease because forse is inversly proportional to mass.

Explanation:

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Partially correct answer iconYour answer is partially correct. A proton initially has and then 2.30 s later has (in meters per s

morpeh [17]

Answer: The question has some missing details. The initial velocity given as u = -6.5i + 17j + 13k and the final velocity v = -2.8i + 17j -9.3k.

a) = (1.82i - 9.69k)m/s2

b) magnitude = 9.85m/s2

c) direction = 280.64 degree

Explanation:

The detailed and step is shown in the attachment.

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3 years ago

Neptune is a planet that...

Ludmilka [50]

It would be option C. It rotates, or spins, on its axis, but it revolves around the sun.

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3 years ago

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An airplane has a speed of 135 mi/h in still air. It is flying straight north so that it is always directly above a north-south

vlabodo [156]

Answer:

The answer is below

Explanation:

Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.

The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph

vₐ = vₙ + vₐₙ

vₐ = vₐₙ

Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.

The direction of the wind θ is:

sin(θ / 2) = vₙ / 2vₐ

sin(θ / 2) = 70/ (2*135)

sin(θ / 2) = 0.2593

θ / 2 = sin⁻¹(0.2593) = 15

θ = 30⁰

2α = 180° - 30°

2α = 150°

α = 75°

a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.

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3 years ago

You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart

hichkok12 [17]

Answer:

$v = 3.5 \times 10^5 m/s$

Explanation:

At some distance from the Earth the force of attraction due to moon is balanced by the force due to Moon

so we will have

$\frac{GM_em}{r^2} = \frac{GM_m}{(d-r)^2}$

now we have

$\frac{d - r}{r} = \sqrt{\frac{M_m}{M_e}}$

$\frac{3.844\times 10^8 - r}{r} = \sqrt{\frac{7.36 \times 10^{22}}{5.9742\times 10^{24}}}$

so we will have

$r = 3.46 \times 10^8 m$

Now by energy conservation

$-\frac{GM_e}{R_e} - \frac{GM_m}{d - (R_e + R_m)} + \frac{1}{2}v^2 = -\frac{GM_e}{r} - \frac{GM_m}{d - r}$

$-6.26 \times 10^{8} - 13046 + \frac{1}{2}v^2 = -1.15 \times 10^6 - 1.28 \times 10^5$

$v = 3.5 \times 10^5 m/s$

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3 years ago

When you set a heavy bag down on the ground, you are doing _______ work on it.

puteri [66]

When you set a heavy bag down on the ground, you are doing negative work on it.

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3 years ago