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ValentinkaMS [17]

3 years ago

11

Is an object moving in uniform circular motion accelerating?

1 answer:

dexar [7]3 years ago

8 0

Yes.

-- 'Acceleration' does NOT mean 'speeding up'.
It means ANY change in the speed OR direction of motion ...
speeding up, slowing down, or turning.

-- If an object is NOT moving in straight line at constant speed,
then its motion is accelerated.

-- In circular motion, or even just going around a curve,
the object is accelerating, because its direction is constantly
changing, even if its speed is constant.

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A motor exerts a 280 N force and does 1220 J

finlep [7]

We know that:

Work = Force × Distance

Solving for Distance:

Distance = Work / Force
Distance = 1220 J / 280 N

<h3> Distance = 4.36 m </h3>

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3 years ago

A scientist that applies the laws of science to the needs of communities is called _____.

s2008m [1.1K]

Answer:

The experimental scientist

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3 years ago

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

a

The orbital speed is $v= 2.6*10^{3} m/s$

b

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

4 years ago

What is a magnetic field?

monitta

A magnetic field is charged electricity acting in the forced of magnatism

4 0

3 years ago

A 13500 kg jet airplane is flying through some high winds. At some point in time, the airplane is pointing due north, while the

Mekhanik [1.2K]

Answer

given,

mass of the jet airplane = 13500 kg

Force on the plane = 35700 N due north

force from wind = 15300 N in direction 80.0° south of west.

Force = $35700 \vec{j} N$

force by wind = $15300(-cos \theta \vec{i}-sin \theta \vec{j})$ N

= $15300(-cos 80^0 \vec{i}-sin 80^0 \vec{j})$ N

net force on the jet airplane(ma)

$m a = 35700 \vec{j} + 15300(-cos 80^0 \vec{i}-sin 80^0 \vec{j})$

$\vec{a} = \dfrac{35700}{13500} \vec{j} + \dfrac{15300}{13500}(-cos 80^0 \vec{i}-sin 80^0 \vec{j})$

$\vec{a} = 2.64\vec{j} -0.197 \vec{i} - 1.116 sin 80^0 \vec{j})$

$\vec{a} = -0.197 \vec{i} + 1.524 \vec{j}$

$a = \sqrt{-0.197^2+1.524^2}$

a = 1.54 m/s²

$\theta = tan^{-1}(\dfrac{-1.524}{0.197})$

$\theta = -82.63^0$

3 0

3 years ago