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harkovskaia [24]

4 years ago

14

What is the name of NASA's special plane?

1 answer:

saveliy_v [14]4 years ago

8 0

Answer:

Boeing

if im wrong sue me ¯\_(ツ)_/¯

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Which of these is not a layer of the skin?

lina2011 [118]

4. hyperdermis is not a layer of skin

3 0

4 years ago

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Suppose a pulley with an IMA of 2 was used with an input force of 50 N to lift a box 45 cm that weighs 100 N. What distance was

IRINA_888 [86]

The ideal mechanical advantage (IMA) is the number of times in which the input force is multiplied under ideal conditions. If the real force was only 50N, the distance at which the rope was pulled will be twice the distance given in this item. The answer is 90 cm.

6 0

4 years ago

A snowboarder starts from rest at the top of a slope. They accelerate to 35 m/s over 9 seconds. What is their acceleration?

leonid [27]

a=v-u/t

v=35m/s

t=9

a=35/9
a=3.88m/s²

4 0

3 years ago

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

$I = \dfrac{m(2gh - v^2)r^2}{v^2}$

$I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}$

$I =mr^2(0.287)$

I = 0.287 MR²

3 0

3 years ago

The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec

saveliy_v [14]

The static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$ .

The given parameters;

<em>mass of the block, = M</em>
<em>coefficient of static friction in section 1, = </em> $\mu_s_1$ <em />
<em>angle of inclination of the plane, = θ</em>

<em />

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

The static friction exerted on the block by the incline is calculated as follows;

$F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta$

Thus, the static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$

Learn more here:brainly.com/question/17237604

3 0

3 years ago