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3 years ago

An airplane weighing 11,000 N climbs to a

Physics

1 answer:

Gennadij [26K]3 years ago

8 0

The power in horsepower is 40.1 hp

Explanation:

We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

$W=(mg)\Delta h$

where

mg = 11,000 N is the weight of the airplane

$\Delta h = 1.6 km = 1600 m$ is the change in height

Substituting,

$W=(11,000)(1600)=17.6\cdot 10^6 J$

Now we can calculate the power delivered, which is given by

$P=\frac{W}{t}$

where

$W=17.6\cdot 10^6 J$ is the work done

$t=9.8 min \cdot 60 = 588 s$ is the time taken

Substituting,

$P=\frac{17.6\cdot 10^6 J}{588}=2.99\cdot 10^4 W$

Finally, we can convert the power into horsepower (hp), keeping in mind that

$1 hp = 746 W$

Therefore,

$P=\frac{2.99\cdot 10^4}{746}=40.1 hp$

Learn more about power:

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A force of 2 newtons is required to stretch a spring 4 cm. The amount of force required to stretch the same spring 8 cm is _____

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Answer will be c for this one!

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4 years ago

023 (part 1 of 2) 10.0 points

Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

$\Delta KE_{rotational} = \dfrac{1}{2} \cdot I \cdot \omega ^2 = \dfrac{1}{2} \cdot m \cdot v ^2$

I = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

$\Delta KE_{rotational}$ = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

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3 years ago

Lunar missions have revealed that the moon has:

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That the moon has soil within its shadowy craters rich and useful material

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3 years ago

A rock is tossed straight up from the ground with a speed of 21 m/s . When it returns, it falls into a hole 10 m deep.a.) What i

Arte-miy333 [17]

(a) 25.2 m/s

Let's take the initial vertical position of the rock as "zero" (reference height).

According to the law of conservation of energy, the speed of the rock as it reaches again the position "zero" after being thrown upwards is equal to the initial speed of the rock, 21 m/s (in fact, if there is no air resistance, no energy can be lost during the motion; and since the kinetic energy depends only on the speed of the rock:

$K=\frac{1}{2}mv^2$

and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)

This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using

$v^2 = u^2 + 2gd$

where

u = 21 m/s is the initial speed of the rock as it enters the hole

g = 9.8 m/s^2 is the acceleration due to gravity

d = 10 m is the depth of the hole

Substituting,

$v=\sqrt{u^2 +2gd}=\sqrt{(21 m/s)^2+2(9.8 m/s^2)(10 m)}=25.2 m/s$

(b) 4.72 s

The vertical position of the rock at time t is given by

$y(t) = v_y t - \frac{1}{2}gt^2$

where

$v_y = 21 m/s$ is the initial vertical velocity

Substituting y(t)=-10 m, we can then solve the equation for t to find the time at which the rock reaches the bottom of the hole:

$-10 = 21 t - \frac{1}{2}(9.8)t^2\\10+21 t -4.9t^2 = 0$

which has two solutions:

t = -0.43 s --> negative, so we discard it

t = 4.72 s --> this is our solution

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4 years ago

An object has rotational inertia I. The object, initially at rest, begins to rotate with a constant angular acceleration of magn

butalik [34]

Answer:

Explanation:

initial angular velocity, ωo = 0

angular acceleration = α

time = t

let the angular velocity after time t is ω.

Use first equation of motion for rotational motion

ω = o + α t

ω = αt

The angular momentum is given by

Angular momentum = moment of inertia x angular velocity

L = I x ω

L = I x αt

L = I α t

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3 years ago