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elena-s [515]
4 years ago
15

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

s an 8.5 N thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 20 m above the launch point. At what horizontal distance left of the loop should you launch?
_____________m
Physics
1 answer:
ollegr [7]4 years ago
5 0

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

F_y=F-mg=8.5-0.46\times 9.8=3.992\ N

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2

Vertical displacement of rocket is same as the height of loop. So, y=20\ m

There is no initial velocity in the vertical direction. So, u_y=0\ m/s

Now, applying equation of motion in vertical direction. we have:

y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

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Answer:

E = 0.1472  J

Explanation:

Given that,

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The length of the solenoid, l = 30 cm = 0.3 m

Current in the solenoid, I = 4 A

We need to find the magnetic energy stored in the solenoid. The expression for the stored energy is :

E=\dfrac{1}{2}LI^2

Where

L is self inductance of the solenoid,

L=\dfrac{\mu_oN^2A}{l}\\\\L=\dfrac{4\pi \times 10^{-7}\times 2100^2\times 0.001}{0.3}\\\\L=0.0184\ H

So,

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So, 0.1472  J of energy is stored in the solenoid.

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3 years ago
A 7.8-kg solid sphere, made of metal whose density is 2500 kg/m3, is suspended by a cord. When the sphere is immersed in water (
Klio2033 [76]

Answer:

Density of the liquid = 1470.43 kg/m³

Explanation:

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Mass of solid sphere(m) = 6.1 kg

Density of the metal = 2600 kg/m³

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The volume of water displaced is equal to the volume of sphere (Archimedes' principle)

Volume displaced = 0.002346 m³

Buoyant force =

Where

is the density of the fluid

g is the acceleration due to gravity

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The free body diagram of the sphere is shown in image.

According to image:

Acceleration due to gravity = 9.81 ms⁻²

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Explanation:

The object stops moving in 5 places (namely B, D, F, H, and position after 28 seconds which isn't marked ).

We know that the object has stopped moving, since its position does not change with the passage of time i.e. at rest.

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Answer:

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The following data were obtained from the question:

Mass (m) = 12.5 kg

Velocity (v) = 1.32 m/s

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Velocity (v) = 1.32 m/s

Kinetic energy (K.E) =?

K.E = ½mv²

K.E = ½ × 12.5 × 1.32²

K.E = 6.25 × 1.7424

K.E = 10.89 J

The kinetic energy of the object is 10.89 J. Hence, the workdone in bringing the object to rest is 10.89 J.

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