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2 years ago

A plant is thrown straight

Physics

1 answer:

olchik [2.2K]2 years ago

4 0

The time taken for the plant to hit the ground from a distance of 7.01m and at a velocity of 8.84m/s is 1.59s.

<h3>How to calculate time?</h3>

The time taken for a motion to occur can be calculated using the following formula:

v² = u² - 2as

Where;

v = final velocity
u = initial velocity
s = distance
a = acceleration

8.84² = 0² + 2 × a × 7.01

78.15 = 14.02a

a = 5.57m/s²

V = u + at

8.84 = 0 + 5.57t

t = 1.59s

Therefore, the time taken for the plant to hit the ground from a distance of 7.01m and at a velocity of 8.84m/s is 1.59s.

Learn more about time at: brainly.com/question/13170991

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Steel wire rope is used to lift a heavy object. We use a 3.1m steel wire that

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Answer:

Young's modulus for the rope material is 20.8 MPa.

Explanation:

The Young's modulus is given by:

$E = \frac{FL_{0}}{A\Delta L}$

Where:

F: is the force applied on the wire

L₀: is the initial length of the wire = 3.1 m

A: is the cross-section area of the wire

ΔL: is the change in the length = 0.17 m

The cross-section area of the wire is given by the area of a circle:

$A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2}$

Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:

$F = T_{w} = W_{o}$

Where:

$T_{w}$ : is the tension of the wire

$W_{o}$ : is the weigh of the object = mg

m: is the mass of the object = 1700 kg

g: is the acceleration due to gravity = 9.81 m/s²

$F = mg = 1700 kg*9.81 m/s^{2} = 16677 N$

Hence, the Young's modulus is:

$E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa$

Therefore, Young's modulus for the rope material is 20.8 MPa.

I hope it helps you!

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3 years ago

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An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the

Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

$(n1)*(sin(theta1))=(n2)*(sin(theta2))$

Then we will find the refraction angle relative to the vertical this way:

$(n1/n2)*(sin(theta1))=sin(theta2)$

$(1/1.33)*(sin(45))=sin(theta2)$

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

$tan(32.12)=(X/30cm)$

$X=(tan(32.12))*(30cm)$

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

$tan(32.12)=(24cm/Y)$