Physics, 02.07.2020 20:01 dasiaflowers85

A puck on the ice travels 20.0 m [25° W of N], gets deflected, and travels 30.0 m [35° N of W]. Determine where the puck will end up with respect to its starting point, which is the puck's total displacement, using the tail-to-tip with math (trigonometry) method. In the same hockey game, a player is positioned 35 m [40° W of S] of the net. He shoots the puck 25 m [E] to a teammate. What second displacement does the teammate shoot the puck with in order to make it to the net? Find the answer using the component method.

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Answer from: tiannahwlit

Given that,

Ice travel = 20.0 m

The direction is 25° W of N

Deflect distance = 30.0 m

The direction is 35° N of W

(I). When ice travel west of north

We need to calculate the displacement

Using formula of displacement of component

$\vec{D}=d\cos\theta+d\sin\theta$

Put the value into the formula

$\vec{D}=20\cos25(j)+20\sin25(-i)$

$\vec{D}=18j-8.44i$

(II). When ice deflect north of west

We need to calculate the displacement

Using formula of displacement of component

$\vec{D'}=d\cos\theta+d\sin\theta$

Put the value into the formula

$\vec{D'}=30\cos35(-i)+30\sin35(j)$

$\vec{D'}=-24.3i+17.1j$

We need to calculate the magnitude puck's total displacement

Using formula for total displacement

$\vec{D''}=\vec{D}+\vec{D'}$

Put the value into the formula

$\vec{D''}=18j-8.44i-24.3i+17.1j$

$\vec{D''}=-32.74i+35.1j$

$|\vec{D''}|=\sqrt{(-32.74)^2+(35.1)^2}$

$D''=47.9\ m$

Hence, The magnitude puck's total displacement is 47.9 m.

(ii). Given that,

He shoot the puck 25 m in east.

The displacement is

$\vec{D}=25i+0j$

A player is positioned 35 m at 40° West of south.

We need to calculate the displacement

Using formula of displacement

$\vec{D'}=d\cos\theta(-i)+d\sin\theta(-j)$

Put the value into the formula

$\vec{D'}=35\cos40(-i)+35\sin40(-j)$

$\vec{D'}=-26.81i-22.49j$

We need to calculate the total displacement

Using formula for total displacement

$\vec{D''}=\vec{D}+\vec{D''}$

Put the value into the formula

$\vec{D''}=25i+0j-26.81i-22.49j$

$\vec{D''}=-1.81i-22.49j$

The magnitude of the total displacement

$|\vec{D''}|=\sqrt{(-1.81)^2+(-22.49)^2}$

$D''=22.56\ m$

Hence, The magnitude of the total displacement is 22.56 m

Answer from: Quest

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

Answer from: Quest

The answer would be a) -80of

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A puck on the ice travels 20.0 m [25° W of N], gets deflected, and travels 30.0 m [35° N of W]. Dete...

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