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3 years ago

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A driver travels 4.1 km west, 17.3 km north, and finally 1.2 km at an angle of 65.4 degree north of west. What is the driver’s d

isplacement?

1 answer:

jek_recluse [69]3 years ago

5 0

Answer:

Explanation:

We shall represent each displacement in vector form .

i will represent east , j will represent north .

D₁ = 4.1 west = - 4.1 i

D₂ = 17.3 north = 17.3 j

D₃ = - 1.2 cos65.4 i + 1.2 sin65.4 j

= - .5 i + 1.09 j

Total displacement

= D₁ + D₂ + D₃

= - 4.1 i + 17.3 j - .5 i + 1.09 j

D = - 4.6 i + 18.39 j

magnitude of D

= √ ( 4.6² + 18.39² )

= √ (21.16 + 338.2 )

= √359.36

= 18.95 km .

Final displacement = 18.95 km .

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Answer:

Explanation:

Diameter of pool = 12 m

radius of pool, r = 6 m

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m = πr²h x d

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m = 113040 kg

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W = 113040 x 9.8 x 5.5

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2 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

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To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

Here,

D is diameter of the eye

$D = \frac{1.22 (539nm)}{5.11 mm}$

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The angle that relates the distance between the lights and the distance to the lamp is given by,

$Sin\theta = \frac{d}{L}$

For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle $sin \theta = \theta$

Therefore,

$L = \frac{d}{sin\theta}$

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Therefore the distance is 5.367km.

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3 years ago

Fatima is watching her pet cat, Winter, napping in the sun. Fatima is curious about the heart rate of Winter when she is napping

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Answer:

Explanation:

There are two hypotheses she could test:

A cat's heart rate changes while it is napping.

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3 years ago

Read 2 more answers

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

vaieri [72.5K]

Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

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

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

$R = 106.4 m$

so the distance from the fence is given as

$d = 106.4 - 97.5$

$d = 8.86 m$

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3 years ago

A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit

lapo4ka [179]

Answer:

The angle is $\theta = 15.48^o$

Explanation:

From the question we are told that

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The horizontal component of the speed of the dart is mathematically represented as

$u_x = ucos \theta$

where u is the the velocity at dart is lunched

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$distance = velocity \ in \ the\ x-direction * time$

substituting values

$3.66 = ucos \theta * (0.455)$

=> $ucos \theta = 8.04 \ m/s$

From projectile kinematics the time taken by the dart can be mathematically represented as

$t = \frac{2usin \theta }{g}$

=> $usin \theta = \frac{g * t}{2 }$

$usin \theta = \frac{9.8 * 0.455}{2 }$

$usin \theta = 2.23$

=> $tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}$

$\theta = tan^{-1} [0.277]$

$\theta = 15.48^o$

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3 years ago