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diamong [38]
3 years ago
10

A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-co

mponent?
Physics
1 answer:
lidiya [134]3 years ago
3 0

The position vector can be transcribed as:

A<span> = 6 i + y j                           </span>

i <span>points in the x-direction and j points in the y-direction.</span>

The magnitude of the vector is its dot product with itself:

<span>|A|2 = A·A</span>

<span>102  = (6 i + y j)•(6 i+ y j)            Note that i•j = 0, and  i•i  = j•j = 1 </span>

<span>100  = 36 + y2       </span>

<span>64    = y2</span>

<span>get the square root of 64 = 8</span>

<span>The vertical component of the vector is 8 cm.</span>

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Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial veloc
svet-max [94.6K]

Incomplete question as time is missing.I have assumed some times here.The complete question is here

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Explanation:

Given data

Vi=10 m/s

S=70 m

(a) t₁=0.5 s

(b) t₂=1 s

(c) t₃=1.5 s

(d) t₄=2 s

(e) t₅=2.5 s

To find

Displacement S from t₁ to t₅

Velocity V from t₁ to t₅

Solution

According to kinematic equation of motion and given information conclude that v is given by

v=v_{i}+gt\\

Also get the equation of displacement

S=v_{i}t+(1/2)gt^{2}

These two formula are used to find velocity as well as displacement for time t₁ to t₅

For t₁=0.5 s

v_{1}=v_{i}+gt\\v_{1}=(10m/s)+(9.8m/s^{2} ) (0.5s)\\v_{1}=14.9m/s\\  And\\S_{1} =v_{i}t+(1/2)gt^{2}\\ S_{1}=(10m/s)(0.5s)+(1/2)(9.8m/s^{2} )(0.5s)^{2} \\S_{1}=6.225m

For t₂

v_{2}=v_{i}+gt\\v_{2}=(10m/s)+(9.8m/s^{2} ) (1s)\\v_{2}=19.8m/s\\  And\\S_{2} =v_{i}t+(1/2)gt^{2}\\ S_{2}=(10m/s)(1s)+(1/2)(9.8m/s^{2} )(1s)^{2} \\S_{2}=14.9m

For t₃

v_{3}=v_{i}+gt\\v_{3}=(10m/s)+(9.8m/s^{2} ) (1.5s)\\v_{3}=24.7m/s\\  And\\S_{3} =v_{i}t+(1/2)gt^{2}\\ S_{3}=(10m/s)(1.5s)+(1/2)(9.8m/s^{2} )(1.5s)^{2} \\S_{3}=26.025m

For t₄

v_{4}=v_{i}+gt\\v_{4}=(10m/s)+(9.8m/s^{2} ) (2s)\\v_{4}=29.6m/s\\  And\\S_{4} =v_{i}t+(1/2)gt^{2}\\ S_{4}=(10m/s)(2s)+(1/2)(9.8m/s^{2} )(2s)^{2} \\S_{4}=39.6m

For t₅

v_{5}=v_{i}+gt\\v_{5}=(10m/s)+(9.8m/s^{2} ) (2.5s)\\v_{5}=34.5m/s\\  And\\S_{5} =v_{i}t+(1/2)gt^{2}\\ S_{5}=(10m/s)(2.5s)+(1/2)(9.8m/s^{2} )(2.5s)^{2} \\S_{5}=55.625m

4 0
4 years ago
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Use the term chemical change in a complete sentence
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The pilot directs the aircraft to fly due north at 600km/h. A side-wind blows at
loris [4]

Answer:

678.2 km/h and 80.54° north of east

Explanation:

From the question,

Using pythagoras theorem,

a² = b²+c²..................... Equation 1

Where a = resultant velocity

Given: b = 600 km/h, c = 100 km/h

Substitute these values into equation 1

R² = 600²+100²

R² = 360000+10000

R² = 460000

R  = √460000

R = 678.2 km/h.

And the direction is

tanθ = 600/100

tanθ = 6

tanθ = 6

θ = tan⁻¹(6)

θ = 80.54°.

Hence the resultant velocity of the aircraft is 678.2 km/h and 80.54° north of east

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