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Alex787 [66]
3 years ago
13

If V has a magnitude of 14 units and the same direction qs a vector 3i+6j+2k find v​

Physics
1 answer:
melomori [17]3 years ago
7 0

Answer:

v = 6i + 12j + 4k

Explanation:

Find the magnitude of the direction vector.

√(3² + 6² + 2²) = 7

Normalize the direction vector.

3/7 i + 6/7 j + 2/7 k

Multiply by the magnitude of v.

v = 14 (3/7 i + 6/7 j + 2/7 k)

v = 6i + 12j + 4k

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A child rides a wagon down a hill. Eventually, the wagon comes to a stop. Which is most responsible for causing the wagon to sto
Dvinal [7]

Answer:

well, the hill isn't constantly going down hill, there's an ending point or goes back up hill making a v/u shape or there's nothing helping the wagon being pushed or pull currently

Explanation:

8 0
3 years ago
On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4
anygoal [31]

Answer:

a) \Delta s=443037.9747\ J.K^{-1}

b) \Delta s=31868131.8681\ J.K^{-1}

Explanation:

a)

Given:

amount of heat transfer occurred,dQ=3.5\times 10^6\ J

initial temperature of car, T_i=36.5+273=309.5\ K

final temperature of car, T_f=44.4+273=317.4\ K

We know that the change in entropy is given by:

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)} (heat is transferred into the system of car)

\Delta s=443037.9747\ J.K^{-1}

b)

amount of heat transfer form the system of house, dQ=5.8\times 10^8\ J

initial temperature of house, T_i=23.5+273=296.5\ K

final temperature of house, T_f=5.3+273=278.3\ K

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{5.8\times 10^8}{278.3-296.5}

\Delta s=31868131.8681\ J.K^{-1}

6 0
3 years ago
A machine has an efficiency of 70%. How much work dose the machine do when 20,000 J of work is done on it ?
Murrr4er [49]

Answer: The machine does 14.000 J

3 0
3 years ago
Two students, Student X and Student Y, stand on a long skateboard that is at rest on a flat, horizontal surface, as shown. In or
OleMash [197]

Answer:

the answer is B.

Explanation:

The claim is correct because Student Y can apply a force that is greater in magnitude than the frictional forces that are exerted on the student-student-skateboard system

6 0
3 years ago
Read 2 more answers
A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angl
weeeeeb [17]

Answer:

b)  k Δx - W cos θ - μ mg cos θ = m a ,  c)  θ = 86.6º, d)  Δx = 1.18 m

Explanation:

a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.

F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring

b) Let's apply Newton's second law for when the spring is compressed

let's use trigonometry to break down the weight

            sin θ = Wₓ / W

            cos θ = W_y / W

             Wₓ = W sin θ

             W_y = W cos θ

Y axis  

               N - W_y = 0

               N = W_y

               N = W cos θ

X axis

           F -Wₓ -fr = ma

the force applied by the spring is given by hooke's law

           F = k Δx

friction force has the expression

           fr = μ N

           fr = μ W cos θ

we substitute

            k Δx - W cos θ - μ mg cos θ = m a           ( 1)

c) If the plane has no friction, what is the angle so that Δx = 0.1m

             

We write the equation 1, with fr = 0 and since the system is still a = 0

            k Δx - W cos θ -0 = 0

            cos θ = \frac{k \Delta x}{ m g}

            cos θ = \frac{880 \ 0.1}{ 150 \ 9.8}

            cos θ = 0.0598

            θ = cos⁻¹ 0.0598

            θ = 86.6º

d) In this part they give the angle θ = 45º and there is no friction, they ask the compression

the acceleration is zero, we substitute in 1

            k Δx - W cos θ - 0 = 0

            Δx = \frac{mg \ cos \  \theta}{k}

            Δx = \frac{ 150 \ 9.8 \ cos45}{880}

            Δx = 1.18 m

7 0
3 years ago
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