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3 years ago

5

Think about a car design. How would you improve efficiency? What new features would your car be able to do? Write at least 5 dif

ferent features of your car.

1 answer:

drek231 [11]3 years ago

3 0

Answer:

fast

solid

big

heavy

loud

Explanation:

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A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,

Alika [10]

For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. <span>

An initial condition has been supplied by the problem:

s=1.80m when t=0.245s

<span>This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2)

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation):

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2

v=8.549m/s

<span>Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t)

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like

0=(v)+(a)(t)

Substituting and solving for t:

0=(8.549m/s)+(-9.81m/s^2)(t)

t=0.8714s

<span>Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m<span>

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window:

3.725m – 1.8m=1.925m</span>

Answer:

<span>1.925m</span>

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3 years ago

How to find acceleration without final velocity?

Andrews [41]

Acceleration is the rate of change of a the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. Calculation of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. We can just use the kinematic equations. However, if we are not given the final velocity, it would not be possible to use the kinematic equations. One possible to calculate this value would be by generating an equation of distance with respect to time and getting the second derivative of the equation.

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3 years ago

Technician A says in a conventional transmission, the speed gears freewheel around the mainshaft until they are locked to it by

Marizza181 [45]

Answer:

Technician B

Explanation:

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3 years ago

Read 2 more answers

A point charge q1 = -4.00 nC is at the point x = 0.60 m, y = 0.80 m , and a second point charge q2 = +6.00 nC is at the point x

makkiz [27]

Answer:

A)Magnitude of the net electric field at the origin due q₁ and q₂

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

β=3.91°

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

Data

q₁ = -4 nC = -4*10⁻⁹ C

q₂ = +6nC =+ 6*10⁻⁹ C

k = 9*10⁹ N*m²/C²

$d_{1} =\sqrt{0.6^{2}+0.8^{2} } =1 m$

d₂ = 0.6 m

Graphic attached

The attached graph shows the field in x=0, y=0, due to the charges q₁ and q₂:

E₁: Total field at point x=0 , y=0 due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge.

E₂: Total field at point x=0 , y=0 due to charge q₂. As the charge q2 is positive (q₂+) ,the field leaves the charge.

Calculation of the electric field at the origin of x-y coordinates due to the charge q₁

E₁=K*q₁/d₁²=9*10⁹*4*10⁻⁹/1²=36N/C

Components (x-y) of the field due to q1:

E₁x=E₁*cosα = 36*(0.6/1) = 36*(0.6) = 21.6 N/C

E₁y=E₁*sinα = 36*(0.8/1) = 36*(0.8) =28.8 N/C

Calculation of the electric field at the origin of x-y coordinates due to the charge q₂

E₂=K*q₂/d₂² = -9*10⁹*6*10⁻⁹/0.6² = -150 N/C

Calculation of the electric field components at the origin of x-y coordinates

E₀ is the electric field at the origin of x-y coordinates (x=0,y=0)

E₀x= E₁x+E₂= 21.6-150 = -128.4 N/C

E₀y= E₁y = 28.8 N/C

A )Magnitude of the net electric field at the origin due q₁ and q₂

$E_{o} =\sqrt{128.4^{2}+28.8^{2} }$

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

$\beta =tanx{-1} \frac{E_{oy} }{Eox}$

$\beta =tan^{-1} \frac{28.8}{128.4}$

β=3.91°

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4 years ago

Which point has the greatest electric potential energy?

tatyana61 [14]

Answer:

D on Edge

Explanation:

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3 years ago