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2 years ago

13

If a plane can travel 450 miles per hour with the wind and 410 miles per hour against the wind, find the speed of the plane with

out a wind and speed of the wind. I need help with this one :(

2 answers:

Vaselesa [24]2 years ago

5 0

the plane is 430 and the wind it 20miles

Citrus2011 [14]2 years ago

3 0

The wind speed is 20
the plane is 430mph in still air

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How much force is needed to keep the bowling ball moving towards the pins once it has

vazorg [7]

Answer:

Explanation:

The amount of force needed needs to be greater than all the forces acting in the opposite direction that the bowling ball was thrown. This includes air resistance, floor friction, gravity, and any other force involved. As long as the force acting on the bowling ball that is causing it to go in the direction of the pins is slightly greater than the opposite acting forces then it will continue in that direction. Since no values are provided we cannot calculate the actual precise value of force needed.

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2 years ago

A two-stage rocket moves in space at a constant velocity of +4300 m/s. The two stages are then separated by a small explosive ch

ololo11 [35]

Answer:

$v_{1f}$ = +3,394 103 m / s

Explanation:

We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.

The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s

The moment before the explosion

p₀ = (m₁ + m₂) v₀

After the explosion

pf = m₁ $v_{1f}$ + m₂ $v_{2f}$

p₀ = [texpv_{f}[/tex]

(m₁ + m₂) v₀ = m₁ $v_{1f}$ + m₂ $v_{2f}$

Let's calculate the final speed (v1f) of the first stage

$v_{1f}$ = ((m₁ + m₂) v₀ - m₂ $v_{2f}$ ) / m₁

$v_{1f}$ = ((2100 +1160) 4300 - 1160 5940) / 2100

$v_{1f}$ = (14,018 10 6 - 6,890 106) / 2100

$v_{1f}$ = 7,128 106/2100

$v_{1f}$ = +3,394 103 m / s

come the same direction of the final stage, but more slowly

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3 years ago

Astronomers use a special kind of instrument called a spectrometer to analyze stars. Spectrometers can determine what stars are

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The approval of the stars

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2 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

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2 years ago

A 9V battery produces a 1.5A current in a piece of wire. What is the resistance of the wire?

MAVERICK [17]

Explanation:

Using Ohm's Law and a bit of substitution, we can use voltage divided by current to solve for resistance. Doing that, we'll get 6 Ohm.

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2 years ago