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V125BC [204]
3 years ago
6

"A toy airplane flies clockwise at a constant speed in a horizontal circle of radius 8.0 meters. The magnitude of the accelerati

on of the airplane is 25 meters per second squared. The diagram shows the path of the airplane as it travels around the circle.
Calculate the speed of the airplane. [Show all work, including the equation and substitution with units.]"

Physics
2 answers:
Citrus2011 [14]3 years ago
6 0

For the answer to the question above, so at the instant, the acceleration of the airplane is southward, the direction of the velocity is also southward. The direction should be the same because it is both a vector quantity and it does not make sense if the direction and acceleration have different direction.
ryzh [129]3 years ago
3 0

Answer:

speed of the airplane: 14.14 m/s

Explanation:

radius, r = 8.0 m

centripetal acceleration, a = 25 m/(s^2)

velocity, v = ?

Centripetal acceleration can be computed as follows:

a = v^2/r

Solving for v and replacing with data:

v = √a*r

v = √25 m/(s^2)*8.0 m

v = 14.14 m/s

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At what rate is thermal energy being generated in the 2R-resistor when  = 12 V and R = 3.0 ?
evablogger [386]

If there's nothing else connected to the battery except the 3-ohm resistor, then the question is defective and the correct answer is not on the list.

With 12v connected across the resistor, the current through the resistor is

I = (E/R) = (12/3) = 4 Amperes.

The power generated inside a resistor is

P = (E x I) = (12v) x (4 A) = 48 watts

8 0
3 years ago
A car mass 600kg starts from rest moving uniform acceleration 0.2 m/s^2 after 60 seconds collides with stationary pick up van of
nlexa [21]

Answer:

The given phenomenon  supports the Principle of Conservation of momentum.

Explanation:

law of conservation of momentum  

Initial Momentum = Final Momentum

So, first we calculate initial momentum of the system:

Initial Momentum  = m1*u1 + m2*u2

and we are given

m1 = mass of car = 600 kg

m2 = mass of van = 400 kg

u1 = Initial Speed of Car  

to find the initial speed we use equation of motion

Vf = Vi + at

Vf = 0 m/s + (0.2 m/s²)(60 s)

Vf = u₁ = 12 m/s  

u₂ = Initial Speed of Van = 0 m/s

Therefore,

Initial Momentum  = (600 kg)(12 m/s) + (400 kg)(0 m/s)

Initial Momentum  = 7200 Ns   .........(1)

final momentum = m₁v₁ + m₂v₂

where,

v₁ = v₂ = final speed of car+van (both locked ) = 7.2 m/s

Therefore,

Final Momentum = (600 kg)(7.2 m/s) + (400 kg)(7.2 m/s)

Final Momentum = 7200 Ns   -------- (2)

on comparing (1) and (2)

Initial momentum = Final Momentum

Hence, the phenomenon of the system supports the principle of conservation of momentum.

6 0
3 years ago
Cells are made of different kinds of tissues that work together true or false
Andru [333]
Cells make up tissues, so the tissue cannot make up cells. Cells are made up of little organelles that help them to function. 
7 0
3 years ago
Read 2 more answers
This is the third time I’m asking, please, On a wet road, is a higher coefficient of friction on the tires safer or a lower one
jenyasd209 [6]

Answer:

higher is safer

Explanation:

because it is a wet slippery surface, you would need more friction on the tires, to get more traction in the slippery wet road, if you had low friction you would not move anywhere and or could swerve off somewhere

4 0
1 year ago
A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thic
Mazyrski [523]

Answer:

The total resistance of the wire is = 1.917\times10^{-3}

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, R_{Total}

\frac{1}{R_{Total}}  =\frac{1}{R_{cu}  }+\frac{1}{R_{al}}

Parameters given:

Length of wire = 1 m

Cross sectional area of copper A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3}  )^{2} =3.142\times10^{-6} m^{2}

Cross sectional area of aluminium wire  

A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3}  )^{2}-(1\times 10^{-3}  )^{2}] =9.42\times10^{-6} m^{2}\\

Resistivity of copper \rho _{cu}=1.7\times 10^{-8}  \Omega .m

Resistivity of Aluminium \rho _{al}=2.8\times 10^{-8}  \Omega .m

Resistance of copper R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} }  =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega

Resistance of aluminium R_{al}= \frac{\rho_{al} \times l}{A_{al} }  =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega

The total resistance of the wire can be obtained as follows;

\frac{1}{R_{Total}}  =\frac{1}{5.41\times10^{-3}  }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}

R_{Total}= 1.917\times 10^{-3}\Omega

∴ The total resistance of the wire = 1.917\times 10^{-3}\Omega

4 0
3 years ago
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