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Alik [6]
3 years ago
6

Teams a and b are in a tug of war challenge. Team a wins. What can be said about team a

Physics
2 answers:
bekas [8.4K]3 years ago
8 0

Answer:

Team A used more force.

Explanation:

PilotLPTM [1.2K]3 years ago
6 0

Answer:

Team a used more aggression  and force towards team be, which left a result  of team A winning

Explanation:

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¿Qué valor tiene la resistencia eléctrica de un cable de cobre, si por él viaja una carga de 100 [C] en 60 segundos, si se encue
solong [7]
I’m sorry but imma need a translater cause I can’t understand you
4 0
3 years ago
A water pipe tapers down from an initial radius of R1 = 0.21 m to a final radius of R2 = 0.11 m. The water flows at a velocity v
Aleks [24]

Answer:

0.116 m^3/s

Explanation:

The volume flow rate of a fluid in a pipe is given by:

Q=Av

where

A is the cross-sectional area of the pipe

v is the speed of the fluid

In this problem, at the initial point we have

v = 0.84 m/s is the speed of the water

r = 0.21 m is the radius of the pipe, so the cross-sectional area is

A=\pi r^2 = \pi (0.21 m)^2 =0.138 m^2

So, the volume flow rate is

Q=(0.138 m^2)(0.84 m/s)=0.116 m^3/s

8 0
3 years ago
A body moving at .500c with respect to an observer
lakkis [162]

Answer:

0.8c and -0.14c

Explanation:

The first fragment will have a speed of +0.5c respect of a frame of reference moving at +0.5c

Lest name v the velocity of the frame of reference, and u' the velocity of the object respect of this moving frame of reference.

The Lorentz transform for velocity is:

u = (u' + v) / (1 + (u' * v) / c^2)

u = (0.5c + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = 0.8c

The other fragment has a velocity of u' = -0.6c respect of the moving frame of reference.

u = (-0.6v + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = -0.14c

6 0
2 years ago
N2 + O2 → 2NO N-N triple bond: 941 kJ/mol O-O double bond: 495 kJ/mol N-O bond: 201 kJ/mol
kiruha [24]

Answer:

\large \boxed{\text{761 kJ}}

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

                        N₂  +   O₂   ⟶         2NO

                     N≡N  + O=O ⟶       2O-N=O

Bonds:         2N≡N   1O=O        2N-O + 2N=O

D/kJ·mol⁻¹:     941      495           201      607

\begin{array}{rcl}\Delta H & = & \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\& = & 2 \times 941 +1 \times 495 - (2 \times 201 + 2\times 607)\\&=& 2377 - 1616\\&=&\textbf{761 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{761 kJ}}$}.

3 0
2 years ago
when approaching the front of an idling jet engine, the hazard area extends forward of the engine approximately
Dominik [7]

when approaching the front of an idling jet engine, the hazard area extends forward of the engine approximately 25 feet.

<h3>What impact, if any, would jet fuel and aviation gasoline have on a turbine engine?</h3>

Tetraethyl lead, which is present in gasoline, deposits itself on the turbine blades. Because jet fuel has a higher viscosity than aviation gasoline, it may retain impurities with greater ease.

Once the gasoline charge has been cleared, start the engine manually or with an electric starter while cutting the ignition and using the maximum throttle.

On the final approach, the aeroplane needs to be re-trimmed to account for the altered aerodynamic forces. A substantial nose-down tendency results from the airflow producing less lift on the wings and less downward force on the horizontal stabiliser due to the reduced power and slower velocity.

Learn more about turbine engine refer

brainly.com/question/807662

#SPJ4

7 0
1 year ago
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