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DENIUS [597]
3 years ago
12

What force would be required to accelerate a 1,100kg car to 0.5 m/s2

Physics
1 answer:
OlgaM077 [116]3 years ago
5 0

Answer:

the force required to accelerate a 1,100kg car is 550N

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W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively
MArishka [77]

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c)  W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

 W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

      W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

    W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

      ΔU = ΔK

7 0
3 years ago
Si un balón tiene una masa aproximada de 0.5 kg ¿con que fuerza hay que patearlo para imprimirle una aceleración de 1.5m/s2?
ExtremeBDS [4]

Answer:

translation

Explanation:

If a ball has an approximate mass of 0.5 kg, with what force must be kicked to give it an acceleration of 1.5m / s2?

8 0
2 years ago
3. What is a manometer used to determine?​
Gemiola [76]

Answer:

It's used to indicate pressure

3 0
2 years ago
Read 2 more answers
Which statement is TRUE?
Salsk061 [2.6K]
Statement B is true.
3 0
3 years ago
Read 2 more answers
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit
lapo4ka [179]

Answer:

The  angle is  \theta  =  15.48^o

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  d  =  3.66  \ m

     The time taken is  t =  0.455 \ s

   

The  horizontal component of the speed of the dart is mathematically represented as

      u_x =  ucos \theta

where u is the the velocity at dart is lunched

  so

      distance =  velocity \ in \ the\  x-direction  *  time

substituting values

      3.66 =   ucos  \theta *  (0.455)

 =>   ucos \theta =  8.04  \ m/s

From projectile kinematics the time taken by the dart can be mathematically represented as

         t  =  \frac{2usin \theta }{g}

=>    usin \theta =  \frac{g  * t}{2 }

       usin \theta =  \frac{9.8  * 0.455}{2 }

      usin \theta = 2.23

=>   tan \theta =  \frac{usin\theta }{ucos \theta }  =  \frac{2.23}{8.04}

       \theta  =  tan^{-1} [0.277]

      \theta  =  15.48^o

     

4 0
3 years ago
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