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aleksklad [387]
2 years ago
5

A toy car that is 0.12 m long is used to model the actions of an actual car that is 6 m long. which ratio shows the relationship

between the sizes of the model and the actual car?
a. 50:1
b. 1:50
c. 5:1
d. 1:5
Physics
1 answer:
GREYUIT [131]2 years ago
7 0

Answer:

its b) 1:50 just did it on apex

Explanation:

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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
PLS ANSWERR THIS QUESTION FOR ME !!
MA_775_DIABLO [31]

Explanation:

20 joule is your answer

Answer:

here

mass m =100kg

distance d=50m

acceleration due to gravity a =10m/s²

work =force×displacement

= ma/d=100×10/50=20joule

3 0
2 years ago
Read 2 more answers
In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas
Anni [7]

Answer:

The phase difference is  \Delta \phi  = 180^o

Explanation:

From the question we are told that

     The distance between the slits is d = 0.2 \ mm = \frac{0.2}{1000}  = 0.2 *10^{-3} \ m

     The distance to the screen is D = 100 cm = \frac{100}{100} = 1 \ m

      The wavelength is  \lambda = 400nm

 The distance of the wave from the  central maximum is L =  5mm = 5*10^{-3} m

   

Generally the path difference of this  waves is mathematically represented as

              y = d sin \theta

Here \theta is the angle between the the line connecting the mid-point of the slits with  the screen and the line  connecting the mid-point of the slits to the central maximum

  This implies that

              tan \theta  = \frac{L}{D}

     =>     \theta = tan ^{-1} \frac{L}{D}

             \theta = tan ^{-1} [\frac{5*10^{-3}}{1}]

           \theta =0.2865

Substituting values into the formula for path difference

       y = 0.2 *10^{-3} sin(0.2864)  

       y = 9.997*10^{-7} \ m  

The phase difference is mathematically represented as

          \Delta \phi = \frac{2 \pi }{\lambda }  * y

Substituting values        

         \Delta \phi = \frac{2 \pi }{400 *10^{-9} }  \ * 9.997*10^{-7}

         \Delta \phi =5 \pi

Converting to degree

         \Delta \phi =5 \pi radians = 5 (180^o)  =  180^o  

the solution is subtracted by 360° in order to get the actual angle

 

             

4 0
3 years ago
How will the positions of the police car and the truck compare when they have the same speed and why?
sdas [7]

Answer:

Two cars of equal weight and braking ability are travelling along the same road but combined with other factors it could mean the difference between life.

7 0
2 years ago
You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body
stellarik [79]

A) The change in internal chemical energy is 1.15\cdot 10^7 J

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

P=Fv

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

P=(80)(8.0)=640 W

The energy output is related to the power by the equation

P=\frac{E}{t}

where:

P = 640 W is the power output

E is the energy output

t = 30 min \cdot 60 = 1800 s is the time elapsed

Solving for E,

E=Pt=(640)(1800)=1.15\cdot 10^6 J

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J

B)

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

E=1.15\cdot 10^7 J

Here we want to find the time t' needed to convert an amount of chemical energy of

E'=3.8\cdot 10^5 J

So we can setup the following proportion:

\frac{t}{E}=\frac{t'}{E'}

And solving for t',

t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min

Learn more about power and energy:

brainly.com/question/7956557

#LearnwithBrainly

3 0
2 years ago
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