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ololo11 [35]
3 years ago
11

Un móvil que viaja a 40 m/s aplica los frenos y desacelera a razón de -0,25 m/seg 2 , ¿Cuánto tiempo le llevo

Physics
1 answer:
aivan3 [116]3 years ago
4 0

Answer:

Where is a picture?

Explanation:

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PLS HELP THIS IS WORTH 180 PTS!!(PLUS I WILL MARK BRAINIEST IF SOMEONE ANSWERS CORRECTY FIRST)
vazorg [7]

\qquad\qquad\huge\underline{{\sf Answer}}

The correct choice is ~ A

Bikes don't require gasoline for their functioning, so it causes less air pollution.

7 0
2 years ago
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If it takes 726 watts of power to move an object 36 m in 14 s, what is the mass of the object?
Pavel [41]

Answer:

i hope the pic helps u dear

4 0
3 years ago
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A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continue
enot [183]

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

7 0
3 years ago
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
lesantik [10]

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

     The power rating  of the first bulb is P_1  = 100 \ W

      The power rating of the second bulb is  P_2  =  50 \ W

     

Generally the power rating of the first bulb is mathematically represented as

      P_1  =  V^2 R

Where  V is the normal household voltage which is constant for both bulbs

  So  

        R_1  =  \frac{V^2}{P_1 }

substituting values

        R_1  =  \frac{V^2}{100}

Thus the resistance of the second bulb would be evaluated as

       R_2  =  \frac{V^2}{50}

From the above calculation we see that

        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

       P_  2  =  I^2_2 R_2

Now given that they are connected in series which implies that the same current flow through them so

       I_1^2 =  I_2^2

This means  that

       P \ \alpha  \  R

So  when they are connected in series

     P_2  >  P_1

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0
4 years ago
If a ball rolls off a 4m high table at 7m/s, how long will it take until it hits the floor
Semenov [28]
It will take at least 0.8
3 0
3 years ago
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