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Mumz [18]
3 years ago
10

PLEASE HELP!!

Physics
1 answer:
butalik [34]3 years ago
8 0
<h2>Answer: Little Bee hits the windshield with a greater force</h2>

According Newton's 2nd Law of Motion  the force F is directly proportional to the mass m and to the acceleration a of a body:

F=m.a    (1)

We already know the mass and the accelaration of both bees. So, we have to find the force which with each bee hits the windshield. As follows:

For<u> Big Bee</u> with a mass m_{B} and acceleration a_{B}, the force F_{B} is:

F_{B}=m_{B}.a_{B}   (2)

F_{B}=(5kg)(2m/{s}^{2})    

Then:

F_{B}=10N>>>Force for Big bee

For <u>Little Bee </u>with a mass m_{L} and acceleration a_{L}, the force F_{L} is:

F_{L}=m_{L}.a_{L}    (3)

F_{B}=(3kg)(4m/{s}^{2})    

Then:

F_{B}=12N>>>Force for Little bee  

At this point we are able to clearly see that Little Bee hits the windshield with a greater force than Big bee.

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Anastaziya [24]

The correct option is

a fire


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6 0
4 years ago
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Under very dim levels of illumination foveas react to increase the sensitivity of the optic nerve. rods are more light-sensitive
Alex777 [14]

Answer:

rods are more light sensitive than cones.

Explanation:

There are two types of photo receptors in retina of our eyes. 1 Rods and 2 Cones. Rods are about 120 million and they are more sensitive then the cones. But the rods are not sensitive to color. Cones help us in seeing the color and there are about 6 to 7 million cones that provide color sensitivity to our eyes. That is why in the dark or where their are dim levels of illumination rods provide us scotopic vision. Because rods are more light sensitive then the cones.

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4 years ago
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If the initial velocity of an object was -2 meters per second
Shalnov [3]
A :-) for this question , we should apply
a = v - u by t
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.:. The acceleration is 0.75 m/s^2.
3 0
3 years ago
¿CUAL ES LA CAUSA DE UNA ONDA ESTACIONARIA?
xxTIMURxx [149]
Fenomeno de resonancia.

Espero que estoy te ayude :)
6 0
3 years ago
In a women's 100-m race, accelerating uniformly, Laura takes 1.82 s and Healan 3.07 s to attain
Sav [38]

Answer:

Laura is ahead and for a distance of 3.22 m

Explanation:

To solve this problem of one-dimensional kinematics, we have to find the acceleration and the final speed of each runner. Let's start with Laura

Lura data is acceleration time 1.82s, total run time 10.4 s and total distance 100m.  In all the races the  rest starts, so the initial speed is zero (Vo = 0)

   Vf1= Vo + a1 t1    

   Vf1 = x/t                

   XT  = X1 + X2

   X1 = Vo t1 + ½ a1 t1²  

   X1 = ½ a1 t1²  

   X2 = Vf1 (t-t1)

This is the remaining time of the race after the acceleration is over.

    XT = ½ a1 t1² + Vf1 (t-t1)

We remplace the expression of Vf1

     XT = ½ a1 t1² + a1 t1 (t-t1)

Laura's aceleration (a1) is

   a1= XT / [ ½  t1² + t1 (t-t1)]

   a1= 100/ [ ½ 1.82²+ 1.82 (10.4 -1.82)]

   a1=  5.79m/s2  

We repeat the same calculation for the other Healan runner, whose data are: total distance 100m, acceleration time 3.07 s and total time 10.4 s

Vf2= Vo + a2 t2    

Vf2 = x/t                

XT  = X3 + X4

X3 = Vo t2 + ½ a2 t2²  

X3 = ½ a2 t2²  

X4 = Vf2 (t-t2)

XT = ½ a2 t2² + Vf2 (t- t2)

XT = ½ a2 t2² + a2 t2 (t-t2)

The aceleration of Healan (a2)

a2 = XT / [½ t2² + t2 (t-t2)]

a2 = 100 / [½ 3,07²+ 3.07 (10.4 -3.07)]

a2 = 3.67 m / s2

We also need the final speeds of each runner

Laura Vf1 = Vo + a1 t1

          Vf1 = 0 + 5.79 1.82

          Vf1 = 10.54 m / s

Healan Vf2 = Vo + a2 t2

            Vf2 = 0 + 3.67 3.07

            Vf2 = 11.27 m / s

Having the acceleration and speed of each runner, you can start answering the questions

a) For t3 = 6.15s

Laura

The time to stop with constant speed is what remains after accelerating

XL= ½ a1 t1² + Vf1 (t3-t1)

XL= ½ 5.79 1.82² + 10.54 (6.15 – 1.82)    

XL= 55.23 m

Healan  

XH= ½ a2 t2² + Vf2 (t3-t2)

             XH= ½  3.67 3.07² + 11.27 (6.15-3.07)

             XH= 52.01 m

             (XL -XH)= 55.23- 52.01

             (XH -XL)=  3.22 m

It is appreciated from these results that Laura is ahead and for a distance of 3.22 m

b) If we analyze the acceleration values ​​of each runner, knowing that they leave the rest and that Healan at the end has a speed greater than Laura, the point of maximum distance difference is when Laura stops accelerating t = 1.82 s

      XL= ½  a1 t12  

      XL= ½ 5.79 1.822

      XL= 9.59 m

      XH = ½ a2 t12

      XH= ½ 3.67 1.822

      XH= 6.08 m

The maximum distance difference is 3.51 m

c) Already analyzed in the previous part 1.82 s, since the Laura stop accelerating and Heala continue with acceleration will travel greater distances in equal time units

6 0
3 years ago
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