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ohaa [14]
3 years ago
6

There is no blank in time between the action and reaction

Physics
1 answer:
Anna11 [10]3 years ago
4 0
I'm not sure what you mean by this...
Please rephrase the question. Thanks!
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Melanie completes a long distance run at an average speed of 6 mph. If it takes her 3 hours, how far did she run?
Rama09 [41]

Answer:

18 miles

Explanation:

The average speed is 6 mph

Melanie ran for 3 hours

Speed × Time = Distance

So, 6 mph × 3 h = 18 miles

6 0
4 years ago
Neutrons can change the charge of an atom<br><br> True<br> False
IceJOKER [234]

Answer:

False. They can’t change the charge but they can change the mass

8 0
3 years ago
Alguien me podria ayudar con esto pliz.. Ahora resuelva el siguiente ejercicio, tenga presente lo visto en clase y aprendido en
creativ13 [48]

Answer:

a.)1.)12m/s

2.)12m/s

3.)7m/s

4.)7m/s

5.)14m/s

6.)0m/s

b.)1.) 3m/s^2

2.)1.71m/s^2

3.).58m/s^2

4.).39m/s^2

5.).70m/s^2

6.)0

c.)1.)48m

2.)84m

3.)84m

4.)126m

5.)280m

6.)suma todos los metros mas 98m y obtendras la distancia en cual el carro se para

Explanation:

vinicial=0m/s

tiempo inicial=0s

Intervalo 1

v=12m/s

t=4s

aceleracion= vf-vi/t

=12-0/4

=12/4=3m/s^2

Intervalo 2

v=12m/s

t=7s

a=12/7

Intervalo 3

v=7m/s

t=12s

a=7/12

Intervalo 4

v=7m/s

t=18s

a=7/18

Intervalo 5

v=14m/s

t=20s

a=14/20

Intervalo 6

v=0m/s

t=27s

a=0/27

7 0
3 years ago
A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after
Delicious77 [7]
\Delta L=  \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

D
5 0
3 years ago
Read 2 more answers
A ball is thrown horizontally to the right, from the top of a vertical cliff of height h. A wind blows horizontally to the left,
Akimi4 [234]

Answer:

 v = \sqrt{\frac{y_o \ g}{2} }

Explanation:

For this exercise we must use the projectile launch ratios, let's start by finding the time it takes to reach the bottom of the cliff, the initial vertical velocity is zero

          y = y₀ + v_{oy} t - ½ g t²

         

at the bottom of the cliff y = 0 and as the body is thrown horizontally the initial vertical velocity is zero

          0 = y₀ + 0 - ½ g t²

          t = \sqrt{2y_o/g}

this time is the same as the horizontal movement.

Let's use Newton's second law to find the acceleration on this x-axis due to the force of the air

           F = m aₓ

they tell us that force is equal to the weight of the body

           -mg = maₓ

           aₓ = -g

the sign indicates that the acceleration is to the left

we write the kinematics equation

          x = x₀ + v₀ₓ t + ½ aₓ t²

They indicate that the final position is the foot of the cliff (x = 0), when it leaves the top it is at x₀ = 0 and has a velocity v₀ₓ = v

we substitute

          0 = 0 + v t + ½ (-g) t²

          v = ½ g t

         

we use the drop time

          v = ½ g \sqrt{\frac{2yo}{g} }

          v = \sqrt{\frac{y_o \ g}{2} }

5 0
3 years ago
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