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lukranit [14]
3 years ago
12

Jenny feels the touch of a feather on her left thumb where is this information registered in the brain?

Physics
1 answer:
Helga [31]3 years ago
4 0
The left sensory strip
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An oil droplet is levitated and held at rest in a region where the electric field is 1.4x104 N/C directed vertically downwards.
Ratling [72]

W = m g     weight of drop

W = q E

m = q E / g = = 3 e E / g

m = 3 * 1.6E-19 * 1.4E4 / 9.8 = 6.96E-16 kg

8 0
2 years ago
What happens to the kinetic energy of a snowball as it rolls across the lawn and gains mass.
LiRa [457]

If the ground is level, then the snowball can never have
any more kinetic energy than it hand when it left your hand.

If more mass sticks to it as it makes its way across the lawn,
then it must slow down, so that its

                 KE = (1/2) (present mass) (present speed)²

never exceeds the KE you gave it when you tossed it.

And we're not even talking yet about all the energy it loses
by scraping through the snow and mashing down the blades
of grass in its path.

5 0
3 years ago
Read 2 more answers
A wave on a string has a wavelength of 0.90 m at a frequency of 600 Hz. If a new wave at a frequency of 300 Hz is established in
kobusy [5.1K]

Answer:

 λ₂ = 1.8 m

Explanation:

given,

wavelength of the string 1 = 0.90 m

frequency of the string 1 = 600 Hz

wavelength of string 2 = ?

frequency of the string 2 = 300 Hz

we now,

f\ \alpha\ \dfrac{1}{\lambda}

now,

\dfrac{f_1}{f_2}=\dfrac{\lambda_2}{\lambda_1}

\dfrac{600}{300}=\dfrac{\lambda_2}{0.9}

λ₂ = 2 x 0.9

 λ₂ = 1.8 m

Hence, the wavelength of the second string is equal to  λ₂ = 1.8 m

8 0
3 years ago
Pls help with this question i’ll give brainliest
bearhunter [10]

Answer:

5m/s to the right

Explanation:

Momentum = mass * velocity

Momentum before = momentum after

m₁u₁+m₂u₂ = m₁v₁+m₂v₂

3000*10 + 1000*0 = 3000*v₁ + 1000*15

30000-15000=3000v₁

15000=3000v₁

v₁=5m/s to the right (to the right because answer is positive)

6 0
3 years ago
If the range of the projectile is 4.3 m, the time-of-flight is T = 0.829 s, and air resistance is negligible, determine the foll
ankoles [38]

Answer

given,

range of the projectile = 4.3 m

time of flight = T = 0.829 s

v =\dfrac{d}{T}

v =\dfrac{4.3}{0.829}

     v = 5.19 m/s

vertical component of velocity of projectile

v_y = gt'

v_y = 9.8 \times {\dfrac{T}{2}}

v_y = 9.8 \times {\dfrac{0.829}{2}}

v_y =4.06\ m/s

a) Launch angle

 \theta = tan^{-1}(\dfrac{v_y}{v})

 \theta = tan^{-1}(\dfrac{4.06}{5.19})

    θ = 38°

b) initial speed of projectile

  v = \sqrt{v_x^2 + v_y^2}

  v = \sqrt{5.19^2 + 4.06^2}

         v = 6.59 m/s

c) maximum height reached by the projectile

     y_{max}=v_{avg}t'

     y_{max}=\dfrac{1}{2}v_y\dfrac{T}{2}

     y_{max}=\dfrac{1}{2}\times(g\dfrac{T}{2})\times\dfrac{T}{2}

     y_{max}=\dfrac{gT^2}{8}          

7 0
3 years ago
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