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yawa3891 [41]
3 years ago
11

The Chair o

Physics
1 answer:
Anvisha [2.4K]3 years ago
5 0

Lol they’re not gonna be able to answer if they can’t watch the video

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A feather is dropped on the moon from the height of 1.4m. The acceleration of gravity on the moon is 1.67ms-1. Determine the tim
Zinaida [17]

1.3s

Explanation:

Given parameters:

Height = 1.4m

Gravity on moon = 1.67ms⁻¹

Unknown:

Time for feather to fall = ?

Solution:

To solve this problem, we are going to use one of the motion equation that relates time, gravity and height.

    H = ut + \frac{1}{2} g t^{2}

Sine the body was dropped from rest, initial velocity is zero;

 H = height

  u = initial velocity

  t = time

  g = acceleration due to gravity

since u = 0;

H = \frac{1}{2} g t^{2}

 1.4 = \frac{1}{2} x 1.67 x t²

  t = 1.3s

learn more:

Gravity brainly.com/question/10934170

#learnwithBrainly

8 0
3 years ago
A ray moving in plastic at 62.9 deg enters water, where it bends to 70.9 deg. What is the index of refraction of the plastic?
DedPeter [7]

Answer:

refractive index of plastic is 1.42

Explanation:

When light ray enters from one medium to other medium then due to transition of light it bends away or towards the normal, this phenomenon is known as refraction of light

So here we know that

n_1 sin i = n_2 sin r

here we have

n_2 = \frac{4}{3}

i = 62.9^o

r = 70.9^o

now we have

n_1 sin62.9 = (\frac{4}{3}) sin70.9

n_1 = 1.42

7 0
3 years ago
What is the current if 4C of charge passes in 2 s?
julia-pushkina [17]

Answer:

I hope 2 amperes of current passes

8 0
3 years ago
When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these
Zigmanuir [339]

Answer:

a)  v = 0.7071 v₀, b) v= v₀, c)  v = 0.577 v₀, d)   v = 1.41 v₀, e)  v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

          v = \sqrt{ \frac{T}{ \mu } }

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

          m = 2m₀

let's find the new linear density

          μ = m / l

iinitial density

          μ₀ = m₀ / l

final density

          μ = 2m₀ / lo

          μ = 2 μ₀

we substitute in the equation for the velocity

initial            v₀ = \sqrt{ \frac{T_o}{ \mu_o} }

with the new dough

                    v = \sqrt{ \frac{T_o}{ 2 \mu_o} }

                    v = 1 /√2  \sqrt{ \frac{T_o}{ \mu_o} }

                    v = 1 /√2 v₀

                    v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

              μ = μ₀

   in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

           μ = 3m₀ / l₀

           μ = 3 m₀/l₀

           μ = 3 μ₀

            v = \sqrt{ \frac{T_o}{ 3 \mu_o} }

            v = 1 /√3  v₀

            v = 0.577 v₀

d) l = 2l₀

            μ = m₀ / 2l₀

            μ = μ₀/ 2

           v = \sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }

           v = √2 v₀

            v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

             μ = 10 m₀/2l₀

             μ = 5 μ₀

we look for speed

             v = \sqrt{ \frac{T_o}{5 \mu_o} }

             v = 1 /√5  v₀

             v = 0.447 v₀

5 0
3 years ago
_____ are formed where bumps from two surfaces come into contact ?
Murrr4er [49]

Answer:

the answer would be microwelds.

3 0
3 years ago
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