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posledela
3 years ago
15

Two people are sitting on playground swings. One is pulled back 4 degrees from the vertical and the other is pulled back 8 degre

es. They are both released at the same instant. Will they both come back to their starting points at the same time
Physics
2 answers:
alex41 [277]3 years ago
5 0

Answer:

The bodies will not come back to their starting point at he same time.

Explanation:

Since they are both pulled back at an angle to the vertical, there is a tangential component of acceleration a = gsinθ

When θ = 4 , a = 9.8sin4 = 0.684 m/s²

When θ = 8 , a = 9.8sin8 = 1.364 m/s²

Using s = ut + 1/2at². Where s is the distance covered and t = time taken, u = initial speed = 0 (assumed since they are both released at the same time)

So s = 0 × t + 1/2at² = 1/2at²

s = 1/2at²

t = √2s/a. Now, since s is the same for both swings, it follows that

t ∝ 1/a. Since their accelerations are different, the bodies will not come back to their starting point at he same time.

lara [203]3 years ago
4 0

Answer:

They will come back at the same time.

Explanation:

The angular velocity equation of ω= \frac{V}{r} where ω is the frequency of the movement, dependent on the angle. But since swings are simple pendulums and their angles of 8 and 4 degrees are small, they will come back to their starting points at the same time.

I hope this answer helps.

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Below is an oscilloscope trace from an AC supply. Calculate the frequency of the supply if each horizontal division represents 0
GuDViN [60]

Answer:

20 hertz of frequency produced.

Explanation:

\boxed{frequency = \frac{1}{period} }

Here we will find frequency and period should be in second, here given: 0.05 seconds

using the formula:

\boxed{frequency = \frac{1}{0.05} }

\boxed{frequency =20}

8 0
3 years ago
I need help on number 4 and 5
Strike441 [17]

4 is G. BB and Bb because that is what you are analyzing the product of.

5 is C. The round shape is obviously a dominant trait since mixing with a wrinkled shape didn't have an effect. The wrinkled shape is recessive.

7 0
4 years ago
if a bullet fired beneath the water, straight up, breaks through the surface at a speed of 25m/s, how high will the bullet be af
Doss [256]

Answer:

32 m and -2.4 m/s

Explanation:

Given:

v₀ = 25 m/s

t = 2.8 s

a = -9.8 m/s²

Find: Δy, v

Δy = v₀ t + ½ at²

Δy = (25 m/s) (2.8 s) + ½ (-9.8 m/s²) (2.8 s)²

Δy = 31.6 m

v = at + v₀

v = (-9.8 m/s²) (2.8 s) + 25 m/s

v = -2.44 m/s

Rounded to two significant figures, the bullet reaches a height of 32 m and a velocity of -2.4 m/s.

8 0
4 years ago
A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help.
uranmaximum [27]

Answer:

532.0725 m

102.17270893 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = g

H = Height of cliff

Distance traveled in 3 seconds

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s=44.145\ m

Distance traveled by sound = 2H-44.145 m

2H-44.145=ut+\dfrac{1}{2}at^2\\\Rightarrow 2H-44.145=340\times 3\\\Rightarrow H=\dfrac{340\times 3+44.145}{2}\\\Rightarrow H=532.0725\ m

The height of the cliff is 532.0725 m

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 532.0725+0^2}\\\Rightarrow v=102.17270893\ m/s

Her speed just before she hits the ground is 102.17270893 m/s

5 0
3 years ago
Part A What is the resistance of a 4.4 m length of copper wire 1.3 mm n diameter? The resistivity of copper is 1.68x 10-8 Ω-m Ex
Natalija [7]

Explanation:

It is given that,

Length of the copper wire, l = 4.4 m

Diameter of copper wire, d = 1.3 mm = 0.0013 m

Radius of copper wire, r = 0.00065 m

The resistivity of the copper wire, \rho=1.68\times 10^{-8}\ \Omega-m

We need to find the resistance of the copper wire. It is given by :

R=\rho\dfrac{l}{A}

R=1.68\times 10^{-8}\ \times \dfrac{4.4\ m}{\pi (0.00065)^2}

R =0.055 ohms

So, the resistance of the copper wire is 0.055 ohms. Hence, this is the required solution.

7 0
3 years ago
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