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3 years ago

Describe a situation where Newton's third law of motion can be observed

1 answer:

3 0

An example would be a swimmer pushes on the wall with her feet, which causes the wall to push back on her feet due to Newton's third law

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John added 150 grams of salt to 2.5 liters of water. what is the concentration of the salt solutions

torisob [31]

Shdnejeiebx ski wiw do

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2 years ago

In the equation vx^2=v0x^2+2ax(x-x0) what does the terms vx, v0x, x, and x0 stand for respectively?

tatuchka [14]

B. velocity at position x, velocity at position x=0, position x, and the original position

In the equation

$v_{x}^{2}$ = $v_{ox}^{2}$ +2 a x (x - x₀)

$v_{x}$ = velocity at position "x"

$v_{ox}$ = velocity at position "x = 0 "

x = final position

$x_{o}$ = initial position of the object at the start of the motion

6 0

3 years ago

Read 2 more answers

Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One pa

Serga [27]

Answer:

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

Explanation:

$v_r$ = Speed of river

$v_c$ = Speed of canoe

$v_r+v_c=2.8\ m/s$

$v_r-v_c=-1.9\ m/s$

Adding the equations we get

$2v_r=0.9\\\Rightarrow v_r=\frac{0.9}{2}\\\Rightarrow v_r=0.45\ m/s$

$0.42+v_c=2.8\ m/s\\\Rightarrow v_c=2.8-0.45\\\Rightarrow v_c=2.65\ m/s$

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

6 0

3 years ago

Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13

lyudmila [28]

Answer:

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

$y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right )$

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

$y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right )$

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$

3 0

2 years ago

In order to open the clam it catches, a seagull will drop the clam repeatedly onto a hard surface from high in the air until the

Vedmedyk [2.9K]

Answer:

2.2 s

Explanation:

Hi!

Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by

$x(t) = 23.1 m - \frac{1}{2}(9.8 m/s^2) t^2$

We are looking for a time t for which x(t) = 0

$0 = 23.1 m - (4.9 m/s^2) t^2$

Solving for t:

$t = \sqrt{\frac{23.1}{4.9}} s = 2.17124 s$

Rounding at the first decimal:

t = 2.2 s

4 0

3 years ago