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Lynna [10]
3 years ago
7

How are newtons third law of motion applies to a system of objects

Physics
1 answer:
11111nata11111 [884]3 years ago
3 0

Answer:

Newton's third law of motion states that whenever a first object exerts a force on a second object, the first object experiences a force equal in magnitude but opposite in direction to the force that it exerts. ... Newton's third law is useful for figuring out which forces are external to a system.

Explanation:

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Indigenous people sometimes cooked in watertight baskets by placing enough hot rocks into the water to bring it to a boil. What
yaroslaw [1]

Answer:

The rock has a mass of 4.02 kg

Explanation:

<u>Step 1: </u>Data given

Mass of the rock = TO BE DETERMINED

Temperature of the rock = 500 °C

Mass of the water  =4.24 kg

⇒ loses 0.044kg as vapor

Initial temperature of the water = 29°C

Final temperature = 100°C

Specific heat of rock = 0.20 kcal/kg °C

Specific heat of water = 1kcal/kg°C

Latent heat of vaporization = 539 kcal/kg

<u>Step 2:</u> formules

Qlost,rock + Qgained,water = 0

Qtotal,water = Qwater +Qvapor

<u>Step 3: </u>Calculate Qvapor

Qvapor = mass of vapor * Latent heat of vapor

Qvapor = 0.044kg * 539 kcal/kg = 23.716 kcal

<u>Step 4: </u>Calculate Qwater

Qwater = mass of water * specific heat * Δtemperature

Qwater = 4.196 kg * 1kcal/kg°C *( 100-29)

Qwater = 297.916 kcal

<u>Step 5:</u> Calculate Qwater,total

Qwater,total = Qwater + Qvapor

Qwater,total = 23.716 kcal + 297.916 = 321.632 kcal

<u>Step 6</u>: Calculate Qrock

Qrock = mass of rock * specific heat rock * Δtemperature

Qrock = mass of rock * 0.20 kcal/kg°C * (100-500)

Qrock = mass of rock * -80 kcal/kg

<u>Step 7:</u> Calculate mass of rock

Qlost,rock + Qgained,water = 0

Qlost,rock = -Qgained,water

mass of rock * -80 kcal/kg = -321.632 kcal

mass of rock = 4.02 kg

The rock has a mass of 4.02 kg

7 0
3 years ago
A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to
Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

p_{bot}=p_{atm}=1.013\cdot 10^5 Pa

So, the net pressure acting on the hatch is

p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa

which acts from above.

The area of the hatch is given by:

A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N

8 0
3 years ago
A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert
pychu [463]

(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

\theta=65.4^{\circ}

The projection of the length of the pendulum along the vertical direction is

L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

\Delta U = mg \Delta y

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

\Delta y = 0.211 m is the vertical displacement of the pendulum

So, the work done by gravity is

W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

\theta=90^{\circ}, cos \theta = 0

and so the work done is zero.

5 0
3 years ago
The equation of a transverse wave traveling along a very long string is y 6.0 sin(0.020px 4.0pt), where x and y are expressed in
zhannawk [14.2K]

Answer:

given

y=6.0sin(0.020px + 4.0pt)

the general wave equation moving in the positive directionis

y(x,t) = ymsin(kx -?t)

a) the amplitude is

ym = 6.0cm

b)

we have the angular wave number as

k = 2p /?

or

? = 2p / 0.020p

=1.0*102cm

c)

the frequency is

f = ?/2p

= 4p/2p

= 2.0 Hz

d)

the wave speed is

v = f?

= (100cm)(2.0Hz)

= 2.0*102cm/s

e)

since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction

f)

the maximum transverse speed is

umax =2pfym

= 2p(2.0Hz)(6.0cm)

= 75cm/s

g)

we have

y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]

= -2.0cm

6 0
3 years ago
Which of kepler’s laws explains why the sun has a slightly larger angular diameter in january than in july?
Bond [772]

Kepler’s three law is the answer. Kepler’s 3 is the amount of time it takes to orbit the sun is related to size and distance.  Kepler’s 3 is one of the planetary motion and can be stated as all planets move in elliptical orbits, having the sun sits at one of the foci.

7 0
3 years ago
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