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AlladinOne [14]
2 years ago
6

There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the differe

nce between these two quantities.
Physics
1 answer:
Usimov [2.4K]2 years ago
5 0

An example that illustrates the difference is the circular motion

Explanation:

Let's start by reminding the definition of the two quantities:

- Speed is a scalar quantity that tells "how fast" an object is moving, regardless of its direction of motion.

Speed can be  calculate as:

speed = \frac{d}{t}

where:

d is the distance travelled

t is the time taken

- Velocity is instead a vector quantity, given by:

velocity = \frac{d}{t}

where;

d is the displacement of the object (displacement is a a vector connecting the initial position to the final position of motion)

t is the time taken

Since it is a vector, velocity has both a magnitude and a direction, therefore it also takes into account the direction of motion of the object.

For an object in motion in a straight line, speed and velocity are the same. However, this is not always the case.

In fact, an example of motion in which the two quantities are different is the circular motion. Consider for example the object making one complete revolution along the circle. Therefore, its average speed is the ratio between the length of the perimeter (the distance) divided by the time taken:

speed = \frac{2\pi r}{t}

where r is the radius of the circle.

However, the displacement of the object is zero (because the object returns to the starting point), and so the average velocity is also zero:

velocity = \frac{0}{t}=0

Learn more about speed and velocity:

brainly.com/question/8893949

brainly.com/question/5063905

brainly.com/question/5248528

#LearnwithBrainly

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A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C
mario62 [17]

Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}

\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}

\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}

x = - 1.4 m

6 0
2 years ago
Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley
kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = \frac{g}{7}

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = \frac{4m}{2} x^2  + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2

T = 4mx^2 + 2my^2 -2mxy  

also the general potential energy can be expressed as

U = -4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant

The Lagrangian of the problem can now be setup as

L =4mx^2 +2my^2 -2mxy +2mgy + constant

next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange  equation = 4x-y =0\\-2y+x +g = 0

solving the equations simultaneously

x ( acceleration of mass 4m ) = \frac{g}{7}

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

8 0
1 year ago
To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch
sergij07 [2.7K]

Answer:

man upward acceleration is 0.14m/s^2

Explanation:

given data:

mass of man = 72 kg

downward force = 360 N

The mass of man of weight 72 kg is hang from two sections of rope, one section pf rope ties around man waist and other section is ties in man hands. when he pulls down the rope  with 360 N force then each section of  rope pulls with 360 N

we know that

Weight= mass × gravity= 72kg × 9.8 = 705.6N

Force = mass× acceleration

Force= -705.6 + (2 × 358) = 10.4 N

acceleration = \frac{10.4}{72} = 0.14m/s^2

4 0
1 year ago
Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3600 Hz . To do this, the car a
Alex17521 [72]

Answer:

The capacitance and the inductance can choose for a car-alarm circuit are

C = 215.27 μF

L = 9.078 μH

Explanation:

V =12.0 V, E = 1.55*10^2 J, f = 3600 Hz

To determine the capacitance can use the equation

U_c= \frac{1}{2}*C*V^2

Solve to C'

C = \frac{U_c*2}{V^2}=\frac{1.55x10^2J*2}{12.0^2V}

C=215.27 uF

To find the inductance can use the frequency of the circuit

f = \frac{1}{2\pi* \sqrt{C*L} }

Solve to L'

L = \frac{1}{4\pi^2*f^2*C}=\frac{1}{4\pi^2*3600^2*215.27 uF}}

L = 9.078 uH

6 0
1 year ago
1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational
sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = \frac{m v^{2} }{r}

==> 1.1 × 10^{19} = (9.4 × 10^{21} × v^{2} ) ÷ 1.5 × 10^{8}

==> 1.1 × 10^{19} = v^{2} × 6.26×10^{13}

==> v^{2} =  1.1 × 10^{19} ÷ 6.26×10^{13}

==> v^{2} = 0.17571885 × 10^{6}

==> v= 0.419188323 × 10^{3} m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5×10^{8} ÷ 0.419188323 × 10^{3}

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0
1 year ago
Read 2 more answers
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