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AlladinOne [14]
3 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]3 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 block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

6 0
3 years ago
For a railgun, determine the current required to accelerate a 1000kg space vehicle to earths escape velocity on 10km of track wi
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Answer:

the current is 15.68 × 10^{5} A

Explanation:

mass = 1000 kg

magnetic fied = 2T

rail separation = 2m

escape velocity is 11.2km/s  = 11.2 × 10^{3} m/s

distance = 10 km = 10^{4} m

to find out

determine the current

solution

we know force F = I×L×B

here I is current and L is rail separation and B is magnetic field

so F = I ×2×2  = 4 I

so

acceleration is a = \frac{F}{mass}

a =  \frac{4I}{1000} m/s²

so equation of motion

v²-u² = 2 a S

here u is initial velocity and S is distance and a is acceleration and v is final velocity

11.2 × 10^{3} - 0 = 2×  \frac{4I}{1000} × 10^{4}

solve we get I

I = 15.68 × 10^{5} A

so the current is 15.68 × 10^{5} A

4 0
3 years ago
The half-life of cobalt-60 is 5.26 years. If 50 g are left after 15.8 years, how many
PtichkaEL [24]

Answer:

400 g

Explanation:

The computation of the number of grams in the original sample is shown below:

Given that

half-life = 5.26 years

total time of decay = 15.8 years

final amount = 50.0 g

Now based on the above information  

number of half-lives past is

=  15.8 ÷ 5.26

= 3 half-lives

Now

3 half-lives = 1 ÷ 8 remains = 50.0 g

So, the number of grams would be

= 50.0 g × 8

= 400 g

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3 years ago
State the formula for calculating power
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A meter stick on earth made from a uniform-density material rests on a fulcrum positioned precisely at its middle. If a box of m
Nadusha1986 [10]

Answer:

A force of 75 N placed at 0.7 m on the meter stick.

Explanation:

The weight of the box is equal to:

W=mg=3*10=30N

The net torque is equal to zero and is equal to:

F(x-0.5)-(30*0.5)=0\\F(x-0.5)=15

For a force with a value of 75 N that is placed at 0.70 m on the meter stick, it would produce a torque of 15 N m

If you replace that values in the equation:

75(0.7-0.5)=15\\15=15

8 0
4 years ago
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