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3 years ago

In one or two sentences, explain the difference between displacement and distance traveled.

Physics

1 answer:

34kurt3 years ago

8 0

Answer: Distance doesn't specify direction but displacement do.

Explanation:

Distance specifies how far an object has gone, it doesn't specify direction of the object compared to displacement that is distance covered by a body in a specified direction.

Distance is therefore a scalar quantity i.e quantity that specifies only magnitude but no direction but displacement is vector quantity that specifies both magnitude and direction.

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As mass increases does potential energy decrease or increase?

timurjin [86]

The equation for potential energy is = mgh (m stands for mass, g stands for gravitational acceleration, and h stands for height)
So if mass increases, it will also increase potential energy

8 0

4 years ago

At take off, a plane flies 100 km north before turning to fly 200 km east. How far is its destination from where the plane took

IRINA_888 [86]

<u>Answer</u>

224 Km

4 Km

<u>Explanation</u>

The plane flies 100 Km north and then 200 Km east. This makes a 90° turn. To get the distance taken by the plane we get the hypotenuse.

c² = a² + b²

c² = 100² + 200²

= 50,000

c = √50,000

= 223.606 Km

The answer to the nearest whole number is 224 Km

If the runner run 15 Km north then 11 Km south, the resultant distance would be,

15 - 11 = 4 Km

4 0

3 years ago

Read 2 more answers

A large, 60 turn circular coil of radius 10.0 cm carries a current of 4.2 A. At the center of the large coil is a small 20 turn

Eddi Din [679]

To solve this problem we apply the concepts related to the electric torque generated by the electromagnetic field. Mathematically this Torque can be written under the following relation

$\tau = NIAB sin\theta$

Here,

N = Number of Turns

I = Current

A = Area

B = Magnetic Field

The maximum torque will be reached when the angle is 90 degrees, then we will have the following relation,

$\theta = 90\°C$

Magnetic Field is given at function of the number of loops, permeability constant at free space at the perimeter, then

$B = \frac{N\mu_0 I}{2\pi r}$

$B = \frac{(60)(4\pi * 10^{-7})(4.2)}{2\pi (0.1)}$

$B = 5.04*10^{-4}T$

Replacing at the first equation we have,

$T = (20)(\pi (0.005)^2)(1)(5.04*10^{-4})$

$T = 7.91*10^{-7}N\cdot m$

4 0

3 years ago

Feeling a bit better about not getting the Nobel Prize, Meitner drove home from the rink. If her 1500kg car accelerated from res

lyudmila [28]

Answer:

<em>The force exerted by the car engine was 3000 N</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

$F=m.a$

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

$v_f=v_o+a.t$

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

The question describes how Meitner drove home taking her car from rest to a speed of 20 m/s in 10 seconds. This provides us the following data:

vf=20 m/s, v0=0 (rest), t = 10 seconds.

From the kinematics equation, we can solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{20\ m/s-0}{10\ s}=2\ m/s^2$

The force exerted by the car engine was:

$F=1500\ kg\cdot 2\ m/s^2$

$F=3000\ N$

The force exerted by the car engine was 3000 N

5 0

3 years ago

An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is

Lady_Fox [76]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}$

$\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

<h3>☯ <u>By using formula of Lens</u> </h3>

$\\$

$\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}$

$\\$

$\dashrightarrow\:\: \sf{ v = 30 \ cm}$

$\\$

<h3>☯ <u>Now, Finding the magnification </u></h3>

$\\$

$\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}$

$\\$

$\dashrightarrow\:\: \sf{m = -2}$

$\\$

<h3>☯ <u>Hence</u>, $\\$ </h3>

$\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}$

$\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}$

3 0

3 years ago