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3 years ago

A student runs 35 m east, then 12 m west. What is the distance run by the student?

Physics

2 answers:

damaskus [11]3 years ago

6 0

Answer:

47 m

Explanation:

Distance is defined as the total path length traveled by the object. It is a scalar quantity. It means it does not depend on the directon.

So, the distance travelled by the student is 35 + 12 = 47 m

lora16 [44]3 years ago

4 0

47m total just add them up 35 + 12 = 47

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How do exons and introns differ?

Norma-Jean [14]

Exons And Introns differ because Exons code for protein and Introns do not.

so your answer would be: Exons code for Proteins

Are you in K12?

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3 years ago

You are In-line skates at the top of a small hill. Your potential energy is equal to 1000 J. The last time we checked, your mass

stira [4]

A) Weight = mass × acceleration (due to gravity)

= 60×9.8

= 588 N

B) Potential energy = mass x gravity x change in height

1,000 = 60.0 x 9.8 x h

h = 1.7 m

C) Kinetic energyF = potential energyI

KEF = 1/2mv2

PEI = mgh = 1,000 J

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3 0

3 years ago

Are light waves longitudinal or transverse

Marysya12 [62]

Answer:

Transverse

Explanation:

There are two types of waves, according to the direction of their oscillation:

- Transverse waves: in a transverse wave, the direction of the oscillation is perpendicular to the direction of motion of the wave. Examples of transverse waves are electromagnetic waves

- Longitudinal waves: in a longitudinal wave, the direction of the oscillation is parallel to the direction of motion of the wave. Examples of longitudinal waves are sound waves.

Light waves corresponds to the visible part of the electromagnetic spectrum, which includes all the different types of electromagnetic waves (which consist of oscillations of electric and magnetic fields that are perpendicular to the direction of propagation of the wave): therefore, they are transverse waves.

6 0

3 years ago

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b

Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

$Q=t^3-2t^2+4t+4$

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4$

At t = 1 s,

Current,

$I=3(1)^2-4(1)+4\\\\I=3\ A$

So, the current at t = 1 s is 3 A.

For lowest current,

$\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s$

Hence, this is the required solution.

7 0

3 years ago

A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220

disa [49]

First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, $\mu m g$ . For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
$ma=-\mu m g$
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
$a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2$

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
$2aS=v_f^2-v_i^2$
where $v_f=0$ is the final speed of the skier and $v_i=4.75 m/s$ is the initial speed. Substituting numbers, we find:
$S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m$

5 0

3 years ago