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3 years ago

How do speed and velocity differ

Physics

1 answer:

Elanso [62]3 years ago

4 0

Answer: Speed is a scalar

LVelocity is a vector quantity

Explanation: Scalar quantity such as speed only have magnitude in them. It only indicates the strength on how an object is moving. For velocity it is a vector quantity it both have magnitude and involves direction where the object is moving.

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10 m/s in 10<br> 3. A car accelerates from 0 m/s to 50 m/s in 10 seconds. What is the acceleration?

son4ous [18]

Answer:

5ms⁻²

Explanation:

Given that,

Initial velocity of car (at t=0) u=0 m/s

Final velocity of car (after 10 sec) v=50 m/s

Time taken (t)= 10 sec.

acceleration(a)=?

now,

v = u + at

at = v-u

a= (v-u) / t

a= (50-0) / 10

a = (50/10)

a=5 ms⁻²

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3 years ago

What would happen if guard cells in a plant stopped working

lions [1.4K]

I am fairly certain that the plant would lose water and that gas flow and such would occur

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3 years ago

An emu moving with constant acceleration covers the distance between two points that are 92 m

nasty-shy [4]

Answer:

a) $V_{o}=14.30 m/s$

b) $a=-0.046 m/s^{2}$

Explanation:

The complete question is written below:

An emu moving with constant acceleration covers the distance between two points that are 92 m apart in 6.5s. Its speed as it passes the second point is 14 m/s. What are (a) its speed at the first point and (b) its acceleration?

Since we are talking about constant acceleration, we can use the following equations:

$d=x-x_{o}=(\frac{V_{o}-V}{2})t$ (1)

$V=V_{o}+at$ (2)

Where:

$d=92 m$ is the distance between the two points

$V_{o}$ is the velocity of the emu at the first point

$V=14 m/s$ is the velocity of the emu at the second point

$t=6.5 s$ is the time it takes to the emu to cover the distance $d$

$a$ is the emu's constant acceleration

Knowing this, let's begin with the answers:

<h2>a) Speed at the first point</h2>

In this situation wi will use equation (1):

$d=(\frac{V_{o}-V}{2})t$ (1)

Finding $V_{o}$ :

$V_{o}=\frac{2d}{t}-V$ (3)

$V_{o}=\frac{2(92 m)}{6.5 s}-14 m/s$ (4)

$V_{o}=14.30 m/s$ (5)

<h2>b) Emu's acceleration</h2>

Now we will substitute (5) in equation (2):

$14 m/s=14.30 m/s+a(6.5 s)$ (6)

Finding $a$ :

$a=-0.046 m/s^{2}$ (7) This means the emu is decreasing its speed at a constant rate.

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3 years ago

Air resistance ____ as you move faster. Question 17 options: increases decreases remains the same eventually disappears

Ket [755]

Explanation:

Air resistance drag as you move faster

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3 years ago

Read 2 more answers

When hydrogen is heated or subjected to an electric discharge, it emits light with specific visible wavelengths. A diffraction g

BartSMP [9]

To solve this problem we will apply the concepts related to the double slit-experiment. For which we will relate the distance between the Slits and the Diffraction Angle with the order of the bright fringe and the wavelength, this is mathematically given as,

$d sin\theta = m\lambda$