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3 years ago

The main difference between distance and displacement is tat

Physics

2 answers:

vladimir2022 [97]3 years ago

8 0

Answer:

Distance is the length of the path taken by an object whereas displacement is the simply the distance between where the object started and where it ended up.

Explanation:

Hope this helps! Good Luck on the rest of your assignments! :)

melomori [17]3 years ago

6 0

Answer:

Distance is quite literally the total amount of distance you have covered. Displacement is the total displacement from your origin. For example, if you were to move 5m north then 2m south, your total displacement from your origin would be 3m north.

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A rider on a bike with the combined mass of 100kgattains aterminal speed of 15m/son a 12% slope. Assuming that the only forces a

Semmy [17]

Answer:

C_d = 0.942

Explanation:

Let's first calculate the angle of inclination.

Formula is;

tan θ = (%slope)

% Slope is given as 12%

Thus;

θ = tan^(-1) (12/100)

θ = 6.843°

Let's now calculate the force due to the weight of the rider and bike combined from;

F = mg sin θ

We are given; m = 100 kg.

Thus;

F = 100 × 9.81 × sin 6.843

F = 116.885 N

The drag force will also be the same as the force due to the weight of the body. Thus;

Drag force; F = C_d(½ρu²A)

Where;

C_d is drag coefficient

ρ is density

U is terminal speed

A is area

We are given;

A = 0.9 m²

U = 15 m/s

From online tables, density of air is approximately 1.225 kg/m³

Thus;

116.885 = C_d(½ × 1.225 × 15² × 0.9)

116.885 = 124.03125C_d

C_d = 116.885/124.03125

C_d = 0.942

4 0

3 years ago

What is a factor that affects the outcome of an experiment?

Elis [28]

The answer is D. Variable. Hope this Helps:)))

6 0

4 years ago

Read 2 more answers

Explain why rolling friction is lesser than sliding friction <br> Pls ans this

larisa [96]

Answer:

As the area of contact is less in the case of rolling than in the case of sliding, rolling friction is less than the sliding friction.

8 0

3 years ago

Read 2 more answers

A spaceship is 1600 m long when it is at rest. When it is traveling at a certain constant speed its length is measured by extern

Firdavs [7]

Answer:

speed of the spaceship is 0.495251 c

Time T = 1.4982 hr

Explanation:

given data

length L1 = 1600 m

length L2 = 1390 m

to find out

speed of the spaceship and time

solution

we know here

L2 = L1 × $\sqrt{1 - V^{2} }$

so we find V by put L1 and L2

1390 = 1600 × $\sqrt{1 - V^{2} }$

V = 0.495251 c

speed of the spaceship is 0.495251 c

and

we know T1 = 1.95 hr

so we find T2

T1 = T2 $\sqrt{1 - v^{2} }$

put here V and T1 so we get T2

1.95 = T2 $\sqrt{1 - 0.495251^{2} }$

Time T = 1.4982 hr

3 0

4 years ago

You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testt

sertanlavr [38]

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

$f_{1}=\dfrac{v}{4l}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times0.11}$

$f_{1}=781.81\ Hz$

$f_{1}=0.782\ kHz$

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

$f_{1}=\dfrac{v}{4(\dfrac{L}{2})}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}$

$f_{1}=1563.63\ Hz$

$f_{1}=1.563\ kHz$

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

6 0

4 years ago