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3 years ago

7

"A hiker starts at point P and walks 2.0 kilometers due east and then 1.4 kilometers due north. The vectors in the diagram below

represent these two displacements.
Using a metric ruler, determine the scale used in the vector diagram"

1 answer:

Law Incorporation [45]3 years ago

3 0

Here, you need to use your "Protractor" as it is given in the question, but we can calculate the value with the help of our mathematical calculation too:
[ Protractor can be use only in real life, not here ]

Draw an imaginary line from initial position to final position.
Now, In that triangle, tan x = P/B
tan x = 1.4 / 2
tan x = 0.70
x = tan⁻¹ (0.70)
x = 35 [ tan 35 = 0.70 ]

In short, Your Answer would be 35 degrees

Hope this helps!

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While traveling along a highway a driver slows from 31 m/s to 15 m/s in 8 seconds. What is the automobile’s acceleration? (Remem

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Answer:

The automobile's acceleration in that time interval is -2 m/s^2

Explanation:

The acceleration is defined as the rate of change of the velocity.

The average acceleration in a given lapse of time is calculated as:

A = (final velocity - initial velocity)/time.

In this case, we have:

initial velocity = 31 m/s

final velocity = 15 m/s

time = 8 seconds.

Then the average acceleration is:

A = (15m/s - 31m/s)/8s = -2 m/s^2

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7 0

2 years ago

Read 2 more answers

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has magnitude of 3.7 m/s^2. We want

denpristay [2]

Answer:

2.32 s

Explanation:

Using the equation of motion,

s = ut+g't²/2............................ Equation 1

Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.

Note: Since Onur drops the basket ball from a height, u = 0 m/s

Then,

s = g't²/2

make t the subject of the equation,

t = √(2s/g')...................... Equation 2

Given: s = 10 m, g' = 3.7 m/s²

Substitute this value into equation 2

t = √(2×10/3.7)

t = √(20/3.7)

t = √(5.405)

t = 2.32 s.

4 0

3 years ago

Read 2 more answers

An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object

tatiyna

Answer:

$D_T=18.567m$

Explanation:

From the question we are told that:

Acceleration $a=8.0 m/s^2$

Displacement $d=1.05 m$

Initial time $t_1=6.0s$

Final Time $t_2=2.5s$

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

$V^2=2as$

$V=\sqrt{2*6*1.05}$

$V=4.1m/s$

Generally the equation for Distance traveled before stop is mathematically given by

$d_2=v*t_1$

$d_2=3.098*4$

$d_2=12.392$

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity $v_3=0 m/s$

Initial velocity $u_3=4.1 m/s$

Therefore

Using Newton's Law of Motion

$-a_3=(4.1)/(2.5)$

$-a_3=1.64m/s^2$

Giving

$v_3^2=u^2-2ad_3$

Therefore

$d_3=\frac{u_3^2}{2ad_3}$

$d_3=\frac{4.1^2}{2*1.64}$

$d_3=5.125m$

Generally the Total Distance Traveled is mathematically given by

$D_T=d_1+d_2+d_3$

$D_T=5.125m+12.392+1.05 m$

$D_T=18.567m$

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3 years ago