How do i turn 36 m/s into mph

A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake. after a short rest at the lake, she hi

3241004551 [841]

Let's choose east as positive x-direction and south as positive y-direction. We can resolve the two displacement along these two axes:

- Displacement 1 (3.5 km, $55^{\circ}$ south of west

$d_{1x}=-(3.5 km)( cos 55^{\circ})=-2.01 km$

$d_{1y}=(3.5 km)( sin 55^{\circ})=2.87 km$

- Displacement 2 (2.7 km, $16^{\circ}$ east of south

$d_{2x}=(2.7 km)( sin 16^{\circ})=0.74 km$

$d_{1y}=(2.7 km)( cos 16^{\circ})=2.60 km$

So, the total components on the two directions are

$d_x = -2.01 km+0.74 km=-1.27 km$

$d_y=2.87 km+2.60 km=5.47 km$

And the magnitude of the hiker's resultant displacement is

$d=\sqrt{(1.27 km)^2+(5.47 km)^2}=5.6 km$

8 0

2 years ago

Read 2 more answers

Calculate the time period of simple pendulum whose length is 98.2cm

alekssr [168]

The time period resulting in oscillations will be 1.986 seconds.

<h3>What is the period of oscillation?</h3>

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

The time period of the oscillation is;

$\rm T = 2 \pi\sqrt{\frac{L}{g}} \\\\ \rm T = 2 \times 3.14 \times \sqrt{ \frac{98.2 \ \times 10^{-2} \ m}{9.81 \ m/s^2}} \\\\ T= 1.986 \ sec$

Hence the time period resulting oscillations will be 1.986 seconds.

To learn more about the time period of oscillation refer to the link;

brainly.com/question/20070798

#SPJ1

5 0

1 year ago

In which one(s) of the following situations will there be an INCREASE in Kinetic Energy? Group of answer choices A block slides

Maksim231197 [3]

Answer:

The Kinetic Energy INCREASE in the following situations:

a)A block slides down a frictionless incline:

e)A merry go round rotates faster due to the push by a person.

Explanation:

The energy kinetics is proportional to the square of the velocity:

$E_k=1/2*mv^2$

We study the cases:

a)A block slides down a frictionless incline:

The gravity made a positive work and the box get a extra velocity, then the kinetics energy INCREASE

b)A box is pulled across a rough floor at constant speed:

The magnitude of the velocity does not change, so the kinetics energy does not change as well

c)A stone at the end of a string is whirled in a horizontal circle at constant speed:

The magnitude of the velocity does not change, so the kinetics energy does not change as well

d)A projectile approaches its maximum height:

If the projectile approaches its maximum height, its velocity approaches to zero. Then the kinetics Energy DECREASE

e)A merry go round rotates faster due to the push by a person.

Thanks to the push, the magnitude of the velocity INCREASE, so the kinetics energy INCREASE as well

6 0

3 years ago

Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on

qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

Δx=760 m
V₀=87 m/s
t=13.6 s
a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

$760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}$

$760=(1183.2)(cos\theta)$

$cos\theta=\frac{760}{1183.2}$

$\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}$

3 0

2 years ago

What are two examples of population distribution?

Alex777 [14]

Three basic types of population distribution within a regional range are (from top to bottom) uniform, random, and clumped.

7 0

3 years ago