E

In reaching her destination, a backpacker walks with an average velocity of 1.32 m/s, due west. This average velocity results, b

Svetach [21]

Answer:

distance in east is 1273.78 m

Explanation:

given data

average velocity = 1.32 m/s west =

hike = 5.21 km = 5.21 × 10³ m

average velocity = 3.49 m/s west

average velocity = 0.687 m/s east

to find out

distance in east

solution

we consider here distance in east is = x

so distance from starting point = 5.21 × 10³ - x ...................1

and we can say time required to reach end

time required = distance / speed

time required = $\frac{5.21 *10^3 - x}{1.32}$ ................2

and

time required for 6.44 km west

time required = $\frac{5.21 *10^3 - x}{3.49}$ ................3

and time required for distance x

time required = $\frac{x}{0.687}$ ................4

so from equation 2 , 3 and 4

$\frac{5.21 *10^3 - x}{1.32}$ = $\frac{5.21 *10^3 - x}{3.49}$ + $\frac{x}{0.687}$

x = 1273.78 m

so distance in east is 1273.78 m

8 0

3 years ago

I NEED HELP! Plzzzz I will do anything.

Leno4ka [110]

Answer:

B

Explanation:

8 0

3 years ago

A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 51.6 m/s at an angle of 37.

Andrei [34K]

Answer:

$v_{y}=35.21m/s$

Explanation:

From the exercise we know the cannonball's initial velocity, the angle which its released with respect to the horizontal and its initial height

$v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m$

If we want to know whats the y-component of velocity we need to use the following formula:

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Knowing that $g=-9.8m/s^2$

$v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s$

So, the cannonball's y-component of velocity is $v_{y}=35.21m/s$

6 0

3 years ago

A spring with a spring constant kk of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. it is then s

tino4ka555 [31]

Given:

k = 100 lb/ft, m = 1 lb / (32.2 ft/s) = 0.03106 slugs

Solution:

F = -kx
mx" = -kx
x" + (k/m)x = 0

characteristic equation:
r^2 + k/m = 0
r = i*sqrt(k/m)

x = Asin(sqrt(k/m)t) + Bcos(sqrt(k/m)t)

ω = sqrt(k/m)
2π/T = sqrt(k/m)
T = 2π*sqrt(m/k)
T = 2π*sqrt(0.03106 slugs / 100 lb/ft)
T = 0.1107 s (period)

x(0) = 1/12 ft = 0.08333 ft
x'(0) = 0

1/12 = Asin(0) + Bcos(0)
B = 1/12 = 0.08333 ft

x' = Aω*cos(ωt) - Bω*sin(ωt)
0 = Aω*cos(0) - (1/12)ω*sin(0)
0 = Aω
A = 0

So B would be the amplitude. Therefore, the equation of motion would be x = 0.08333*cos[(2π/0.1107)t]

5 0

4 years ago

When Pluto was classified as a planet it was known as a oddball planet why? Why is it less if an oddball now?

hichkok12 [17]

Answer:

a strange world that has baffled scientists ever since it was discovered in 1930. It is not the large gas giant that one might expect to find in the outer reaches of the solar system.

Explanation:

Explanation

7 0

3 years ago