8a/46=92 a=x explain

Ilya and Anya each can run at a speed of 8.90mph and walk at a speed of 3.80mph . They set off together on a route of length 5.0

jekas [21]

Answer:

a) t = 56.34 minutes

b) v = 5.32mph

c) t = 47.22 minutes

d) v = 6.35 mph

Explanation:

Given;

The speed of Ilya and Anya are

Run = 8.90mph

Walk = 3.80mph

distance = 5.00 miles

a) anya walk half the distance and run half the distance.

Speed = distance/time

time = distance/speed

t1 = 5×0.5/3.80 = 0.658hr

t2 = 5×0.5/8.90 = 0.281hr

t = t1 + t2 = 0.658+0.281 = 0.939hr × 60minutes/hr

t = 56.34 minutes

b) average speed = total distance/total time taken

v = 5.00miles/0.939hr = 5.32mph

c) Ilya walk half the time and run half the time.

t1 = t2 = t/2

Distance = speed × time

Substituting the values.

5 = 3.80(t/2) + 8.90(t/2)

5 = 12.70(t/2)

t = 10/12.70

t = 0.787 hr × 60 minutes/hr

t = 47.22 minutes

d) average speed = 5miles/0.787hr

v = 6.35 mph

5 0

3 years ago

In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of

antoniya [11.8K]

Answer : 2446 years.

Explanation :

Length of semi major axis is, $a=84\ au= 1.496\times 10^{11}\ m$

According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.

i.e $T^2=\dfrac{4\pi^2}{GM}a^3$

where G is gravitational constant

M is mass of sun, $M=1.98\times 10^{30}\ Kg$

So, $T^2=\dfrac{4\times (3.14)^2}{6.6\times 10^{-11}Nm^2/Kg\times 1.98\times 10^{30}\Kg}$

$T^2=3\times 10^{-19}\times(84\times 1.496\times 10^{11})^3$

$T^2=3\times 10^{-19}\times 1984415.6\times 10^{33}$

$T^2=59532469.8\times 10^{14}\ s$

$T=7715.7\times 10^7\ s$

since, $1\ sec=3.17\times 10^{-8}\ years$

So, orbital period is approximately 2446 years.

7 0

4 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

Help with velocity math pls help asap

pychu [463]

2 is the answer have a nice day <3

5 0

3 years ago

Otion

Scorpion4ik [409]

Answer:

SKID

Explanation:

In general, airplane tracks are flat, they do not have cant, consequently the friction force is what keeps the bicycle in the circle.

Let's use Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis.

Y axis y

N- W = 0

N = W

X axis (radial)

fr = m a

the acceleration in the curve is centripetal

a = $\frac{v^2}{r}$

the friction force has the expression

fr = μ N

we substitute

μ mg = m v²/r

v = $\sqrt{\mu g r}$

we calculate

v = $\sqrt{0.1 \ 9.8 \ 3}$

v = 1,715 m / s

to compare with the cyclist's speed let's reduce to the SI system

v₀ = 18 km / h (1000 m / 1 km) (1 h / 3600 s) = 5 m / s

We can see that the speed that the cyclist is carrying is greater than the speed that the curve can take, therefore the cyclist will SKID

5 0

3 years ago