All categories

3 years ago

A race car circles 10 times around a circular 8.0-km track in 20 min.

2 answers:

6 0

Find the average speed and the average velocity.

Average speed = distance / time

distance = 10 x 8000 m = 80,000 m
time = 20 min * 60 s/min = 1200 s

Average speed = 80,000 m / 1200 s = 66.67 m/s

Average velocity = displacement / time

Given that the race car made complete circles the final poin is the same initial point, then its displacement is zero and the average velocity is zero too.

fredd [130]3 years ago

5 0

Assuming that you're asking for the average speed,

Average speed = distance / time

since he circle 8 km track 10 times, the distance is 80,000 m and the time is 20 min or 1200 seconds
Average speed = 80,000 / 1200 = 66.67 m/s

Hope this helps

You might be interested in

A cube of an unknown element has a shiny, silvery color. The side of the cube measures 2.0 cm and the cube has a mass of 14.56 g

shepuryov [24]

Answer: beryllium

Explanation: i took the test i got it right!!

8 0

2 years ago

A positively charged particle of mass 7.2 x 10-8 kg is traveling due east with a speed of 88 m/s and enters a 0.6-T uniform magn

Marianna [84]

Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

F = ma

where force is magnetic force

F = q v x B

the bold are vectors, if we write the module of this expression we have

F = qv B sin θ

as the particle moves perpendicular to the field, the angle is θ= 90º

F = q vB

the acceleration of the particle is centripetal

a = v² / r

we substitute

qvB = m v² / r

qBr = m v

q = $\frac{m\ v}{B\ r}$

The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

v = d / t

the distance is ¼ of the circle,

d = $\frac{1}{4} \ 2\pi r$

d = $\frac{\pi }{2r}$

we substitute

v = $\frac{\pi r}{2t}$

r = $\frac{2 \ t \ v}{\pi }$

let's calculate

r = $\frac{2 \ 2.2 \ 10^{-3} \ 88}{\pi }$ 2 2.2 10-3 88 /πpi

r = 123.25 m

let's substitute the values

q = $\frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}$ 7.2 10-8 88 / 0.6 123.25

q = 8.57 10⁻⁸ C

Let's reduce to mC

q = 8.57 10⁻⁸ C (10³ mC / 1C)

q = 8.57 10⁻⁵ mC

4 0

2 years ago

How many times does the earth spin in a year

sp2606 [1]

It spins one bc it takes 365 days to make a year and to make a year the earth has to mov some

7 0

2 years ago

Read 2 more answers

A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a

Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

Fc = f

m×(v^2)/(R) = μ×m×g

(v^2)/(R) = g×μ

μ = (v^2)/(R×g)

= ((25)^2)/((100)×(9.8))

= 0.64

Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.

4 0

2 years ago

a stone is projected vertically up from the top of a tower 73.5m with velocity 24.5 m/s . find the time taken by the stone to re

hoa [83]

The stone's altitude at time $t$ is given by