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love history [14]
1 year ago
5

A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *

Physics
1 answer:
Oksi-84 [34.3K]1 year ago
7 0

Given that,

Radius of track, r = 50 m

time , t = 9 s

velocity, v = ?

Distance covered by car in one lap around a track is equal to the circumference of the track.

C = 2 π r = 2 * 3.14 * 50

C = 314.159 m

Distance covered by car, s = 314.159 m

Velocity = distance/ time

V = 314.159 / 9

V = 34.9 m/s

The average velocity of car is 34.9 m/s.

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f a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance D40 compared to the stopping distance
Oksana_A [137]

Answer:

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

Explanation:

given data

speed = 40 miles / hour

distance = D40

speed limit = 25 miles / hour

distance = D25

to find out

express number a multiple of stopping distance @ 25 mph

solution

we know here stopping distance is directly proportional to (speed)²

so here speed ratio is

initial speed = \frac{40}{25}

so initial speed = 1.6

so

stopping distance increase = (1.6)²

\frac{D40}{D25} = (1.6)²

\frac{D40}{D25} = 2.56

so here

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

5 0
2 years ago
Suppose that you purchased a water bed with the dimensions 2.55 m à 2.53 dm à 245 cm. what mass of water does this bed contain
Nadya [2.5K]

dimensions of the bed is given as

length = 2.55 m

thickness = 2.53 dm = 0.253 m

width = 245 cm = 2.45 m

now the volume of the bed is given as

V = 2.55 * 0.253 * 2.45 m^3

V = 1.581m^3

now the mass of water in it is given as

mass = density * volume

mass = 1000* 1.581

mass = 1581 kg

<em>so it will contain 1581 kg mass in it</em>

6 0
2 years ago
A bicyclist travels the first 800 m of a trip 1.4 minutes, the next 500 m in 1.6 minutes, and finishes up the final 1200 m in 2
soldier1979 [14.2K]

Answer:

 v_average = 500 m / min

Explanation:

Average speed is defined

         v = (x_{f} -x₀) / Δt

let's look in each section

section 1

the variation of the distance is 800 in a time of 1.4 min

         v₁ = 800 / 1.4

         v₁ = 571.4 m / min

section 2

distance interval 500 in a 1.6 min time interval

         v₂ = 500 / 1.6

         v₂ = 312.5 m / min

section 3

distance interval 1200 m in a time 2 min

         v₃ = 1200/2

         v₃ = 600 m / min

taking the speed of each section we can calculate the average speed

         

the distance traveled

        Δx = 800 + 500 + 1200

        Δx = 2500 m

the time spent

        Δt = 1.4 + 1.6+ 2

        Δt = 5 min

         v_average = Δx / Δt

         v_average = 2500/5

         v_average = 500 m / min

7 0
2 years ago
A violin with string length 32 cm and string density 1.5 g/cm resonates in its fundamental with the first overtone of a 2.0-m or
love history [14]

Answer:

T=1022.42 N

Explanation:

Given that

l = 32 cm ,μ = 1.5 g/cm

L =2 m  ,V= 344 m/s

The pipe is closed so n= 3 ,for first over tone

f=\dfrac{nV}{4L}

f=\dfrac{3\times 344}{4\times 2}

f= 129 Hz

The tension in the string given as

T = f²(4l²) μ

Now by putting the values

T = f²(4l²) μ

T = 129² x (4 x 0.32²)  x 1.5 x  10⁻³ x 100

T=1022.42 N

6 0
2 years ago
A split highway has a number of lanes for traffic. For traffic going in one direction, the radius for the inside of the curve is
Arte-miy333 [17]
Cuál es tu pregunta
7 0
2 years ago
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