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2 years ago

12

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

ately 57.0 m57.0 m . If the track is completely flat and the race car is traveling at a constant 31.5 m/s31.5 m/s (about 7.0×101 mph7.0×101 mph ) around the turn, what is the race car's centripetal (radial) acceleration?

1 answer:

crimeas [40]2 years ago

5 0

Answer:

Centripetal acceleration of the car is 17.4 m/s²

Explanation:

It is given that,

Radius of the track, r = 57 m

Speed of car, v = 31.5 m/s

We need to find the centripetal acceleration of the race car. The formula for the centripetal acceleration is given by :

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

$a=\dfrac{(31.5\ m/s)^2}{57\ m}$

$a=17.4\ m/s^2$

So, the centripetal acceleration of the race car is 17.4 m/s². Hence, this is the required solution.

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A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,

Annette [7]

Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

$d_x = 10 + 2= 12 blocks$

Total displacement towards South

$d_y = 5 block$

now to find the net displacement we can use vector addition

$d = \sqrt{d_x^2 + d_y^2}$

$d = \sqrt{12^2 + 5^2}$

$d = 13 blocks$

<em>so magnitude of net displacement will be equal to 13 blocks</em>

6 0

1 year ago

A (1.25+A) kg bowling ball is hung on a (2.50+B) m long rope. It is then pulled back until the rope makes an angle of (12.0+C)o

daser333 [38]

Answer:

F = 0.535 N

Explanation:

Let's use the concepts of energy, at the highest and lowest point of the trajectory

Higher

Em₀ = U = mg y

Lower

$Em_{f}$ = K = ½ m v²

Emo = $Em_{f}$

mg y = ½ m v2

v = √ 2gy

y = L - L cos θ

v = √ (2g L (1-cos θ))

Now let's use Newton's second law n at the lowest point where the acceleration is centripetal

F = ma

a = v² / r

In turning radius is the cable length r = L

F = m 2g (1-cos θ)

Let's calculate

F = 2 1.25 9.8 (1 - cos 12)

F = 0.535 N

7 0

2 years ago

A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope

pogonyaev

Answer:

The speed of transverse waves in the wire is 234.26 m/s.

Explanation:

Given that,

Length of wire = 129 cm

Speed of sound in air = 340 m/s

First position of resonance = 31.2 cm

We need to calculate the wavelength

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance.

At 1st resonance,

$L=\dfrac{\lambda}{4}$

$\lambda=4\times L$

Put the value into the formula

$\lambda=4\times31.2\times10^{-2}$

$\lambda=1.248\ m$

We need to calculate the frequency of sound in pipe

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{340}{1.248}$

$f=272.4\ Hz$

We need to calculate the node distance

For the wire, there are 3 segments, so 4 nodes

Node-node distance ,

$L=\dfrac{l}{3}$

$L = \dfrac{129}{3}$

$L=43\ cm$

We need to calculate the wavelength of the wire

Using formula of length

$L=\dfrac{\lambda}{2}$

$\lambda=L\times 2$

Put the value into the formula

$\lambda=43\times2$

$\lambda=86\ cm$

We need to calculate the speed of the wave

Using formula of frequency

$f=\dfrac{v}{\lambda}$

$v=f\times\lambda$

Put the value into the formula

$v=272.4\times86\times10^{-2}$

$v=234.26\ m/s$

Hence, The speed of transverse waves in the wire is 234.26 m/s.

4 0

2 years ago

James Joule (after whom the unit of energy is named) claimed that the water at the bottom of Niagara Falls should be warmer than

Molodets [167]

Answer:

0.12 K

Explanation:

height, h = 51 m

let the mass of water is m.

Specific heat of water, c = 4190 J/kg K

According to the transformation of energy

Potential energy of water = thermal energy of water

m x g x h = m x c x ΔT

Where, ΔT is the rise in temperature

g x h = c x ΔT

9.8 x 51 = 4190 x ΔT

ΔT = 0.12 K

Thus, the rise in temperature is 0.12 K.

7 0

2 years ago

3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

aleksandr82 [10.1K]

Answer:

(a) $W=-19.25J$

(b) $W=-52.8J$

Explanation:

Hello.

(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:

$W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J$

Which is negative as it expands against the given pressure.

(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:

$W=nRTln(\frac{V_1}{V_2} )$

Whereas the moles are computed from the given mass of argon:

$n=6.56g*\frac{1mol}{39.95g}=0.164mol$

Thus, the work is:

$W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J$

Regards.

4 0

2 years ago