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3 years ago

A bus traveling at 24 m/s slows down to 12 m/s in 5.0 seconds. What is the acceleration?

Physics

1 answer:

Contact [7]3 years ago

8 0

Answer:

a = -2.4 m/s²

Explanation:

Given,

The initial speed of the bus, u = 24 m/s

The final speed of bus, v = 12 m/s

Time taken to reach final speed is, t = 5.0 s

The acceleration of the body is given by the change in velocity by time

a = (v - u) / t

= (12 - 24) / 5

= -2.4 m/s²

The negative sign in the acceleration indicates that the bus is decelerating.

Therefore, the acceleration of the bus is, a = -2.4 m/s²

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An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into th

Black_prince [1.1K]

Answer:

a) R = ρ₀ L /π(r_b² - R_a²) , b) ρ₀ = V / I π (r_b² - R_a²) / L

Explanation:

a) The resistance of a material is given by

R = ρ l / A

where ρ is the resistivity, l is the length and A is the area

the length is l = L and the resistivity is ρ = ρ₀

the area is the area of the cylindrical shell

A = π r_b² - π r_a²

A = π (r_b² - r_a²)

we substitute

R = ρ₀ L /π(r_b² - R_a²)

b) The potential difference is related to current and resistance by ohm's law

V = i R

we subsist the expression of resistance

V = I ρ₀ L /π (r_b² - R_a²)

ρ₀ = V / I π (r_b² - R_a²) / L

6 0

3 years ago

A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What

Crank

Answer:

$\mu = \frac{r (2\pi f)^{2}}{g}$

Explanation:

$N$ = normal force acting on the coin

Normal force in the upward direction balances the weight of the coin, hence

$N = mg$

$f$ = frequency of rotation

Angular velocity of turntable is hence given as

$w = 2\pi f$

$r$ = distance from the axis of rotation

$\mu$ = minimum coefficient of static friction

static frictional force is given as

$f = \mu N\\f = \mu mg$

The static frictional force provides the necessary centripetal force , hence

Centripetal force = Static frictional force

$m r w^{2} = \mu mg\\r w^{2} = \mu g\\\\\mu = \frac{r w^{2}}{g} \\\mu = \frac{r (2\pi f)^{2}}{g}$

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3 years ago

A cellist tunes the C string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.5

Dahasolnce [82]

Answer:

$\lambda = 1.18 \ m$

$v = 77.172 \ m/s$

$T = 151.41 \ N$

Explanation:

From the question we are told that

The frequency is $f = 65.4 \ Hz$

The length of the vibrating string is $L = 0.590 \ m$

The mass is $m = 15.0 \ g = 0.015 \ kg$

Generally the wavelength is mathematically represented as

$\lambda = 2 * L$

=> $\lambda = 2 * 0.590$

=> $\lambda = 1.18 \ m$

Generally the wave speed is

$v = \lambda * f$

=> $v = 1.18 * 65.4$

=> $v = 77.172 \ m/s$

Generally the tension on the wire is mathematically represented as

$T = v^2 * \frac{ m }{L }$

=> $T = 77.172 ^2 * \frac{ 0.015 }{0.590}$

=> $T = 151.41 \ N$

7 0

3 years ago

What mistake did Dominic make?

Tatiana [17]

Answer:

Im not sure who you are talking about, but if you are talking about Dominic Raab, well, he failed to make a crucial phone call to seek urgent help airlifting translators out of Afghanistan. if you are not talking about this Dominic, then please add more information to your question.

7 0

2 years ago

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op

Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

$f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz$

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0

3 years ago