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2 years ago

A cyclist rides in a circle with speed 8.1 m/s. What is his centripetal

Physics

1 answer:

Alina [70]2 years ago

5 0

Explanation:

We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:

Vt = ωr with ω being the angular velocity.

We now calculate ω:

ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s

Now that we have ω we can calculate the centripetal aceleration:

a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2

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1) A person having a mass of 59.1 kg stands before a flight of 30 stairs each of which is 25.0 cm high. He runs up 20 stairs, tu

poizon [28]

Answer:

P = 147,75 W

Explanation:

A man whose mass is 59.1kg climbs up 30 steps of a stair each step is 25 cm high

Height at 30 steps , h=30×2.5= 7.5 m

Change in potential energy , =mgh=59.1×10×7.5 = 4432.5 J

So, Work done by the man , W= 4432.5J

Power used , P= $\frac{W}{T}$

P = 4432.5 /30

P = 147,75 W

Solve any question of Work, Energy and Power with:-

brainly.com/question/3854047

#SPJ1

3 0

1 year ago

What is the current if a charge of 25 coulombs passes in 5 seconds?

Jobisdone [24]

Answer: 5 amps

Explanation: I = Q/t

8 0

2 years ago

A 1,000 kg car is travelling at 6.5 m/s to the North. A 3,500 kg truck is travelling South at the same velocity. What is the tot

Molodets [167]

Answer:

16250 kgm/s due south

Explanation:

Applying,

M = mv................. Equation 1

Where M = momentum, m = mass, v = velocity.

From the car,

Given: m = 1000 kg, v = 6.5 m/s

Substitute these values into equation 1

M = 1000(6.5)

M = 6500 kgm/s

For the truck,

Given: m = 3500 kg, v = 6.5 m/s

Substitute these values into equation 1

M' = 3500(6.5)

M' = 22750 kgm/s.

Assuming South to be negative direction,

From the question,

Total momentum of the two vehicles = (6500-22750)

Total momentum of the two vehicles = -16250 kgm/s

Hence the total momentum of the two vehicles is 16250 kgm/s due south

3 0

2 years ago

A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the

Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

$qvB = m\frac{v^2}{r}$

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

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

Re-arranging for r

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

And substituting into the previous equation

$qvB = m \frac{Tv^3}{2\pi}$

Solving for T,

$T=\frac{2\pi q B}{m v^2}$

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0

3 years ago

On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced

Korolek [52]

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

$v =f\lambda$

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

$\mu = \frac{m}{l}$

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

$\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m$

Now, for the tension we use the formula:

$v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T$

T = 0.015 N

(b)

The mass to be hung is:

$T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\$

M = 1.53 x 10⁻³ kg = 1.53 g

4 0

3 years ago