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Viefleur [7K]
3 years ago
13

A racing car is traveling on a circular race course of radius 569 m at a speed such that it makes 1.46 rev in 1 minute. the mass

of the car is 1550 kg. what is the centripetal force acting on the car? answer in units of n.
Physics
1 answer:
HACTEHA [7]3 years ago
8 0
1.46rev in 1min = 2pi*1.46revs/60s=0.152891 rad/s=ω
Fc=m*r*ω^2
Fc= 1550kg*(0.152891)^2 * 569m = 19951N 

hope this helps! Thanks!
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As a ball rolls ... even if there were no air resistance ... the ball
has to push down and climb over every one of those little fibers
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7 0
3 years ago
Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0
Tema [17]

Answer:

841.5 Hz

Explanation:

Given

y = 50 cm = 0.5 m

d = 5.00 m

L = 12.0 m away from the wall

v = speed of sound = 343 m/s

The image of the scenario is presented in the attached image.

When destructive interference is being experienced from 50 cm (0.5 m) parallel to the wall, the path difference between the distance of the two speakers from the observer is equal to half of the wavelength of the wave.

Let the distance from speaker one to the observer's new position be d₁

And the distance from the speaker two to the observer's new position be d₂

(λ/2) = |d₁ - d₂|

d₁ = √(12² + 3²) = 12.3693 m

d₂ = √(12² + 2²) = 12.1655 m

|d₁ - d₂| = 0.2038 m

(λ/2) = |d₁ - d₂| = 0.2038

λ = 0.4076 m

For waves, the velocity (v), frequency (f) and wavelength (λ) are related thus

v = fλ

f = (v/λ) = (343/0.4076) = 841.5 Hz

Hope this Helps!!!

7 0
3 years ago
Read 2 more answers
A 1.13 kg skateboard is coasting along the pavement at a speed of 4.28 m/s when a 0.93
faust18 [17]

By conservation of momentum,

Pinitial = Pfinal

m1v1 + m2v2 = (m1 + m2)*vf

m1 = mass of skateboard = 1.13 kg

m2 = mass of cat = 0.93 kg

v1 = initial velocity of skateboard = 4.28 m/s

v2 = initial velocity of cat = 0 m/s

vf = final velocity of skateboard-cat combo

So plug in the values and solve for vf,

1.13(4.28) + 0.93(0) = (1.13 + 0.93)vf

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5 0
3 years ago
Planet with the most extreme temperature range
Leokris [45]

Answer:

Mercury

Explanation:

3 0
3 years ago
Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms o
ipn [44]

Answer:

U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

Explanation:

Given:

- The extension in spring @ equilibrium = x m

- The spring constant = k

- The amount of distance pulled down = d

- mass of the toy = m

Find:

- The total mechanical energy E_top at the top position h_max in terms of the available variables.

Solution:

- First we need to determine the types of Energy that are in play:

- The Elastic potential Energy E_p in a spring is given:

                              E_p: 0.5 * k * (ext)

- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:

             ext = Equilibrium + distance pulled - h_max = (x + d - h_max)

Hence, the elastic potential energy will be:

                              E_p = 0.5 * k *(x + d - h_max)^2

- The gravitational potential energy E_g is given by:

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Where, bottom most position is taken as reference (datum).

- The kinetic Energy E_k is given by:

                              E_k = 0.5*m*v_top^2

- Since we know that the maximum height is reached when velocity is zero

Hence,                   E_k = 0.5*m*0^2 = 0.

The total Energy of the system U is sum of all energies and play:

                               U = E_p + E_k + E_g

                               U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

8 0
4 years ago
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