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gulaghasi [49]
4 years ago
6

A 1300-kg car initially has a velocity of 22.2 m/s due south. It brakes to a stop over a 180 m distance.

Physics
1 answer:
Vaselesa [24]4 years ago
4 0
The acceleration of the object which moves from an initial step to a full halt given the distance traveled can be calculated through the equation,
                                     d = v² / 2a
where d is distance, v is the velocity, and a is acceleration
Substituting the known values,
                                     180 = (22.2 m/s)² / 2(a)
The value of a is equal to 1.369 m/s²
The force needed for the object to be stopped is equal to the product of the mass and the acceleration.
                                      F = (1300 kg)(1.369 m/s²) 
                                            F = 1779.7 N
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3 0
3 years ago
A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i
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(a) Wavelength is 0.436 m

(b) Length is 0.872 m

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As per the question:

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y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

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Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

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\omega = 166\ rad/s

where

A = Amplitude

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Now, to calculate the string's wavelength,

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(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

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Answer:

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Explanation:

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