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andrew11 [14]
2 years ago
14

A car travels in a straight line for 4.4 h at a constant speed of 81 km/h. What is its acceleration?

Physics
2 answers:
Fed [463]2 years ago
7 0

The speed and length of time don't matter.

Constant speed in a straight line is the DEFINITION of ZERO acceleration.

irga5000 [103]2 years ago
5 0

Answer:

a = 0m/s^2

Explanation:

we can use the equation:

v_f = a(t) + v_i

And plug in our numbers:

81km/h = am/s^2(4.4h) + 81hm/h

0 = am/s^2(4.4h)

0 = am/s^2

a = 0m/s^2

Because the speed was constant, there was no difference between the final and initial velocities, meaning that once you do the math you see there is no acceleration affecting the speed.

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A ball is dropped and falls with an acceleration of 9.8 m/s2 downward. It hits the ground with a velocity of 49 m/s downward. Ho
il63 [147K]
Hey!

NOTE-:

u= initial velocity
v= final velocity
g= acceleration due to gravity
t= time

u= 0
v= 49 m/s
t=?
g= 9.8 m/s^2

Using first equation of motion -

v-u=at
49-0= 9.8×t
49 = 9.8t
49/9.8= t
t= 5 second


Hope it helps...!!!
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A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down
rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

V = \sqrt{\frac{T}{\rho}}

Where,

T = Tension

\rho = Linear density

Our data are given by

Tension , T = 70 N

Linear density , \rho = 0.7 kg/m

Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

V = \sqrt{\frac{T}{\rho}}

V = \sqrt{\frac{70}{0.7}

V = 10m/s

Speed can also be expressed as

V = \lambda f

Re-arrange to find \lambda

\lambda = \frac{V}{f}

Where,

f = Frequency,

Which is also described in function of the Period as,

f = \frac{1}{T}

f = \frac{1}{0.35}

f = 2.86 Hz

Therefore replacing to find \lambda

\lambda = \frac{10}{2.86}

\lambda = 3.49m

Therefore the wavelength of the waves created in the string is 3.49m

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