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4 years ago

a car accelerates from rest at 3.6m/s^2. how much time does it need to attain a speed of 7 m/s? awnser in units of s.

Physics

2 answers:

Ne4ueva [31]4 years ago

7 0

General formula is acceleration= velocity / time
acceleration / velocity = time
3.6 / 7 = time
0.514 seconds = time

Ivahew [28]4 years ago

6 0

Initial velocity: 0
final velocity: 7 m/s
a = 3.6
t = ?
x = ?

(7-0)/3.6 = t
t = 1.94 s

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A woman is standing in the ocean, and she notices that after a wave crest passes, five more crests pass in a time of 54.0 s. The

timama [110]

Answer:

a) f=0.1 Hz ; b) T=10s

c)λ= 36m

d)v=3.6m/s

e)amplitude, cannot be determined

Explanation:

Complete question is:

Determine, if possible, the wave's (a) frequency, (b) period, (c) wavelength, (d) speed, and (e) amplitude.

Given:

number of wave crests 'n'= 5

pass in a time't' 54.0s

distance between two successive crests 'd'= 36m

a) Frequency of the waves 'f' can be determined by dividing number of wave crests with time, so we have

f=n/t

f= 5/ 54 => 0.1Hz

b)The time period of wave 'T' is the reciprocal of the frequency

therefore,

T=1/f

T=1/0.1

T=10 sec.

c)wavelength'λ' is the distance between two successive crests i.e 36m

Therefore, λ= 36m

d) speed of the wave 'v' can be determined by the product of frequency and wavelength

v= fλ => 0.1 x 36

v=3.6m/s

e) For amplitude, no data is given in this question. So, it cannot be determined.

5 0

3 years ago

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

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4 years ago