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matrenka [14]
3 years ago
6

two skateboarders of mass 50 kg and 60 kg push each other with force 70N.what is the acceleration of each skaters.step by step.

Physics
1 answer:
Keith_Richards [23]3 years ago
6 0

Answer:

f=ma

f=70N

m1=50kg

m2= 60kg

a1= f/m

a1= 70/50

a1= 1.4N/kg

a2= 70/60

a2=1.17N/kg

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alekssr [168]

Answer:

31.2 m/s

Explanation:

f_{app} = Frequency of approach = 480 Hz

f_{aw} = Frequency of going away = 400 Hz

V = Speed of sound in air = 343 m/s

v = Speed of truck

Frequency of approach is given as

f_{app} = \frac{Vf}{V - v}                           eq-1

Frequency of moving awayy is given as

f_{aw} = \frac{Vf}{V + v}                          eq-2

Dividing eq-1 by eq-2

\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}

\frac{480}{400} = \frac{343 + v}{343 - v}

v = 31.2 m/s

7 0
3 years ago
Yea, gonna need some help. Thanks
natulia [17]

Answer:

t = 3.48 s

Explanation:

The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:

h(t) = -16t^2+v_ot+h_o\\\\\frac{dh(t)}{dt}=0=-32t+v_o\\\\v_o = 32t

where,

v₀ = initial speed = 110 ft/s

Therefore,

110 = 32t\\\\t = \frac{110}{32}\\\\

<u>t = 3.48 s</u>

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