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WITCHER [35]
3 years ago
9

A ticker timer makes 50 dots per second. When a body is pulled by a tap through the timer, the distance between the third and fo

urth dot is 3cm and the distance between the fourth and fifth dot is 4cm.what was the acceleration of the body?
Physics
1 answer:
fenix001 [56]3 years ago
8 0

Answer:

The acceleration is 1 cm/s^2.

Explanation:

The acceleration is defined as the rate of change of velocity.

Here, initial velocity, u = 3/1 = 3 cm/s

final velocity, v = 4/1 = 4 cm/s

time, t = 1 s

Let the acceleration is a.  

Use first equation of motion

v =  u + at

4 = 3 + 1 x a

a = 1 cm/s^2

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A 15 kg kangaroo jumps with an upward acceleration of 3 m/s2 by pushing hard off the ground.
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ANSWER FAST PLEASE HELP
KengaRu [80]

Answer:

B. 175 N

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

Fnet = Fapp + Fg

Where;

Fnet is the net force

Fapp is the applied force

Fg is the force due to gravitation

In this scenario, we observed that both forces are acting in the same direction.

Therefore:

Net force = 100 N + 75 N

Net force = 175 Newton

6 0
3 years ago
Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te
seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant \gamma =1.4

specific heat is given as =\frac{\gamma R}{\gamma -1}

gas constant =287 J⋅kg−1⋅K−1

Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}

specific heat at constant volume

Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}

change in internal energy = Cv(T_2 -T_1)

                            \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}

change in enthalapy \Delta H = Cp(T_2 -T_1)

                                 \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}

change in entropy

\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})

\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})

\Delta S = 382.79 J kg^{-1} K^{-1}

7 0
3 years ago
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