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Karo-lina-s [1.5K]
2 years ago
14

a body of mass 4 kg moves around a circle of radius 6 m with a constant speed of 12m/sec². calculate tw angular velocity and the

force towards the centre​
Physics
1 answer:
NeTakaya2 years ago
3 0

Answer:

• Angular velocity is 2 m/s

• Centripetal force is 96 N

Explanation:

For angular velocity:

{ \rm{v =  \omega r}}

  • v is velocity
  • w is angular velocity
  • r is radius

{ \rm{ \omega =  \frac{v}{r} }} \\  \\ { \rm{ \omega =  \frac{12}{6} }} \\  \\  { \boxed{ \rm{ \omega = 2 \:  {ms}^{ - 1} }}}

For the centripetal force:

{ \rm{force = m { \omega}^{2}r }} \\  \\ { \rm{force = (4 \times  {2}^{2}  \times 6)}} \\  \\ { \rm{force = 96 \: newtons}}

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Your heart pumps blood at a pressure of 100 mmHg and flow speed of 60 cm/s. At your brain, the blood enters capillaries with suc
Savatey [412]

Answer:

1.28 m

Explanation:

Generally, pressure of fluid is given by

P=\rho g h where g is acceleration due to gravity, h is the height and \rho is the density

Considering that the pressure for mercury is same as for blood only that the height and density of fluid are different then

\rho_b g h_b= \rho_m g h_m

Since g is constant, then

\rho_b h_b= \rho_m h_m

Making h_b the subject of the formula then

h_b=\frac {\rho_m h_m}{\rho_b}

Where subscripts m and b denote mercury and blood respectively

Assuming density of blood is 1060 Kg/m3, density of mercury as 13600 Kg/m3 and substituting height of mercury for 0.1 m then

h_b=\frac {13600*0.1}{1060}=1.283018868  m \approx 1.28 m

7 0
3 years ago
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
liq [111]

Answer:

A) total time = 55.5 seconds

B) average velocity = 25.27 m/s

Explanation:

It starts from rest, so initial velocity, u = 0 m/s

We are given;

acceleration; a = 2 m/s²

Final velocity; v = 31 m/s

From Newton's first law of motion,

v = u + at

So, 31 = 0 + 2t

t = 31/2

t = 15.5 sec

We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;

(a) Total time in which car is in motion = 15.5 + 35 +5 = 55.5 seconds

b)Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;

S = ut + ½at²

S1 = 0 + ½(2 * 15.5²)

S1 = 240.25 m

Distance traveled in 35 sec with with velocity of 31 m/sec is;

S2 = velocity x time

S2 = 35 × 31 = 1085 m

Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity(u) will be 31 m/s.

From the first equation of motion,

a = (v - u)/t

a = (0 - 31)/5

a = -6.2 m/s²

So, distance travelled is;

S3 = ut + ½at²

S3 = (31 × 5) + ½(-6.2 × 5²)

S3 = 155 - 77.5

S3 = 77.5 m

So overall total distance = S1 + S2 + S3

Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m

Average velocity = total distance/total time

Average velocity = 1402.75/55.5 = 25.27 m/s

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Can't say anything 'bout it.....

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During an experiment, Ellie records a measurement of 0.0034 m. How would
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Answer:

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