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REY [17]
2 years ago
14

A hummingbird flies in circles around a bush. if the velocity of the hummingbird is 720 cm/s and the radius of the circular path

is 9.0 cm, what is the centripetal acceleration of the hummingbird?
Physics
2 answers:
Rashid [163]2 years ago
5 0
57,600 cm/s ^2 ...........
denis-greek [22]2 years ago
4 0

Answer:

57600cm/s²

Explanation:

centripetal accleration best describes the motion of the humming bird. it is the motion inwards towards the centre of a circle.

centripetal acceleration (a_{c}) = \frac{v^{2} }{r}

where v= velocity = 720cm/s

r=9cm

r= radius

a_{c}= 720² ÷ 9

a_{c}= \frac{720 * 720}{9}= 57600cm/s²

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A volcano erupts and launches a chunk of hot magma horizontally with a speed of 252 m/s. The magma travels a horizontal distance
ArbitrLikvidat [17]

Answer:

The value is v_y  =  -48.61 \ m/s

Explanation:

From the question we are told that

   The horizontal speed is  u_x  = 252 \  m/s

    The horizontal distance is  d = 1250 \ m

Generally the time taken by the hot magma in air before landing is mathematically represented as

       t = \frac{d}{u_x}

=>    t = \frac{ 1250 }{252}

=>    t = 4.96 \  s

Generally the initial vertical velocity of the magma when it was lunched is  

    u_y = 0 \ m/ s

Then the final velocity of the magma when it hits the ground is mathematically represented s

       - v_y  =  u_y + gt

Here the negative sign mean that the direction of the velocity is towards the negative y -axis

So  

        - v_y  =  48.61 \ m/s

=>     v_y  =  -48.61 \ m/s

7 0
2 years ago
Which state of matter takes both the shape and volume of its container?
Tresset [83]

The <em>gaseous state</em> of matter does that.  A gas expands to take the shape and volume of whatever you put it into.

6 0
3 years ago
car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw
sergejj [24]

Answer:

The force of car 3 on car 2 ≈ 1810.82 N

Explanation:

The equation for the change in momentum of the two cars are;

Conservation of linear momentum

150( 2.2 - v) = 265(1.5-v)

150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = m \times  v_f - m \times  v_0

Where;

v_f = The final velocity

v_0 = The initial velocity

For the the 265 kg mass, we have;

v_f = 0.1627 m/s

v_0 = 1.5 m/s

Which gives the impulse a s F×Δt =  265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

Whereby the distance moved in one second is 0.1627 m, we have;

Work done = Force × Distance = Force × 0.1627 = 294.62

Force = 294.62/0.1627 = 1810.82 N.

8 0
3 years ago
The format specifier ________ is a placeholder for an int value. %d %n %int %s
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