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3 years ago

A pumpkin is dropped after 5 seconds its velocity is 47m/s what is its acceleration

Physics

2 answers:

ANEK [815]3 years ago

8 0

9.81m/s Hope i helped

Vesna [10]3 years ago

6 0

Answer:

Acceleration, $a=9.4\ m/s^2$

Explanation:

It is given that,

Initial speed of the pumpkin, u = 0

Final speed of the pumpkin, v = 47 m/s

Time taken, t = 5 seconds

Acceleration of an object is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{47\ m/s}{5\ s}$

$a=9.4\ m/s^2$

So, the acceleration of the pumpkin is $9.4\ m/s^2$ . Hence, this is the required solution.

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Which of the following occurs when the fight-or-flight response is triggered?

kiruha [24]

Answer:

Explanation:

The autonomic nervous system has two components, the sympathetic nervous system and the parasympathetic nervous system. The sympathetic nervous system functions like a gas pedal in a car. It triggers the fight-or-flight response, providing the body with a burst of energy so that it can respond to perceived dangers.

4 0

3 years ago

If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?

ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: $(m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})$

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

$(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})$

$\frac{8.6}{2.5} = \log_{10}(b_{1}/b_{2})$

$\log_{10}(b_{1}/b_{2}) = 3.44$

Thus,

$(b_{1}/b_{2}) = 2754.22$

It means star A is 2754.22 time brighter than Star B.

3 0

3 years ago

Why is the same side of the Moon always visible from Earth?

son4ous [18]

Answer:

Explanation:

Answer

The true fact is that C is what happens in outer space. Both rotations take 27.3 days.

A: The exact opposite is true. It does rotate about it's axis.

B: Again this is just plain false. Given the way we observe it, the moon must be rotating around the earth.

D. they don't. 27.3 hours and 24 hours are not the same.

8 0

2 years ago

A 0.350kg bead slides on a curved fritionless wire,

LuckyWell [14K]

Answer:

h2 = 0.092m

Explanation:

From a balance of energy from point A to point B, we get speed before the collision:

$m1*g*h-\frac{m1*V_B^2}{2}=0$ Solving for Vb:

$V_B=\sqrt{2gh}=6.56658m/s$

Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:

$V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s$

Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:

$m1*g*h2-\frac{m1*V_{B'}^2}{2}=0$ Solving for h2:

h2 = 0.092m

6 0

3 years ago

A child walks due east on the deck of a ship at 2 miles per hour. the ship is moving north at a speed of 18 miles per hour. find

garik1379 [7]

Refer to the figure shown below.

The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.

The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph

The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.

Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.

8 0

3 years ago