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Dimas [21]
4 years ago
5

Falling objects drop with an average acceleration of 9.8 m/s2. An arrow is shot with a velocity of 11.76 m/s straight down from

the top of a tall castle hitting an armored knight with a downward velocity of 49 m/s. How long was the arrow in flight?
Physics
1 answer:
In-s [12.5K]4 years ago
6 0

Answer:

3.8 secs

Explanation:

Parameters given:

Acceleration due to gravity, g = 9.8 m/s^2

Initial velocity, u = 11.76 m/s

Final velocity, v = 49 m/s

Using one of Newton's equations of linear motion, we have that:

v = u + gt

where t = time of flight of arrow

The sign is positive because the arrow is moving downward, in the same direction as gravitational force.

Therefore:

49 = 11.76 + 9.8*t\\\\\\\49 - 11.76 = 9.8t\\\\=> 9.8t = 37.24\\\\\\t = \frac{37.24}{9.8} \\\\\\t = 3.8 secs

The arrow was in flight for 3.8 secs

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A step-up transformer has 50 turns on its primary coil and 2000 turns on its secondary
ryzh [129]

Answer:

Vs= 6000 v

Ip=9.6 A

Explanation:

Ns/Np = Vs/Vp

2000/50 = Vs/150

40 x 150 = Vs

Vs= 6000 v

Vs/Vp=Ip/Is

6000/150=Ip /0.24

Ip=9.6 A

7 0
3 years ago
A portable basketball set has a base and a post arrangement. The post arrangement consists of a post, backboard, hoop and net. T
Ray Of Light [21]

The rotational equilibrium condition allows finding the response to the minimum force of the wind and what happens when changing the water for sand, in the system

  a) The minimum force of the wind that turns the system is Fw = 17.64 N

  b) The system resists much greater forces because the base has more mass

 Newton's Second Law can be applied to rotational motion in this case when the angular acceleration is zero we have the special case of rotational equilibrium

               Σ τ = 0

Where τ is the torque  

The reference system is a coordinate system with respect to which the torques are measured, in this case we will fix the system at the turning point, the junction of the base and the pole, we will assume that the counterclockwise rotations are positive.

For the torque the distance used is the perpendicular distance from the direction of the force to the axis of rotation, let's find this distance for each force

Wind force

         cos 15 = \frac{y_w}{2.35}

         y_w = 2.35 cos 15

Post Weight

        sin 15 = \frac{x_p}{2.00}

         xp = 2.0 sin 15

Base weight

         cos (90-15) = \frac{x_b}{0.25}

         xB = 0.25 cos 75

Let's substitute in the rotational equilibrium equation

     

          F_w \ y_w  + W_p \ x_p - W_b \ x_b = 0

a) To calculate the minimum wind force we substitute the given values

They indicate the weight of the post is W_p = 26.0 N and the weight of the base with water is W_b = 810 N

     F_w = \frac{W_b \ x_b - W_p \ x_p }{y_w}

     F_w = \frac{W_b \ 0.25 cos75 \ - W_p \ 2 sin 15}{2.35 cos 15}

       

Let's  calculate

     F_w = \frac{810 \ 0.25 \ cos75 \ - 26.0 \ 2 \ sin 15}{2.35 cos15}\\F_w = \frac{52.41 - 10.30}{2.3699}

     F_w = 17.64 N

b) The water is exchanged for sand.

In this case, as the density of the sand is greater than that of the water, the base will have more weight, so it will resist stronger winds before turning over.

Using the rotational equilibrium condition we can find the response to the minimum force of the wind and what happens when changing the water for sand,

  a) the minimum force of the wind that turns the system is Fw = 17.64 N

  b) the system resists much greater forces because the base has more mass

Learn more  here: brainly.com/question/7031958

3 0
3 years ago
How does atmosphere behave as a system
boyakko [2]
I hope this helps! I added an example but there are many more!

5 0
4 years ago
What does this mean??
Fudgin [204]
I need help on this too lol
7 0
3 years ago
Velocity is a: a) Fundamental quantity b) MKS system c) Standard system d) Derived quantity​
Naya [18.7K]

Velocity is a derived quantity, which is derived from two fundamental quantities ilength ( displacement ) and time

it can be depicted as :

  • \boxed{ \boxed{velocity =  \dfrac{displacement}{time} }}

4 0
3 years ago
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