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3 years ago

9

What element has 5 valence electrons and is in period 3.

1 answer:

Maru [420]3 years ago

5 0

Answer:

Phosphorous

Explanation:

it is the fifth element from the left and it is in the third period so it has 5 valence electrons

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A helicopter carries a 1000-kg-car suspended from a rope below it. The helicopter flies horizontally at a constant speed of 25 m

kolbaska11 [484]

Answer:

(a) Tension, T = 28.653 kN

(b) Wind resistance force, $26.925\ kN$

Solution:

As per the question:

Mass of the car, m = 1000 kg

Speed of the helicopter, v = 25 m/s

Angle made by the rope, $theta = 20^{\circ}$

Now,

(a) To calculate the tension, T in the car:

Tension along the direction of motion, $T_{h} = Tcos20^{\circ}$

Tension along the vertical direction, $T_{v} = Tsin20^{\circ}$

Now, let the force due to the wind directed in the opposite direction of the motion be $F_{W}$ and it balances the horizontal component of the tension, T.

The vertical component is balance by the weight of the car, i.e., mg that acts vertically downwards.

Now,

$T_{v} = mg$

$Tsin20^{\circ} = 1000\times 9.8$

T = 28653 N = 28.653 kN

(b) The force of the wind resistance:

$F_{W} = T_{h}$

$F_{W} = 2cos20^{\circ} = 26925\ N = 26.925\ kN$

(c) Now,

If the angle made by the rope with the vertical is $0^{\circ}$ :

$mg = Tsin(90^{\circ} - 0^{\circ})$

$Tsin90^{\circ} = mg = 9800\ N$

The tension in the rope will be equal to the weight the car.

Wind resistance force, $F_{W} = Tcos90^{\circ} = 0\ N$

If the angle made by the rope with the vertical is $90^{\circ}$ :

$mg = Tsin(90^{\circ} - 90^{\circ})$

T = 0 N

Wind resistance force, $F_{W} = Tsin0^{\circ}$

$Tsin0^{\circ} = mg$

$F_{W} = \infty$

There will be no tension in the rope and wind resistance will be infinite.

3 0

3 years ago

A telescope can be used to enlarge the diameter of a laser beam and limit diffraction spreading. The laser beam is sent through

Novay_Z [31]

Answer:

Angular spread, $\theta=1.472\times 10^{-7}\ rad$

Explanation:

It is given that,

Wavelength of the light, $\lambda=613\ nm=613\times 10^{-9}\ m$

Diameter of the telescope, D = 5.08 m

The minimum angular spread is given by :

$\theta=\dfrac{1.22\lambda}{D}$

$\theta=\dfrac{1.22\times 613\times 10^{-9}}{5.08}$

$\theta=1.472\times 10^{-7}\ rad$

So, the minimum angular spread of the beam is $1.472\times 10^{-7}\ radian$ . Hence, this is the required solution.

5 0

3 years ago

Which sport requires the least amount of agility? HELLLLPPP

ZanzabumX [31]

Answer:

golf

Explanation:

it's less physical strength in more of you hitting the ball with whatever the stick is called

5 0

3 years ago

What are the characteristics and phases of the moon

svetoff [14.1K]

Crescent, gibbous, waxing, and waning.

4 0

3 years ago

A cat walks along a plank with mass M= 7.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di

solong [7]

Answer:

Before the plank will tip cat will walk 1.652 m

Explanation:

Mass of the cat along with plank $m_1=7kg$

Center of mass of the plank $d_1=0.850m$

Mass of cat $m_2=3.6kg$

We have to find how far right of sawhorse B.

Plank will tip when weight of the cat about B is greater than the torque by the weight of the plank.

Balancing the the torque

$m_1gd_1=m_1gd_2$

$7\times 9.8\times 0.850=3.6\times 9.8\times d_2$

$d_2=1.652m$

5 0

3 years ago