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ladessa [460]
3 years ago
14

On the periodic table , the vertical columns that extend down the periodic table are called ?

Physics
1 answer:
Darina [25.2K]3 years ago
3 0

Answer:

groups

Explanation:

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A jet plane lands with a speed of 100 m/s and can
kiruha [24]

Answer:

a) t = 20 [s]

b) Can't land

Explanation:

To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.

a)

v_{f}=v_{i}-(a*t)

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = desacceleration = 5 [m/s^2]

t = time [s]

Note: the negative sign of the equation means that the aircraft slows down as it stops.

0 = 100 - 5*t

5*t = 100

t = 20 [s]

b)

Now we can find the distance using the following kinematics equation.

x -x_{o}=(v_{o}*t)+\frac{1}{2}*a*t^{2}

x - xo = distance [m]

x -xo = (0*20) + (0.5*5*20^2)

x - xo =  1000 [m]

1000 [m] = 1 [km]

And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land

4 0
4 years ago
Hello guys! Can u please help me with physics. I translated it in English. Can yall help me please how much u can!!
DedPeter [7]

1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula

v² - u² = 2a ∆x

where

u = initial speed

v = final speed

a = acceleration

∆x = displacement

At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

0² - (20 m/s)² = 2 (-9.8 m/s²) y

Solve for y :

y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m

2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.

3. At any time t ≥ 0, the body's vertical velocity is given by

v = 20 m/s - gt

At the highest point, we have

0 = 20 m/s - (9.8 m/s²) t

and solving for t gives

t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

4. The body's height y above the ground at any time t ≥ 0 is given by

y = 60 m + (20 m/s) t - 1/2 gt²

Solve for t when y = 0 :

0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²

Using the quadratic formula,

t = (-b + √(b² - 4ac)) / (2a)

(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with

t ≈ 6.09 s

5. At the time found in (4), the body's velocity is

v = 20 m/s - g (6.09 s) ≈ -39.7 m/s

Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.

6 0
3 years ago
Two parallel copper rods supply power to a high-energy experiment, carrying the same current in opposite directions. The rods ar
mart [117]

Answer:

the maximum allowable current is 7302.967  amperl

Explanation:

The computation of the maximum allowable current is shown below;

Force F = mean ÷ 4π 2 I_1 I_2 ÷d  × ΔL

200 N = (10)^-7 (2I × I) ÷ 0.08 × 1.5

200 = 3.75 × 10^-6 I^2

I = √200 ÷ √ 3.75 × 10^-6

= 7302.967  amperl

Hence, the maximum allowable current is 7302.967  amperl

Basically we applied the above formula

6 0
3 years ago
In an experiment replicating Millikan’s oil drop experiment, a pair of parallel plates are placed 0.0200 m apart and the top pla
lianna [129]

Answer: See below

Explanation:

<u>Given:</u>

The potential between plates, V = 240 V

Distance between plates, d = 0.02 m

The mass of drop, m = 2x10^-11

Charge on electron, e = 1.6x10^-19

Part (a)

The free-body diagram is attached below

Part (b)

The electric field is given by,

E=\frac{V}{d}

On applying force balance, the force on oil drop is equal to the weight of the oil,

$$\begin{aligned}F_{E} &=m g \\q E &=m g \\q \frac{V}{d} &=m g \\q &=\frac{m g d}{V}\end{aligned}$$

Substituting the given values in the above equation,

\begin{aligned}&q=\frac{2 \times 10^{-11} \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} \times \frac{1 \mathrm{~N}}{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}} \times 0.02 \mathrm{~m}}{240 \mathrm{~V} \times \frac{1 \mathrm{~N} \cdot \mathrm{m} / \mathrm{C}}{1 \mathrm{~V}}} \\&q=1.63 \times 10^{-14} \mathrm{C}\end{aligned}

Therefore, the charge on the oil drop is 1.63x10^-14 C

Part (c)

There will be an excess of electrons on the oil drop.

The number of electrons on oil drop can be calculated as,

\begin{aligned}q &=n e \\1.63 \times 10^{-14} \mathrm{C} &=n \times 1.6 \times 10^{-19} \mathrm{C} \\n &=1.01 \times 10^{5}\end{aligned}

Therefore, the number of excess electrons is 1.01x10^5

3 0
2 years ago
Which three characteristics do mechanical waves and electrocmagnetic
almond37 [142]
B the waves transfer energy from its source
8 0
3 years ago
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