Looking at the periodic table which element would you think is the best conductor of electricity?

In the 2016 Olympics in Rio, after the 50 m freestyle competition, a problem with the pool was found. In lane 1 there was a gent

Dmitry_Shevchenko [17]

Answer:

Explanation:

A )

speed of swimming in still water is given by the expression

distance / time

= 50 / 25

= 2 m /s

In lane 1 , 1.2 cm/s current is flowing in the direction that the swimmers are going so swimmers will cover distance at the rate of 2 + 1.2 = 3.2 m /s.

time to cover distance of 50 m in lane 1

= distance / speed

= 50 / 3.2 = 15.625 s

In lane 8 , 1.2 cm/s current is flowing against the direction that the swimmers are going so swimmers will cover distance at the rate of 2 - 1.2 = .8 m /s.

time to cover distance of 50 m in lane 1

= distance / speed

= 50 / .8 = 62.5 s

8 0

3 years ago

During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal

weeeeeb [17]

The kinematics of the uniform motion and the addition of vectors allow finding the results are:

The Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º

The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

Vectors are decomposed into a coordinate system

The components are added

The resulting vector is constructed

Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

v = $\frac{\Delta d}{t}$

Δd = v t

Where v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

t₁ = 2 min = 120 s

Δd₁ = v₁ t₁

Δd₁ = 600 120

Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s

time t₂ = 1 min = 60 s

Δd₂ = v₂ t₂

Δd₂ = 400 60

Δd₂ = 24 103 m

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

cos (180 -35) = $\frac{x_1}{\Delta d_1}$

sin 145 = $\frac{y_1}{\Delta d1}$

x₁ = Δd₁ cos 125

y₁ = Δd₁ sin 125

x₁ = 72 103 are 145 = -58.98 103 m

y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

cos (90+ 30) = $\frac{x_2}{\Delta d_2}$

sin (120) = $\frac{y_2}{\Delta d_2}$

x₂ = Δd₂ cos 120

y₂ = Δd₂ sin 120

x₂ = 24 103 cos 120 = -12 10³ m

y₂ = 24 103 sin 120 = 20,78 10³ m

The component of the resultant vector are

Rₓ = x₁ + x₂

R_y = y₁ + y₂

Rx = - (58.98 + 12) 10³ = -70.98 10³ m

Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

R = $\sqrt{R_x^2 +R_y^2}$

R = $\sqrt{70.98^2 + 62.08^2}$ 10³

R = 94.30 10³ m

We use trigonometry for the angle

tan θ ’= $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{62.08}{70.98}$

θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

θ = 180 - θ'

θ = 180 -41.2

θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

v = $\frac{\Delta R}{t_3}$

v = $\frac{94.30}{120} \ 10^3$

v = 785.9 m / s

in the opposite direction, that is, the angle must be

41.2º to the South of the East

In conclusion, using the kinematics of the uniform motion and the addition of vectors, results are:

To find the initial Barry trajectory is 94.30 10³ m with n angles of 138.8º

The return trajectory and speed is v = 785.9 m / s, with an angle of 41.2º to the South of the East

Learn more here: brainly.com/question/15074838

4 0

2 years ago

A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each

irinina [24]

Answer:

1.24 m/s

Explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

$\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s$

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

$\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s$

So the flow speed of each of the narrower pipe is:

$v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}$

$v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s$

8 0

3 years ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$