What is the purpose of zigzag line on a periodic table ?

The top of the pool table is 0.810 m from the floor. the placement of the tape is such that 0 m is aligned with the edge of the

8090 [49]

Compute first for the vertical motion, the formula is:

y = gt²/2

0.810 m = (9.81 m/s²)(t)²/2

t = 0.4064 s

whereas the horizontal motion is computed by:

x = (vx)t

4.65 m = (vx)(0.4064 s)

4.65 m/ 0.4064s = (vx)

(vx) = 11.44 m / s
So look for the final vertical speed.

(vy) = gt

(vy) = (9.81 m/s²)(0.4064 s)

(vy) = 3.99 m/s

speed with which it hit the ground:

v = sqrt[(vx)² + (vy)²]

v = sqrt[(11.44 m/s)² + (3.99 m/s)²]

v = 12.12 m / s

6 0

3 years ago

Which step should you take before you begin a scientific investigation? Print the directions after you complete the experiment.

agasfer [191]

Answer:

Clarify any confusing information in the instructions.

Purchase the equipment and chemicals you will need.

Explanation:

Steps of a scientific investigation include identifying a research question or problem, forming a hypothesis, gathering evidence, analyzing evidence, deciding whether the evidence supports the hypothesis, drawing conclusions, and communicating the results.

8 0

2 years ago

Read 2 more answers

A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040

Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force ( $F_{B}$ ), as follows:

$F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB$ (1)

Where:

m: is the proton's mass = 1.67*10⁻²⁷ kg

v: is the proton's velocity

r: is the radius of the proton's orbit

q: is the proton charge = 1.6*10⁻¹⁹ C

B: is the magnetic field = 0.040 T

Solving equation (1) for r, we have:

$r = \frac{mv}{qB}$ (2)

By conservation of energy, we can find the velocity of the proton:

$K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V$ (3)

Where:

K: is kinetic energy

U: is electrostatic potential energy

ΔV: is the potential difference = 1.0 kV

Solving equation (3) for v, we have:

$v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s$

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

$r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m$

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!

5 0

2 years ago

What happens to the spring when we sit on a sofa

irinina [24]

Answer:

Explanation:

bites your as crack

5 0

2 years ago

A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d

andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x² - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0

3 years ago