In which part of the scientific method do you sum up your results

While standing on a bridge 40.0 m above the ground, you drop a stone from rest. when the stone has fallen 3.80 m, you throw a se

Norma-Jean [14]

Given: distance 1 d₁ = 40 m; distance 2 d₂ = 3.8 m g = -9.8 m/s²

Initial Velocity Vi = 0 Final Velocity of stone 2 is unknown = ?

Total distance dₓ = d₁ - d₂ = 40 m - 3.8 m = 36.2 m

Formula: a = Vf² - Vi²/2d derive for Final Velocity Vf

acceleration is now due to gravity, therefore a = g

Vf = √2gd Vf = √2(9.8 m/s²)(36.2 m)

Vf = 26.64 m/s

Reason: The second stone will still start from rest.

3 0

4 years ago

A car travels 30.0 km to the south and then turns around and travels 20.0 km to the north in 1.5 hours time. What is the car's t

Gelneren [198K]

Explanation:

Total displacement =30+20=50 Km

Average velocity =displacement /time

V=50/1.5=33.33km/hr

7 0

3 years ago

Read 2 more answers

Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el

lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

$F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}$

The electrostatic force between them is

$F_{e}=\frac{Kq^{2}}{d^{2}}$

According to the question

1 % of Fg = Fe

$0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}$

$2.927 \times 10^{35}=9 \times10^{9}q^{2}$

$3.25 \times 10^{25}=q^{2}$

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0

3 years ago

Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) .

jek_recluse [69]

Explanation:

Given that,

Force, $F=((-8i)+6j)\ N$

Position of the particle, $r=(3i+4j)\ m$

(a) The toque on a particle about the origin is given by :

$\tau=F\times r$

$\tau=((-8i)+6j) \times (3i+4j)$

Taking the cross product of above two vectors, we get the value of torque as :

$\tau=(0+0-50k)\ N-m$

(b) Let $\theta$ is the angle between r and F. The angle between two vectors is given by :

$cos\theta=\dfrac{r.F}{|r|.|F|}$

$cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }$

$cos\theta=\dfrac{0}{50}$

$\theta=90^{\circ}$

6 0

3 years ago

A toroidal solenoid has 600 turns, cross-sectional area 6.90 cm2, and mean radius 4.30 cm.

Dahasolnce [82]

(a) The coil's self-inductance is 7.26 mH.

(b) The self-induced emf in the coil is 7.26 V

(c) The direction of the induced emf is from b to a.

<h3>Coil's self-inductance</h3>

L = N²μA/I

L = (600² x 4π x 10⁻⁷ x 6.9 x 10⁻⁴)/(0.043)

L = 7.26 x 10⁻³ H

L = 7.26 mH

<h3>Self-induced emf in the coil</h3>

emf = N(ΔBA)/t

where;

B is magnetic field
A is area
N is number of turns
t is time

B = μNI/L

B1 = (4π x 10⁻⁷ x 600 x 5)/0.043

B1 = 0.0876 T

B2 = (4π x 10⁻⁷ x 600 x 2)/0.043

B2 = 0.035 T

emf = NΔBA/t

emf = (600)(0.0876 - 0.035)(6.9 x 10⁻⁴) / (3 x 10⁻³)

emf = 7.26 V

The direction of the induced emf is always opposite to the direction of the applied current.

Thus, the direction of the induced emf is from b to a.

Learn more about induced emf here: brainly.com/question/13744192

#SPJ1

3 0

2 years ago