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3 years ago

34000×0.4 ---------------- 0.02×200 In scientific notation

Physics

1 answer:

VashaNatasha [74]3 years ago

4 0

Answer:

the simplified expression is written as 3.4 x 10³

Explanation:

Given expression;

$\frac{34000\times 0.4}{0.02 \times 200}$

in scientific notation, the expression is simplified as;

$\frac{34000\times 0.4}{0.02 \times 200} = \frac{13600}{4} = 3400 = 3.4 \times 10^3$

Therefore, in scientific notation, the simplified expression is written as 3.4 x 10³

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A pendulum is in a spacecraft to measure

Tcecarenko [31]

Answer:

For small angles (✓ < 30 degrees), the period of a simple pendulum4can be approximated by: 3A radian is an angle measure based upon the circumference of a circle C =2⇡r where r is the radius of the circle. A complete circle (360 ) is said to have 2⇡ radians. Therefore, a 1/4 circle (90 )is⇡/2 radians.

Explanation:

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3 years ago

A hydrogen bond forms by the electrostatic interaction of opposite charges in two molecules. If

Darya [45]

The force between the molecules involved in the bond is 6. 426 *10^-11 Newton

<h3>How to determine the force</h3>

Using the formula:

F = K[q1 x q2]/D^2

where K is coulombs constant =9 *10 ^9 Nm^2/C^2.

q1 and q2 = charges = 1.60x10 -20C

d = distance between the charges = 2x10 -10 m

Substitute the values into the formula

F = $9 * 10^9\frac{ 1.60*10^ -20 * 1.60 *10^ -20}{2x10^ -10^{2} }$

F = $9 *10^9\frac{2. 856* 10^-40}{4* 10^-20}$

F = $9* 10^9 * 7. 14* 10^-21$

F = $6. 426 * 10^-11$ Newton

Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton

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brainly.com/question/8424563

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5 0

2 years ago

A concrete piling of 50 kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch?

pentagon [3]

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, $Y=20\times 10^{10}\ N/m^2$

The young modulus of a wire is given by :

$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$

$Y=\dfrac{F.L}{A\Delta L}$

$\Delta L=\dfrac{F.L}{A.Y}$

$\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}$

$\Delta L=0.034\ m$

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

8 0

4 years ago

Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot

Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² = 2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ = m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁ = 4.427 - 2v₂ ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂ = 8.854

v₂ = 8.854 / 3

v₂ = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁ = - 1.477 m/s

v₁ = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

$mgh_{max} = \frac{1}{2}mv_{max}^2$

(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max} = 0.11 \ m$

(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max} = 0.44 \ m$