Suppose you ride your bicycle to the library traveling at 0.5 km/min. It takes you 25 minutes to get to 1.3

In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam of intensity 8

BlackZzzverrR [31]

Answer:

The correct answer is B

Explanation:

To calculate the acceleration we must use Newton's second law

F = m a

a = F / m

To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface

P = I / c absorbent surface

P = F / A

F / A = I / c

F = I A / c

The area of area of a circle is

A = π r²

We replace

F = I π r² / c

Let's calculate

F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸

F = 8.375 10⁻²³ N

Density is

ρ = m / V

m = ρ V

m = ρ (4/3 π r³)

m = 4500 (4/3 π (1 10⁻⁶)³)

m = 1,885 10⁻¹⁴ kg

Let's calculate the acceleration

a = 8.375 10⁻²³ / 1.885 10⁻¹⁴

a = 4.44 10⁻⁹ m/s² absorbent surface

The correct answer is B

4 0

3 years ago

Why is a protective apron or lab coat important to use when working with acids?

Kipish [7]

Answer:

Acids break down fabrics and can cause burns if the acids are strong.

Explanation:

A protective apron or lab coat is important when working with acids because acids break down fabrics and can cause burns if the acids are strong.

An acid is a substance that interacts with water to produce excess hydroxonium ions in an aqueous solution.
A strong acid ionizes completely in solution.
When they come in contact with a fabric, they break them down violently.
So, if they come in contact with the skin, it causes a violent break down of body tissues.
The apron acts a protective layer.

4 0

3 years ago

Read 2 more answers

If the time of impact in a collision is extended by 3 times, by hour much is

vodomira [7]

Answer:

The force of the impact would be smaller

Explanation: Examples:

If the force is big then the time would be small (2500N of Force = 10 seconds)

If the force is small then the time would be big (250N of Force = 50 seconds)

Impulse/Collision -> [Ft] = [M (vf-vo)] <- Change in momentum

4 0

3 years ago

A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o

masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, $F_{f2}$ = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = $F_{f2}$ × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

$T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27$

$Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92$