Ask question

Ask question

All categories

3 years ago

10

A 29 foot ladder leans against a building so that the angle between the ground and the ladder is 75 ∘ . How high does the ladder

reach up the side of the building? Round to 2 decimal places.

1 answer:

Aloiza [94]3 years ago

3 0

Answer:

28.01m

Explanation:

Opp/Hyp = Sin

Sin 75 = x/29

x = 29 sin 75

x = 28.01m

You might be interested in

Explain why ventilation is very important if there is risk of exposure to random gas in your home school

bonufazy [111]

Ventilation is very important because it helps remove the gas form people’s homes and schools and it redirects the random gas outside so it is less likely to hurt people

3 0

3 years ago

If a cart is accelerating downhill under a net force of 25 N, what additional force would cause the cart to have a constant velo

Vladimir [108]

No additional force is required because it's already going downhill

7 0

3 years ago

(b) The distance of mass from mass A if there is no gravitational force acted on C

shepuryov [24]

Answer:

(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons

(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

Explanation:

The given parameters are;

The mass of particle, A, m₁ = 2 kg

The mass of particle, B, m₂ = 0.3 kg

The mass of particle, C, m₃ = 0.05 kg

The distance between particle 'A' and particle 'B', r₁ = 0.15 m

The distance between particle 'B' and particle 'C', r₂ = 0.05 m

(a) The gravitational force, 'F', is given as follows;

$F =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}$

Where;

F = The force between the two masses

G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m₁ = The mass of object 1

m₂ = The mass of object 2

If 'C' is placed at 0.05 m from 'B', we have;

F₂₃ = 6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰

The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)

F₁₃ = 6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)

The force, 'F', acting on object 'C' = F₁₃ - F₂₃

F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N

The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N

(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x

We have;

F₂₃ = F₁₂

$F_{23} =G \times \dfrac{m_{1} \times m_{2}}{r_1^{2}} = F_{13} =G \times \dfrac{m_{1} \times m_{3}}{r_2^{2}}$

By plugging in the values and removing like terms, we get;

$\dfrac{0.3 \times 0.05}{(1.15 - x)^{2}} = \dfrac{2 \times 0.05}{x^2}$

(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²

0.1·x² - 0.23·x + 1.3225 = 0.015·x²

0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

0.085·x² - 0.23·x + 0.13225= 0

x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))

x ≈ 0.829, or x ≈ 1.877

Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'

7 0

2 years ago

Give examples of not useful high friction

Tatiana [17]

Friction can be bad by being too strong or too weak.

<span>Sometimes, when it is too strong, it decreases efficiency since some energy is wasted and turns to heat. Friction can also d</span><span>amage equipment/objects like when you slide it on the floor.
</span>
When friction is too weak, like for instance when there is black ice- our center of gravity is displaced too quickly and we can fall. Likewise, if there is a lot of slush on the ground, cars can slip and slide.

6 0

3 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

3 years ago