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melamori03 [73]
4 years ago
13

A ramp is needed to allow vehicles to climb a 2 foot wall. The angle of elevation in order for the vehicles to safely go up must

be 30 o or less, and the longest ramp available is 5 feet long. Can this ramp be used safely
Physics
1 answer:
stepan [7]4 years ago
5 0

Answer:

Yes the ramp can be safely used

Explanation:

Here, we have

Length of longest ramp = 5 ft

Height of wall = 2 ft

Therefore, the sine of the angle adjacent to the ramp which is equal to the angle of elevation is given by;

Sin\theta = \frac{Opposite \, side \, to\,  angle}{Hypothenus\, side \, of\,  triangle}

Where:

The opposite side to angle = 2 ft wall and

Hypotenuse side = Ramp = 5 ft

Therefore,

Sin\theta = \frac{2}{5} = 0.4 and θ = sin⁻¹0.4 = 23.55 °

The ramp can be safely used as the angle it is adjacent to is less than the specified 30°.

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A ray of red light in air is incident at an angle of 30. on a
Vlad [161]

Answer:

20 degrees.

Explanation:

From Snell’s law of refraction:

sinθ1•n1 = sinθ2•n2

where θ1 is the incidence angle, θ2 is the refraction angle, n1 is the refraction index of light in medium1, and n2 is the refraction index for virgin olive oil. The incidence angle of the red light is θ1 = 30 degrees.

The red light is in air as medium1, so n1 (air) = 1.00029

So, to find θ2, the refracted angle:

sinθ1•1.00029 = sinθ2•1.464

sin(30)•1.00029 / 1.464 = sinθ2

0.5•1.00029 / 1.464 = sinθ2

sinθ2 = 0.3416291

θ2 = arcsin(0.3416291)

θ2 = 19.976 degrees

To the nearest degree,

θ2 = 20 degrees.

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3 years ago
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SashulF [63]
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7 0
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An object, initially at rest, moves 250 m in 17 s. What is its acceleration?
mash [69]

Answer:

1.73 m/s²

Explanation:

Given:

Δx = 250 m

v₀ = 0 m/s

t = 17 s

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5 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.

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_____________ forces are always attractive and the mass of an object determines how strong the ____________ pull is.
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Gravitational, gravitational ! both the option are same
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