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Radda [10]
3 years ago
5

A shell if fired from the ground with an initial velocity of 1,700 m/s at an initial angle of 55 degrees to the horizontal. Negl

ecting air resistance, what is the shell's horizontal range?
Physics
1 answer:
Triss [41]3 years ago
8 0

Answer:

Therefore the horizontal range = 294897.96 m.

Explanation:

Range of a projectile: The range is defined as the horizontal distance from the point of projection to the point where the projectile hit the projection plane again. The S.I unit of range is Meter (m).

It can be expressed mathematically as

R = u²sin2∅/g............................. Equation 1

Where R = Horizontal range, ∅ = angle of projection, u = initial velocity, g = acceleration due to gravity.

<em>Given: u = 1700 m/s, </em>∅ = 55°,

Constant: g = 9.8 m/s²

Substituting these values into equation 1

R = (1700²sin55)/9.8

R = 2890000/9.8

R = 294897.96 m.

Therefore the horizontal range = 294897.96 m.

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An electron in an atom's orbital shell, labeled X in the model below, released enough energy to move to a different orbital shel
Delicious77 [7]

Answer:

Lower energy shell which will be nearer to the nucleus.

Explanation:

When electron move from one energy level to another, an electron must gain or lose just the right amount of energy.

When atoms releases energy, electrons move into lower energy levels.  The electrons in the shells aways from the nucleus have more energy as compared to the electrons in the nearer shells.

Electrons with the lowest energy are found closest to the nucleus, where the attractive force of the positively charged nucleus is the greatest. Electrons that have higher energy are found further away

7 0
3 years ago
ASAPP PLS HELP MEE
Anastasy [175]

Answer:

B) 2.7 g of aluminium has a volume of 1 cm^3

Explanation:

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

Density = \frac{mass}{volume}

If the density of aluminum is 2.7 g/cm³, it simply means that 2.7 g of aluminium has a volume of 1 cm³

Check:

Given the following data;

Mass = 2.7 grams

Volume = 1 cm³

Substituting into the formula, we have;

Density = \frac{2.7}{1}

Density = 2.7 g/cm³

7 0
3 years ago
One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a s
DedPeter [7]

Complete Question

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 2 cm across, and you estimate that the distance from the window shade to the wall is about 5 m.

Required:

Estimate the diameter of the pinhole.  

Answer:

The diameter is  d =0.000336 m

Explanation:

     From the question we are told that

            The central maxima is D= 2cm = \frac{2}{100} = 0.02m

            The distance from the window shade is L = 5m

     The  average wavelength of the  sun is mathematically evaluated as

                         \lambda_{ave } = \frac{\lambda_i  + \lambda_f}{2}

 Generally the visible light spectrum  has a wavelength  range  between  400 nm  to 700 nm  

        So  the initial wavelength of the sun is \lambda _i = 400nm

           and the final wavelength is  \lambda_f = 700nm

  Substituting this into the above equation

                 \lambda_{sun} = \frac{400nm  +700nm}{2}

                        = 550nm

The diameter is evaluated as

              d = \frac{2.44 \lambda_{sun} L}{D}

substituting values

              d = \frac{2.44 * 550*10^{-9} * 5 }{0.02}

                d =0.000336 m

5 0
3 years ago
What is necessary to designate a position? A. a reference point B. a direction C. fundamental units D. motion E. all of these
max2010maxim [7]

Answer:

E. all of these

Explanation:

The designation of a point in space all the points that necessary

- reference point

- a direction

- fundamental units

- a direction

- motion

all are necessary to designate a point in space. Hence option E is correct.

For example in simple harmonic motion we need to specify all the above factors of the object in order to designate the position of the object.  

8 0
3 years ago
P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar
creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

3 0
3 years ago
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