PHYSICS COLLEGE

An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's altitude is 532 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
(b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (a) Number Units (b) Number Units

Answers

Answer 1

Answer:

Answer:

$D = 1162.7 \ m$

$\beta =- 65.55^o$

Explanation:

From the question we are told that

The speed of the airplane is $u = 92.3 \ m/s$

The angle is $\theta = 51.1^o$

The altitude of the plane is $d = 532 \ m$

Generally the y-component of the airplanes velocity is

$u_y = v * sin (\theta )$

=> $u_y = 92.3 * sin ( 51.1 )$

=> $u_y = 71.83 \ m/s$

Generally the displacement traveled by the package in the vertical direction is

$d = (u_y)t + (1)/(2)(-g)t^2$

=> $-532 = 71.83 t + (1)/(2)(-9.8)t^2$

Here the negative sign for the distance show that the direction is along the negative y-axis

=> $4.9t^2 - 71.83t - 532 = 0$

Solving this using quadratic formula we obtain that

$t = 20.06 \ s$

Generally the x-component of the velocity is

$u_x = u * cos (\theta)$

=> $u_x = 92.3 * cos (51.1)$

=> $u_x = 57.96 \ m/s$

Generally the distance travel in the horizontal direction is

$D = u_x * t$

=> $D = 57.96 * 20.06$

=> $D = 1162.7 \ m$

Generally the angle of the velocity vector relative to the ground is mathematically represented as

$\beta = tan ^(-1)[(v_y)/(v_x ) ]$

Here $v_y$ is the final velocity of the package along the vertical axis and this is mathematically represented as

$v_y = u_y - gt$

=> $v_y = 71.83 - 9.8 * 20.06$

=> $v_y = -130.05 \ m/s$

and v_x is the final velocity of the package which is equivalent to the initial velocity $u_x$

$\beta = tan ^(-1)[-130.05}{57.96 } ]$

$\beta =- 65.55^o$

The negative direction show that it is moving towards the south east direction

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Answers

Answer:

1.2 s

Explanation:

Given:

v₀ = 8.0 m/s

v = -4.0 m/s

a = -10 m/s²

Find: t

v = at + v₀

(-4.0 m/s) = (-10 m/s²) t + (8.0 m/s)

t = 1.2 s

We can calculate the force that the atmospheric pressure produces on a surface. Consider a living room that has a 4.0m×5.0m floor and a ceiling 3.0m high. What is the total force on the floor due to the air above the surface if the air pressure is 1.00 atm?

Answers

Answer:

Force, $F=2.02* 10^6\ N$

Explanation:

It is given that,

Length of the room, l = 4 m

breadth of the room, b = 5 m

Height of the room, h = 3 m

Atmospheric pressure, $P=1\ atm=101325\ Pa$

We know that the force acting per unit area is called pressure exerted. Its formula is given by :

$P=(F)/(A)$

$F=P* l* b$

$F=101325* 4* 5$

$F=2.02* 10^6\ N$

So, the total force on the floor due to the air above the surface is $2.02* 10^6\ N$ . Hence, this is the required solution.

A plank 2.00 cm thick and 13.0 cm wide is firmly attached to the railing of a ship by clamps so that the rest of the board extends 2.00 m horizontally over the sea below. A man of mass 68.4 kg is forced to stand on the very end. If the end of the board drops by 5.20 cm because of the man's weight, find the shear modulus of the wood.

Answers

Answer:

9.93 MPa

Explanation:

Given:

- mass of the man = 68.4 kg

- Deflection dx = 5.2 cm

- thickness of plank t = 2.0 cm

- width of plank w = 13.0 cm

- Length subtended L = 2.0 m

Find:

Shear Modulus of Elasticity S :

S = shear stress / shear strain

Shear stress = F / A

Shear stress = 68.4*9.81 / 0.02*0.13

Shear stress = 258078.4615 Pa

Shear strain = dx / L

Shear Strain = 0.052 / 2

Shear Strain = 0.026

Hence,

S = 258078.4615 / 0.026

S = 9.93 MPa

1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?

Answers

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = $\sqrt{3^(2) +2^(2) }$ = $√(13)$ = 3.61cm = 0.036m

r₂ = $\sqrt{4^(2) + 3^(2) }$ = $√(25)$ = 5cm = 0.05m

electric potential V = $(kq)/(r)$

change in potential ΔV = $V_(1) - V_(2)$

ΔV = $(2kq_(1) )/(r_(1)) - (2kq_(2) )/(r_(2) )$ , where $q_(1) = q_(2)=$ 2.00μC

ΔV = $2kq((1)/(r_(1)) - (1)/(r_(2) ))$

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × $((1)/(0.036) - (1)/(0.05) )$

ΔV= 2.789×10⁵

$(1)/(2)mv^(2)$ = ΔV × q₃

$(1)/(2)$ ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

Read more about point charges at;brainly.com/question/13914561

A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational force on it is 600lb, what is the size of the force when it is at each of the following distances from the Earths surface?a. 20,000 miles
b. 30,000 miles
c. 100,000 miles

Answers

Grav Force = GMm/r squared. M is earth's mass. m is the probe. r is distance to centre of earth.Given that 600 = GMm/(10,000^2).GMm = 600x(10,000^2) for 10, 000 milesGMm = ? x (20,000^2) for 20, 000 milestherefore600x(10,000^2)=? x (20,000^2)therefore [600x(10,000^2)]/(20,000^2)=? [600x(1x1)]/(2x2)=? [600/4]=? ?=150lb.Same procedure for b and c.Please ask if you want further help.

gravitational force varies based on 1/r^2

when you're double the distance =10,000 to 20,000, the force is 4 times smaller so on and so forth.

As force is proportional to 1 / {distance squared}, the force will be 1 / 2^2 (i.e. 1/4) of the force at the reference distance (i.e. 1/4 * 600 = 150 lb)

Energy can be transferred from a closed system to the surroundings by: (A) Internal chemical reactions (B) Heat (C) Shaft work (D) Change in pressure without changing volume (E) Mass transfer

Answers

Answer:

option the correct is B

Explanation:

Let's analyze the different options, for a closed system

- an internal reaction changes the system, but does not affect the surrounding environment

- Heat, is a means of transfer that occurs when two bodies are in contact, one of the body can be a closed system since the only thing that happens is thermal transfer, without movement of the system itself. This is the correct result.

- Work implies a movement whereby the system must be mobile, it is not an option

- Pressure change. change in the system, but does not affect the environment

- Mass transfer is not possible in a closed system

After analyzing each option the correct one in B

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