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pychu
3 months ago
14

At a football game, an air gun fires T-shirts into the crowd. The gun is fired at an angle of 46 degree from the horizontal with

an initial speed of 27 m/s. A fan who is sitting 60 m horizontally from the gun, but high in the stands, catches a T-shirt. A) How long does it take for the T-shirt to reach the fan? B) At what height h is the fan from the ground?
Physics
1 answer:
Keith_Richards [3.2K]3 months ago
8 0

Response:

a) t=3.199 seconds

b) h = 11.97 m

Clarification:

As this situation pertains to projectile motion,

a) we know that

Vcos\theta=\frac{R}{t}

Where,

V = initial velocity

Θ = angle with respect to the horizontal

R = horizontal distance

t = duration to traverse the distance 'R'

Given values:

V = 27m/s

R = 60m

Θ = 46°

Thus,

the formula transforms to

27\times cos46^o=\frac{60}{t}

or

t=\frac{60}{27\times cos46^o}

t=3.199 seconds

b) The equation for height can be expressed as:

h = Vsin\theta \times t-\frac{1}{2}\times gt^2\\

where,

g = acceleration due to gravity = 9.8m/s²

Substituting the values into the above equation results in

h = 27\times sin46^o\times 3.199-\frac{1}{2}\times 9.8\times 3.199^2\\

or

h = 62.124-50.14

or

h = 11.97 m

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An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground
Ostrovityanka [3204]

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = the position of the ladder's center of mass = (0.5) L = (0.5) 8 = 4 m

AC = distance of the climber from the bottom of the ladder = x

W = weight of the ladder = 240 N

F_{g} = weight of the climber = 710 N

F = force exerted by the wall on the ladder

N = normal force acting on the ladder from the ground =?

By applying force equilibrium in the vertical direction

N = F_{g} + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f = static friction force on the ladder

Static friction force can be expressed as

f = μ N

f = (0.55) (950)

f = 522.5 N

The equation for force along the horizontal axis reads

F = f

F = 522.5 N

using torque equilibrium around point A

F Sin50 (AD) = W Cos50 (AB) + (F_{g} Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

x = 5.7 m

7 0
2 months ago
A book rests on the shelf of a bookcase. The reaction force to the force of gravity acting on the book is 1. The force of the sh
serg [3582]

Answer:

1. The force applied by the shelf supporting the book.

Explanation:

The free body diagram for the book is represented as follows:

1 - The weight of the book acting downward

2 - The normal force exerted by the shelf upward on the book.

As the book remains stationary, these two forces balance each other, and in accordance with Newton's Third Law, the reactive force equivalent to gravity is opposite and equal to the weight of the book. This reaction force prevents the book from falling off the shelf.

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3 months ago
Read 2 more answers
Compare the time period of two simple pendulums of length 4m and 16m at a place.
Ostrovityanka [3204]

Answer:

The period of the pendulum measuring 16 m is double that of the 4 m pendulum.

Explanation:

Recall that the period (T) of a pendulum with length (L) is defined by:

T=2\,\pi\,\sqrt{ \frac{L}{g} }

where "g" denotes the local gravitational acceleration.

Since both pendulums are positioned at the same location, the value of "g" will be consistent for both, and when we compare the periods, we find:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

Thus, the duration of the 16 m pendulum is two times that of the 4 m one.

5 0
3 months ago
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
kicyunya [3294]

Response:

Clarification:

The force between two charges, q₁ and q₂ at a distance d is represented by the formula

F = k q₁ q₂ / d²

Here, the force between charge q₁ = -15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C with distance d = (1.66 - 1.24) = 0.42 mm

k = 1 / (4π x 8.85 x 10⁻¹²)

Substituting the values into the equation

F = 1 / (4π x 8.85 x 10⁻¹²) x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 9 x 10⁹ x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 35969.4 x 10⁻³ N.

For the force between charge q₂ = 34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C at a distance d = (1.24 - 0) = 1.24 mm.

Substituting the values into the expression

F = 1 / (4π x 8.85 x 10⁻¹²) x 34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 9 x 10⁹ x -34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 82729.6 x 10⁻³ N

Both forces direct towards the left (away from the origin, towards the negative x-axis)

Total force = 118699 x 10⁻³

= 118.7 N.

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2 months ago
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