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DochEvi [55]
2 years ago
10

An object is dropped from rest and falls through height h. It travels 0.5h in the last 1 second of fall. Find the total time &am

p; height of the fall. (Hint: use two triangles!)
Physics
1 answer:
STALIN [3.7K]2 years ago
4 0

Answer:

3.41 s

114 m

Explanation:

The object is falling in free fall, accelerated by the surface gravity of Earth. We can use the equation for position under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

We set up a frame of reference with the origin at the point the object was released and the X axis pointing down. Then X0 = 0. Since the problem doesnt mention an initial speed we assume V0 = 0.

It travels 0.5h in the last 1 second of the fall. This means it also traveled in the rest of the time of the fall. t = t1 is the moment when it traveled 0.5*h.

0.5*h = 1/2 * a * t1^2

h = a * t1^2

It travels 0.5*h in 1 second.

h = X(t1 + 1) = 1/2 * a * (t1+1)^2

Equating both equations:

a * t1^2 = 1/2 * a * (t1+1)^2

We simplify a and expand the square

t1^2 = 1/2 * (t1^2 + 2*t1 + 1)

t1^2 - 1/2 * t1^2 - t1 - 1/2 = 0

1/2 * t1^2 - t1 - 1/2 = 0

Solving electronically:

t1 = 2.41 s

total time = t1 + 1 = 3.41.

Now

h = a * t1^2

h = 9.81 * 3.41^2 = 114 m

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allsm [11]

Answer:

The charge on the third object is − 21.7nC

Explanation:

From Gauss's Law

Φ = Q/ε₀

where;

Φ is the total electric flux through the shell = − 533 N⋅m²/C

Q is the total charge Q in the shell = ?

ε₀ is the permittivity of free space = 8.85 x 10⁻¹²

From this equation; Φ = Q/ε₀

Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²

Q =  −4.7 X 10⁻⁹ C = -4.7nC

Q = q₁ + q₂ + q₃

− 4.7nC = − 14.0 nC + 31.0 nC + q₃

− 4.7nC − 17nC = q₃

− 21.7nC = q₃

Therefore, the charge on the third object is − 21.7nC

8 0
3 years ago
The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared
velikii [3]

Answer:

\displaystyle a(5)=-32

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

\displaystyle v(t)=\frac{df}{dt}

And the acceleration is

\displaystyle a(t)=\frac{dv}{dt}

Or equivalently

\displaystyle a(t)=\frac{d^2f}{d^2t}

The given height of a projectile is

f(t)=-16t^2 +238t+3

Let's compute the speed

\displaystyle v(t)=-32t+238

And the acceleration

\displaystyle a(t)=-32

It's a constant value regardless of the time t, thus

\boxed{\displaystyle a(5)=-32}

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3 years ago
An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop-the-loop maneuver. The acceleration
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Answer:

The radius of the loop is  20.66 km

Explanation:

let the radius of the loop be r

mass of airplane is m

At the top, the pilot experiences two radial forces, which are

1) Gravitational force is  mg

2) Centrifugal forces mv²/r out of the center

When the pilot experiences no weight,

then, mg = mv²/r

r = v² / g

 = 450² / 9.8

 = 20.66 x 10³3

 = 20.66 km

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