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SVEN [57.7K]
3 years ago
7

During a hurricane in 2008, the Westin Hotel in downtownNew Orleans suffered damage. Suppose a piece of glass dropped near the t

op of the hotel falling from the 300 meter building takes 6.62 s. Ignoring air resistance, what will the velocity of the piece of glass be when it strikes the ground?
Physics
1 answer:
tangare [24]3 years ago
8 0

Answer:

  • <u>77.8 m/s, downward</u>

Explanation:

For uniform acceleration motion, the average speed is equal to half the soum of the initial velocity, Vi, and the final velocity, Vf

  • Average speed = (Vf + Vi)/2

Also, by definition, the average speed is the distance divided by the time:

  • Average speed = distance / time

Then:

  • (Vf + Vi)/2 = 300m/6.62s

Other kinematic equation for uniform acceleration is:

  • Vf = Vi + a×t

Since the window is falling and the air resistance is ignored, a = g (gravitational acceleration ≈ 9.8m/s²)

Replacing the known values we can set a system of two equations:

From (Vf + Vi)/2 = 300m/6.62s

(Vf + Vi) = 2 × 300m/6.62s

  • Vf + Vi = 90.634      equation 1

From Vf = Vi + a×t

Vf - Vi = 9.8 (6.62)

  • Vf - Vi = 64.876     equation 2

Adding the two equations:

  • 2Vf = 155.510

  • Vf = 77.8 m/s downward (velocities must be reported with their directions)
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A car is initially traveling at 12 m/s when the driver sees a yellow light ahead. He accelerates at a constant 7 m/s^2 for 6 s i
Vitek1552 [10]

Answer:

v = 54m/s

Explanation:

a =  \frac{v - u}{t}

a = 7m/s²

u = 12m/s

t = 6s

7 = (v-12)/6

v - 12 = 42

v = 54m/s

(Correct me if i am wrong)

3 0
3 years ago
A yoyo with a mass of m = 150 g is released from rest as shown in the figure.
avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

  • I is moment of inertia
  • α is angular acceleration
  • T is tension in the rope
  • r is inner radius
  • R is outer radius
  • f is frictional force

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

  • a is the linear acceleration of the yoyo

Torque equation for frictional force;

f = (\frac{r}{R} T) - (\frac{I}{R^2} )a

solve (1) and (2)

a = \frac{TR(R - r)}{I + MR^2}

since the yoyo is pulled in vertical direction, T = mg a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

#SPJ1

3 0
2 years ago
Light-rail passenger trains that provide transportation within and between cities are capable of modest accelerations. The magni
Mkey [24]
During the first phase of acceleration we have:
v o = 4 m/s;  t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?;  v o = 13 m/s; a = 1.125 m/s² ;  t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
v = 13 m/s + 18 m/s = 31 m/s 
3 0
3 years ago
What is the momentum of a 750kg car traculing at a velocity of 24m/s
m_a_m_a [10]
The answer is 18000 kgm/s

Momentum is mass times velocity so just do 750•24.
7 0
3 years ago
when an element tends to lose its valence electrons in chemical reactions , does it behave more like a metal or nonmetal
Juli2301 [7.4K]

It behaves more like a metal

Explanation:

When an element tends to lose its valence electrons in chemical reactions, they behave more like a metal.

Metals are electropositive.

Electropositivity or metallicity is the a measure of the tendency of atoms of an element to lose electrons.

This is closely related to ionization energy and the electronegativity of the element.

  • The lower the ionization energy of an element, the more electropositive or metallic the element is .

Metals are usually large size and prefers to be in reactions where they can easily lose their valence electrons.

When most metals lose their valence electrons, they attain stability.

Non-metals are electronegative. They prefer to gain electrons.

learn more:

Reactivity brainly.com/question/6496202

#learnwithBrainly

4 0
3 years ago
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