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wolverine [178]
3 years ago
12

A meterstick is initially standing vertically on the floor. If the meterstick falls over, with what angular velocity will it hit

the floor? Assume that the end in contact with the floor experiences no friction and slips freely.
Physics
1 answer:
notsponge [240]3 years ago
8 0

L = length of the meter stick = 1 m

h = height of center of mass of stick from bottom end on the floor = L/2 = 1/2 = 0.5 m

m = mass of the meter stick

I = moment of inertia of the meter stick about the bottom end

w = angular velocity as it hits the floor

moment of inertia of the meter stick about the bottom end is given as

I = m L²/3

using conservation of energy

rotational kinetic energy of meter stick as it hits the floor = potential energy when it is vertical

(0.5) I w² = m g h

(0.5) (m L²/3) w² = m g h

( L²) w² =  6g h

( 1²) w² =  6 (9.8) (0.5)

w = 5.4 rad/s

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Answer:

1. A.) The moment of the truck before it hits the brakes is 214,400 kg·m/s

B.) The force it takes to stop the truck is approximately 17,290.4 N

Explanation:

1. A.) The given parameters are;

The mass of the truck, m = 8,000 kg

The velocity of the truck when it hits the brakes, u = 26.8 m/s

Momentum = Mass × Velocity

The moment of the truck = The mass of the truck × The velocity of the truck

Therefore;

The moment of the truck before it hits the brakes = 8,000 kg × 26.8 m/s = 214,400 kg·m/s

B.) The amount of momentum lost when the truck comes to a stop = The initial momentum of the truck

The time it takes the truck to come to a complete stop, t = 12.4 s

The deceleration, "a" of the truck is given by the following kinematic equation of motion

v = u - a·t

Where;

v = The final velocity of the truck = 0 m/s

u = The initial velocity = 26.8 m/s

a = the deceleration of the truck

t = The time of deceleration of the truck = 12.4 s

Substituting the known values gives;

0 = 26.8 - a × 12.4

Therefore;

26.8 = a × 12.4

a = 26.8/12.4 ≈ 2.1613

The deceleration (negative acceleration) of the truck, a ≈ 2.1613 m/s²

Force = Mass × Acceleration

The force required to stop the truck = The mass pf the truck × The deceleration (negative acceleration) given to the truck

∴ The force it takes to stop the truck = 8,000 kg × 2.1613 m/s² ≈ 17,290.4 N.

8 0
3 years ago
3. Compare the slope of the velocity-time graph to the average of all your acceleration values. Are they close? What does the sl
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The slope of a speed-time graph is the acceleration represented by the graph.

All other parts of this question refer to a lab experiment or exercise
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6 0
3 years ago
A 5.0-ohm resistor, a 10.0-ohm resistor, and a 15.0-ohm resistor are connected in parallel with a battery. The current through t
Kaylis [27]

Answer:

E = 3456 J

Explanation:

The electrical energy expended in a resistor can be easily calculated by using the following formula:

E = Pt

where,

E = Energy Expended = ?

I = current through 5 ohm resistor = 2.4 A

R = Resistance = 5 ohms

P = Electrical Power = VI

Since,

V = IR (Ohm's Law)

Therefore,

P = (IR)(I) = I²R = (2.4 A)²(5 ohms) = 28.8 Watt

t = time taken = (2 min)(60 s/1 min) = 120 s

Therefore,

E = (28.8 Watt)(120 s)

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8 0
3 years ago
Is a sound wave mechanical or an electromagnetic wave
NNADVOKAT [17]
It is a mechanical wave and cannot travel through a vacuum.
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3 years ago
A spaceship accelerates uniformly for 1220km how much time is needed for the spaceship to increase its speed from 11.1km/s to 11
snow_lady [41]

The time taken for the spaceship to increase its speed from 11.1 km/s to 11.7 km/s is 107 s

<h3>Data obtained from the question</h3>

The following data were obtained from the question given above:

  • Initial velocity (u) = 11.1 Km/s
  • Final velocity (v) = 11.7 Km/s
  • Distance (s) = 1220 Km
  • Time (t) =?

<h3>How to determine the time</h3>

The time taken for the spaceship to increase its speed from 11.1 km/s to 11.7 km/s can be obtained as illustrated below:

s = (u + v)t / 2

Cross multiply

(u + v)t = 2s

Divide both sides by (u + v)

t = 2s / (u + v)t

t = (2 × 1220) / (11.1 + 11.7)

t = 2440 / 22.8

t = 107 s

Thus, the time taken for the spaceship to change its speed is 107 s

Learn more about speed:

brainly.com/question/680492

Learn more about velocity:

brainly.com/question/3411682

#SPJ1

6 0
1 year ago
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