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Makovka662 [10]
3 years ago
6

An elevator of massmis initially at rest onthe first floor of a building. It moves upward,and passes the second and third floors

witha constant velocity, and finally stops at thefourth floor.The distance between adjacentfloors is h.What is the net work done on the elevatorduring the entire trip, from the first floor tothe fourth floor?
1. None of these.
2. W=-3m g h
3. W=-4m g h
4. W= 3m g h
5. W= 0correct
6. W= 4m g h
Physics
1 answer:
Varvara68 [4.7K]3 years ago
7 0

To solve this problem it is necessary to apply the concepts of Work. Work is understood as the force applied to travel a determined distance, in this case the height. The force in turn can be expressed by Newton's second law as the ratio between mass and gravity, as well

W = mgh

Where,

m = mass

h = height

g = Gravitational constant

When it ascends to the second floor it has traveled the energy necessary to climb a height, under this logic, until the 4 floor has traveled 3 times the height h of each of the floors therefore

h = 3h

Replacing in our equation we have to

W = mgh\\W = mg(3h)\\W = 3mgh

The correct answer is 4.

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A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 40 m/s when i
Alenkinab [10]

Given that,

Initial velocity , Vi = 0

Final velocity , Vf = 40 m/s

Acceleration due to gravity , a = 9.81 m/s²

Distance can be calculated as,

2as = Vf² - Vi²

2 * 9.81 *s = 40² - 0²

s = 81.55 m

For half height, that is, s = 40.77m

Vf= ??

2as = Vf² - Vi²

2 * 9.81 * 40.77 = Vf² - 0²

Vf² = 800

Vf = 28.28 m/s

Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.  

5 0
3 years ago
The head of a rattlesnake can accelerate at 49 m/s2 in striking a victim. If a car could do as well, how long would it take to r
Anettt [7]
<h2>Time taken is 0.459 seconds</h2>

Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = 81 km/hr = 22.5 m/s    

     Time, t = ?

     Acceleration, a = 49 m/s²

     Substituting

                      v = u + at  

                      22.5 = 0 + 49 x t

                      t = 0.459 seconds

   Time taken is 0.459 seconds

3 0
3 years ago
A catapult with a spring constant of 10,000 N/m is used to launch a target from the deck of a ship. The spring is compressed a d
Mnenie [13.5K]

Answer:

(C) 40m/s

Explanation:

Given;

spring constant of the catapult, k = 10,000 N/m

compression of the spring, x = 0.5 m

mass of the launched object, m = 1.56 kg

Apply the principle of conservation of energy;

Elastic potential energy of the catapult = kinetic energy of the target launched.

¹/₂kx² = ¹/₂mv²

where;

v is the target's  velocity as it leaves the catapult

kx² = mv²

v² = kx² / m

v² = (10000 x 0.5²) / (1.56)

v² = 1602.56

v = √1602.56

v = 40.03 m/s

v ≅ 40 m/s

Therefore, the target's velocity as it leaves the spring is 40 m/s

6 0
2 years ago
Based on experimental observations, the acceleration of a particle is defined by the relation a = -(0.1 + sin x/b), where a and
Fiesta28 [93]

Answer:

Velocity,v = 0.323 m/s

Explanation:

The acceleration of a particle is given by :

a=-(0.1+sin\dfrac{x}{b})

b = 0.8 m when x = 0

Since, a=v\dfrac{dv}{dx}  

v\dfrac{dv}{dx}=-(0.1+sin\dfrac{x}{b})  

\int{v.dv}=\int{-(0.1+sin\dfrac{x}{b})}.dx

\dfrac{v^2}{2}=-[0.1x-0.8cos\dfrac{x}{0.8}]+c

At x = 0, v = 1 m/s

\dfrac{1}{2}=0.8+c

c=-0.3

\dfrac{v^2}{2}=-[0.1x-0.8cos\dfrac{x}{0.8}]-0.3

At x = -1 m

\dfrac{v^2}{2}=-0.1(-1)+0.8cos\dfrac{(-1)}{0.8}-0.3

{v^2}=0.1045

v = 0.323 m/s

So, the velocity of the particle is 0.323 m/s. Hence, this is the required solution.

8 0
3 years ago
Compared to microwaves, ultraviolet waves have ___________________________ frequency.
vampirchik [111]
I think its a higher frequency
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