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Margaret [11]
1 year ago
13

A woman living in a third-story apartment is moving out. Rather than carrying everything down the stairs, she decides to pack he

r belongings into crates, attach a frictionless pulley to her balcony railing, and lower the crates by rope.
Required:
How hard must she pull on the horizontal end of the rope to lower a 49 kg crate at steady speed?
Physics
1 answer:
Flura [38]1 year ago
8 0

Answer:

T = 480.2N

Explanation:

In order to find the required force, you take into account that the sum of forces must be equal to zero if the object has a constant speed.

The forces on the boxes are:

T-Mg=0      (1)

T: tension of the rope

M: mass of the boxes 0= 49kg

g: gravitational acceleration = 9.8m/s^2

The pulley is frictionless, then, you can assume that the tension of the rope T, is equal to the force that the woman makes.

By using the equation (1) you obtain:

T=Mg=(49kg)(9.8m/s^2)=480.2N

The woman needs to pull the rope at 480.2N

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a student wants to push a box of books with the mass of 50 kg in 3 m horizontally towards the location of the shelves where the
irina1246 [14]

Answer:

The work done is 360 J.

Explanation:

Given that,

Mass = 50 kg

Distance =3 m

We need to calculate the work done

The work done is equal to the product of force and displacement.

Using formula of work done

W = F\cdot d

W = Fd\cos\theta

Where, F = force

D = distance

θ = Angle between force and displacement

Put the value into the formula

W=120\times3\cos0^{\circ}

W=360\ J

Hence, The work done is 360 J.

8 0
2 years ago
A rubber ball with a mass 0.20 kg is dropped vertically from a height of 1.5 m above the floor. The ball bounces off of the floo
Digiron [165]
Potential Energy = mass * Hight * acceleration of gravity
PE=hmg
PE = 1.5 * .2 * 9.81
PE = 2.943
it lost .6 so 2.943 - .6 = 2.343
now your new energy is 2.343 so solve for height
2.343 = mhg
2.334 = .2 * h * 9.81
h = 1.194
the ball after the bounce only went up 1.194m
8 0
1 year ago
There are lots of examples of ideal gases in the universe, and they exist in many different conditions. In this problem we will
elena-14-01-66 [18.8K]

Answer:

P = ρRT/M

Explanation:

Ideal gas equation is given as follows generally:

PV = nRT (1)

P = pressure in the containing vessel

V = volume of the containing vessel

n = number of moles

R = gas constant

T = temperature in K

n = m/M

m = mass of the gas contained in the vessel in g

M = molar mass in g/mol

ρ = m/V

Density of the gas = ρ

Substituting for n in (1)

PV = mRT/M. (2)

Dividing equation (2) through by V

P = m/V ×RT/M

P = ρRT/M

5 0
1 year ago
An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini
arsen [322]
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

  m₁:  KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

  m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

   m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

   (0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

   V = 0.2977 m/s

The kinetic energy is,
   KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
   KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J. 
7 0
2 years ago
The speed of sound in seawater is 1470 m/s. A dolphin sends out a click that reflects off of an
Nitella [24]

Answer: 0.204 s

Explanation:

The speed of sound V is defined as the distance traveled d in a especific time t:  

V=\frac{d}{t}  

Where:  

V=1470 m/s is the speed of sound  in seawater

t is the time the sound wave travels from the dolphin and then returns after the reflection

d=2(150 m) is twice the distance between the dolphin and the object to which the sound waves are reflected

Finding t:

t=\frac{d}{V}  

t=\frac{2(150 m)}{1470 m/s}

<u>Finally:</u>

t=0.204 s

3 0
2 years ago
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