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Kryger
2 months ago
11

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. How hard must she pull on the horizontal end of the rope to lower a 25 kg crate at steady speed?
Physics
1 answer:
inna [3.1K]2 months ago
5 0
Let T be the force exerted on the rope by her. This force induces tension in the rope, which exerts an upward force on the crates, while the weight of the crate pulls downward. Thus, the net force acting on the crate can be expressed as mg - T, acting in the downward direction. According to Newton's law, we can set up the equation: mg - T = ma. Given that a = 0 (the speed remains constant), this simplifies our equation to mg - T = 0, which leads to T = mg. Therefore, T = 25 x 9.8 = 245 N, indicating that the force she needs to apply is 245 N.
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ \frac{du}{dy}

therefore

= µ \frac{v}{h}............1

substituting the values

= 1.75 ×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

the area between the air and puck is given by

Area = \frac{\pi }{4} d^{2}

area = \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v

now applying Newton's second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

solving for t = \frac{0.1 v * 0.03}{1.37*10^{-3} v}

the time needed after impact for the puck is 2.18 seconds

3 0
3 months ago
A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "
kicyunya [3294]

Answer:

a) Q = C*emf

b)  Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let V_{R}represent the Voltage across the Resistor

V_{c}and

represent the Voltage across the capacitor

Implementing KVL;

emf - V_{R} - V_{c} = 0\\

.........................(1)

Since this is a series connection, the same current traverses through the circuit

V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C

Integrating V_{c}and V_{R}into equation (1)

emf - IR - Q/C = 0

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

I = 0A\\emf = Q/C\\Q = C* emf

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\

Substituting the values of t and I₀ into the aforementioned formula for I

I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf

4 0
2 months ago
Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat
Yuliya22 [3333]

Answer:

Explanation:

The distance between the electrodes is denoted as d.

The kinetic energy of the electron is represented as Ek when the electrodes are positioned at a distance of "d" apart.

Our goal is to determine the kinetic energy when they are separated by a distance of d/3.

K.E = ½mv²

It’s important to note that the mass remains constant; only velocity varies.

Additionally,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Assuming constant acceleration

Hence, m and a are fixed,

therefore,

K.E is directly related to d

Thus, as d increases, K.E increases, and conversely, when d decreases, K.E decreases.

Consequently,

K.E_1 / d_1 = K.E_2 / d_2

With K.E_1 equating to E_k

and d_1 being d

while d_2 is represented as d/3

This leads to K.E_2 = K.E_1 / d_1 × d_2

Thus, K.E_2 = E_k × ⅓d / d

Finally,

K.E_2 = ⅓E_k

Therefore, the resultant kinetic energy is one third of the original E_k

7 0
2 months ago
Which phrase describes an atom?
kicyunya [3294]
<span>an atom is described as having a negatively charged electron cloud surrounding a positively charged nucleus, which is the correct choice.</span><span>

The nucleus contains electrically neutral neutrons and positively charged protons, establishing its positive charge. In contrast, electrons carry a negative charge. The electromagnetic force keeps the atoms bound to the nucleus.
</span>
4 0
3 months ago
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