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igor_vitrenko [27]

3 years ago

6

A box of Thanksgiving presents slides across a waxy floor at 20m/s. It comes to a stop in 4 seconds. What distance did the box t

ravel across the floor?

1 answer:

mars1129 [50]3 years ago

6 0

Answer:

V = d/t = 20

20=d/4 so d = 80

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What could be the possible answer to the question ?<br><br>thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$

F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0

2 years ago

Read 2 more answers

A 180 cm length of string has a mass of 5.0 g. it is stretched with a tension of 8.6 n between fixed supports. (a) what is the w

Svetlanka [38]

P U S S Y

<span>Joy is planning to purchase a sweater that costs $30 dollars at her local department store. The sweaters are on sale for 20% off. Which steps are needed to find the sale price of the sweater?</span>

5 0

3 years ago

three condensers are connected in series across a 150 volt supply. The voltages across them are 40,50 and 60 volts respectively,

ioda

Explanation:

Given that,

The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.

(a) Capacitance of capacitor 1,

$C_1=\dfrac{Q}{V_1}\\\\C_1=\dfrac{6\times 10^{-8}}{40}\\\\C_1=1.5\times 10^{-9}\ F\\\\C_1=1.5\ nF$

Capacitance of capacitor 2,

$C_2=\dfrac{Q}{V_2}\\\\C_2=\dfrac{6\times 10^{-8}}{50}\\\\C_2=1.2\times 10^{-9}\ F\\\\C_2=1.2\ nF$

Capacitance of capacitor 3,

$C_3=\dfrac{Q}{V_3}\\\\C_3=\dfrac{6\times 10^{-8}}{60}\\\\C_3=1\times 10^{-9}\ F\\\\C_3=1\ nF$

(b) The equivalent capacitance in series combination is :

$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\\\\dfrac{1}{C}=\dfrac{1}{1.5}+\dfrac{1}{1.2}+\dfrac{1}{1}\\\\C=0.4\ nF$

Hence, this is the required solution.

5 0

3 years ago

The _____ of a mechanical wave is a direct measure of its energy.

KiRa [710]

I think it's amplitude

6 0

3 years ago

What type of relationship exists between the length of a wire and the resistance, if all other factors remain the same? A. Resis

ZanzabumX [31]

<h3>Answer</h3>

(A) Resistance is directly related to length.

<h3>Explanation</h3>

Formula for resistance

R = p(length) / A

where R = resistance

p = resistivity(material of wire)

A = cross sectional area

So it can be seen that resistance depends upon 3 factors that are length of wire , resistivity of wire and the cross sectional area of the wire.

If two of the factors, resistivity and cross sectional area, are kept constant then the resistance is directly proportional to the length of wire.

<h3> R ∝ length</h3>

This means that the resistance of the wire increases with the increase in length of the wire and decreases with the decrease of length of the wire.

5 0

3 years ago