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jolli1 [7]
3 years ago
5

Can someone please write a summery about Yes day the movie from 2021 and it has to have something about Hygiene in it and this i

s for 100 points so can you please help me I will give you $100 and 100 points for a 2 page essay

Physics
2 answers:
qwelly [4]3 years ago
8 0

Answer:                           This book is about a mom that is always saying no to her kids and is always pushing them to focus on what they are doing, but one day the mom or dad runs into a man that does what he calls "A Yes Day" with his kids and what this is, is for 24 hours you say yes to whatever you want to do and you aren't able to say no. So the mom tries yes day with her kids and they go through many crazy/extreme events. This has to deal with hygiene because the kids trash the house and get very dirty in a few parts of the movie. They always try to be as clean as possible, but the limits have been broken... This is why I think this movie has to deal with hygiene.

ella [17]3 years ago
3 0

Answer:

I can't do 2 pages but I'll try

Explanation:

Yes day is a movie about parents and kids who usually fought, until someone suggested a yes day a day when parents can't say no. In the beginning of the movie, you'll first see that the parents used to say yes to everything, until they has kids they started saying no to everything. When someone suggested yes day they had to earn it, they did all the chores and they got a yes day. They made a list of 5 things they wanna do during yes day. i don't know what else to put for hygiene but yea.

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A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800.0 N on him. Th
ipn [44]

Answer:

a) F_{fric} = 692 N

b) F_{applied} = 932 N

Explanation:

a)

According to newton's second law of motion, acceleration of an object is directly proportional to the net force acting on it. When there is no net force force acting on the body, there is no acceleration. A force is a push or a pull, and the net force ΣF is the total force, or sum of the forces exerted on an object  in all directions.

F_{net}  ∝ a

F_{net} =  ma

F_{applied} - F_{fric} = ma

Given data:

F_{applied} = 800 N

Mass = m = 90 kg

acceleration = a = 1.2 m/s²

F_{fric} = ?

800 - F_{fric} = (90)(1.2)

F_{fric} = 692 N

b)

According to newton's second law of motion,

F_{net}  ∝ a

F_{net} =  ma

F_{applied} - F_{fric} = ma

Given data:

If we assume the same friction and acceleration between player's feet and ground as calculated in part a

F_{fric} = 692 N

acceleration = a = 1.2 m/s²

We take the equal mass to the total mass of both the players because when the winning player push losing player backward, he exert force on the ground not only due to his mass but also due to the mass of losing player.

Mass = M = m₁ + m₂ = 110 kg + 90 kg

= 200 kg

F_{applied} = ?

F_{applied} - 692 N = (200)(1.2)

F_{applied} = 692 + 240

F_{applied} = 932 N

7 0
4 years ago
Metal sphere having an excess of +5 elementary charges has a net electric charge of
Natasha2012 [34]

Answer:

q=+8.01\cdot 10^{-19}\ coulombs

Explanation:

<u>Elementary charge</u>

The elementary charge, denoted by the symbol e is the electric charge carried by a proton or, equivalently, the magnitude of a negative electric charge carried by an electron, which has charge −e.

The value of the elementary charge is a fundamental constant in physics:

\mathbf{e}=1.60217662 \cdot 10^{-19}\ coulombs

If a metal sphere has an excess of +5 elementary charge, then it has a net charge of:

q=5*\mathbf{e}=+5*1.60217662 \cdot 10^{-19}\ coulombs

\boxed{q=+8.01\cdot 10^{-19}\ coulombs}

5 0
3 years ago
How was the theory of plate tectonics developed?
stepan [7]
Answer: A conflict between established theory, and empirical evidence, resulting in compromise.



Explanation;

Plate tectonic theory had its beginnings in 1915 when Alfred Wegener proposed his theory of "continental drift." Wegener proposed that the continents plowed through crust of ocean basins, which would explain why the outlines of many coastlines look like they fit together like a puzzle.
6 0
3 years ago
8-14 A Cu-30% Zn alloy tensile bar has a strain-hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm an
77julia77 [94]

Answer:

The true stress at true strain 0.05cm/cm is 80MPa

Explanation:

Given that

the strength coefficient is K

true strain is ε

strain hardening exponent is n

initial diameter of bar is d = 1cm, (10mm)

tensile force is F

engineering stress(S) = 120

the engineering stress(S) = \frac{F}{\frac{\pi }{4}(d^2) }

To find force (F) =

                    120 = \frac{F}{\frac{\pi }{4}(100^{2} )}

                     F = 120 * (π/4) * (100)

                     F = 9425N

Calculate the true strain  (ε) = In (l₀ / l₁)

where

l₀ =  initial length of the metallic bar = 3cm

l₁ = final length of metallic bar = 3.5cm

ε = In (3.5 / 3)

  = In 1.1667

  = 0.154cm/cm

Calculate the true stress (σ) at fracture point

          = \frac{F}{\frac{\pi }{4}(d^2) }}

tensile force is F and final diameter of bar is d₁ (d in the eqn)

Substitute 9425 N for F and 0.926 cm (9.26mm) for d₁ (d in the eqn)

σ = \frac{9425}{\frac{\pi }{4}(9.26^2) }}

   = 140MPa

To find the strength coefficient (K) of the material bar

K = \frac{140}{\sqrt{0.154} }

K = \frac{140}{0.3925}

   = 356.75MPa

To calculate the true stress σ true strain of 0.05cm/cm

K  = 356.75MPa

σ = 356.75(0.05)^0^.^5

  = 356.75 ( 0.2236)

  = 80MPa

The true stress at true strain 0.05cm/cm is 80MPa

6 0
3 years ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
Tcecarenko [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

3 0
3 years ago
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