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3 years ago

An Apple falls from a tree and one-half second later hits the ground

1 answer:

8 0

There isnt enough information to answer the question, the missing variable is "distance from said falling spot and ground"

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How do noise and vibration affect you when operating a boat?

kogti [31]

The noise and vibration could affect your body and mood in many ways while operating a boat. The noise and vibration could make you moody or uncomfortable, because it will also distract you in doing what you're supposed to do. Your body will also be affected, making your body tired because of the things surrounding you and the noise and vibration, it could affect your body in many ways.

3 0

3 years ago

Read 2 more answers

According to the barometer in the attachment below, which statement is true?

mixer [17]

Your answer is A

Pls mark me brainiest and I sure hope this helps you

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2 years ago

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A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.7 m/s^2.

Komok [63]

Answer:

The coefficient of static friction between the car and the track

u=0.572

Explanation:

We don't know the mass of the car or any other information so the acceleration is the reason to solve the friction coefficient

∑ $F=F_{f}=m*a_{t}$

As we know

$F_{f}=u*F_{N}=u*m*g$

Also the center ward direction forces

$F_{fc}=m*a_{c}$

$a_{c}=\frac{v_{t}^2}{r}$

$F_{fc}=m*\frac{v_{t}^2}{r}$

But now vt relation with the tangential acceleration

$v_{t}=2*a_{t}*\frac{\pi }{r}$

replacing

$F_{fc}=m*a_{t}*\frac{2\pi*r}{2r}$

$F_{fc}=m*a_{t}*\pi$

So magnitude of the force can get by

$F_{f}=\sqrt{(m*a_{t}*\pi)^{2}+(m*a_{t})^{2}}$

Get the factor to simplify

$F_{f}=a_{t}*m*\sqrt{(1+\pi^2)}$

$u*m*g=m*a_{t}*(\sqrt{1+\pi^2})$

Solve to u'

$u=\frac{a_{t}}{g}*(\sqrt{1+\pi^2})$

$u=\frac{17\frac{m}{s^2} }{9.8\frac{m}{s^2}}*(\sqrt{1+\pi^2})=0.572$

5 0

3 years ago

330 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass i

Alexus [3.1K]

Answer:

T = 74°C

Explanation:

Given Mw = mass of water = 330g, Ma = mass of aluminium = 840g

Cw = 4.2gJ/g°C = specific heat capacity of water and Ca = 0.9J/g°C = specific heat capacity of aluminium

Initial temperature of water = 100°C.

Initial temperature of aluminium = 29°C

When the boiling water is poured into the aluminum pan, heat is exchanged and after a short time the water and aluminum pan both come to thermal equilibrium at a common temperature T.

Heat lost by water equal to the heat gained by aluminium pan.

Mw × Cw×(100 –T) = Ma × Ca × (T–29)

330×4.2×(100– T) = 890×0.9×(T–29)

1386(100 – T) = 801(T –29)

1386/801(100 – T) = T – 29

1.73(100 – T) = T – 29

173 –1.73T = T –29

173+29 = T + 1.73T

202 = 2.73T

T = 202/2.73

T = 74°C

6 0

3 years ago

A grocery cart with a mass of 15 kg is pushed at constant speed along an aisle by a force fp = 12 n which acts at an angle of 17

Korolek [52]

Given:

Mass of cart: 15kg

Aisle length = 14m

Angle = 17° below the horizontal

Force fp = 12 N

So, the solution would be like this for this specific problem:

1) W(by applied force) = F(applied) x s x cosθ 
=>W(a) = 12 x 14 x cos17* = 160.66 J 

2) By F(net) = F(applied) - F(friction) 
=>As v = constant => a = 0 => F(net) = 0
=>F(applied) = F(friction)
=>W(friction) = -F(friction) x s
=>W(friction) = -F(applied) x s
=>W(f) = -12 x 14 x cos17* = -160.56 J 

3) 0, as the displacement is perpendicular to Force 

4) 0, as the displacement is perpendicular to Force

To add, the force that is applied to an object by a person or another object is called the applied force.

8 0

3 years ago