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Alex Ar [27]
1 year ago
13

Two students, Jenny and Cho, are investigating motion.

Physics
1 answer:
IgorLugansk [536]1 year ago
8 0

Answer:

1: a measuring instruments the students should use for time is a stopwatch

2: a measuring instruments the students should use for distance is a measuring tape

Explanation:

pls mark brainliest

You might be interested in
A 30-cm-diameter, 4-m-high cylindrical column of a house made of concrete ( k = 0.79 W/m⋅K, α = 5.94 × 10 −7 m2/s, rho = 1600kg
PilotLPTM [1.2K]

Answer:

a) Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b) Amount of heat transfer is 5320 kJ  

c) Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

Explanation:

Given that;

Diameter D = 30 cm

Height H = 4m

heat transfer coeff h = 14 W/m².°C

thermal conductivity k = 0.79 W/m.°C

thermal diffusivity α  = 5.94 × 10⁻⁷ m²/s

Density p = 1600 kh/m³

specific heat Cp = 0.84 Kj/kg.°C

a)

the Biot number is

Bi = hr₀ / k

we substitute

Bi = (14 W/m².°C × 0.15m) / 0.79 W/m.°C

Bi = 2.658

From the coefficient for one term approximate of transient one dimensional heat conduction The constants λ₁ and A₁ corresponding to this Biot number are,  

λ₁ = 1.7240

A₁ = 1.3915

Once the constant J₀ = 0.3841 is determined from corresponding to the constant λ₁

the Fourier number is determined to be  

[ T(r₀, t) -T∞ ] / [ Ti - T∞]  = A₁e^(-λ₁²t') J₀ (λ₁r₀ / r₀)

(27 - 28) / (14 - 28)   = (1.3915)e^-(17240)²t (0.3841)  

t' = 0.6771

Which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes  

t =  t'r₀² / ₐ

= (0.6771 × 0.15 m)² /  (5.94 x 10⁻⁷ m²/s)

= 23,650 s

= 7.1 hours

Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b)

The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C.  

Maximum heat transfer between the ambient air and the column is

m = pV

= pπr₀²L

= (1600 kg/m³ × π × (0.15 m)² × (4 m)

= 452.389 kg

Qin = mCp [T∞ - Ti ]

= (452.389 kg) (0.84 kJ/kg.°C) (28 - 14)°C

= 5320 kJ  

Amount of heat transfer is 5320 kJ  

(c)

the amount of heat transfer until the surface temperature reaches to 27°C is

(T(0,t) - T∞) / Ti - T∞  = A₁e^(-λ₁²t')

= (1.3915)e^-(1.7240)² (0.6771)

= 0.1860

Once the constant J₁ = 0.5787 is determined from Table corresponding to the constant λ₁, the amount of heat transfer becomes  

(Q/Qmax)cyl = 1 - 2((T₀ - T∞) / ( Ti - T∞)) ((J₁(λ₁)) / λ₁)

= 1 - 2 × 0.1860 × (0.5787  / 1.7240)  

= 0.875

Q = 0.875Qmax

Q = 0.875(5320 kJ)  

Q = 4660 kJ

Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

6 0
2 years ago
The parietal lobe is the portion of the cerebral cortex located below the temporal lobe.
jasenka [17]

Answer:

False

Explanation:

Please see the attached file

5 0
3 years ago
Read 2 more answers
On arrival at a vehicle collision, you observe a small fire in the engine compartment. A bystander is attempting to smother the
mina [271]

Answer:

<em>Aim at the base of the fire and use short bursts until the fire is out.</em>

<em></em>

Explanation:

Fire extinguishers use CO2 (Carbondioxide) as the extinguishing agent. This is because CO2 is denser than air, and does not support combustion.

Aiming at the base of the fire causes the CO2 to fall on the base of the fire, where the source of the fire is, trapping it, and preventing it from further reacting with air in a combustion reaction. Also, the short burst creates a strong wind that forces the flame to blow out.

3 0
3 years ago
5
Tanya [424]

Answer:

Reset

Explanation:

Digital methods are the methods that are uses methodological outlook to study societal change and cultural condition of online data. Reset is use to disguise data In digital methods. It is use to set again and conceal data by giving the data a different form. It restores the device to the original manufacture's settings.

8 0
3 years ago
In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

3 0
3 years ago
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