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kap26 [50]
4 years ago
11

What is the mean of the data set 3cm, 7cm, 7cm, 7cm, 9cm, 11cm, 12cm

Physics
2 answers:
vesna_86 [32]4 years ago
7 0
Mean = Sum of the items ÷ No of items

Mean = 3+7+7+7+9+11+12 ÷ 7

Mean = 56 ÷ 7

Mean = 8
zzz [600]4 years ago
6 0
The correct answer for the mean would be 8

I hope that this helps !
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Which statements are part of the safety protocol for this lab experiment? Check all that apply:
vagabundo [1.1K]

Answer:

Always wear a lab coat and safety goggles when performing an experiment

7 0
2 years ago
¿que es hipótesis?¿que es teoría?​
Ludmilka [50]

Answer:

The hypothesis is only an assumption that has not yet been proven, when it has been tested, and it cannot be shown that it is false, then it is no longer a simple hypothesis, it is a Theory. Hypothesis is when we raise a supposed result. Theory is the result of an experiment.

Explanation:

4 0
3 years ago
Read 2 more answers
What is the longest wavelength that can be observed in the third order for a transmission grating having 5200 slits/cm?
Anarel [89]

Answer:

641 nm.

Explanation:

Given that,

A transmission grating has 5200 slits/cm.

We need to find the longest wavelength that can be observed in the third order. Using grating equation as follows :

d\sin\theta=m\lambda ...(1)

d is slit spacing

No fo slit per unit length :

N={5200}\ slit/cm\\\\=520000\ slits/m

We know that, N = 1/d

For longest wavelength, θ = 90°

From equation (1)

\dfrac{\sin\theta}{m\lambda}=\dfrac{1}{d}\\\\520000=\dfrac{\sin(90)}{3\lambda}\\\\\lambda=\dfrac{1}{520000\times 3}\\\\=6.41\times 10^{-7}\ m\\\\=641\ nm

Hence, the longest wavelength in third order for a transmission grating is 641 nm.

3 0
3 years ago
Help please!!!
TiliK225 [7]

Answer:

Cu 8.92

Explanation:

The formula for density is mass/volume. If you were to divide 62.44 by 7, you would get 8.92. Since copper is the only metal in this table that has a density of 8.92, that is the answer.

7 0
3 years ago
How much heat must be removed from 200 pounds of blanched vegetables at 189 degrees F to freeze them to a temperature of 22 degr
Virty [35]

To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.

By definition the exchange of heat is given by

Q =mc\Delta T

where,

m = mass

c = specific heat

\Delta T = Change in temperature

Therefore the total heat exchange is given as

\Delta Q = Q_w+Q_v

\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)

Our values are given as,

Total mass is M_T = 200lb ,however the mass of solid vegetable and water is given as,

m_v= 0.4*200lb = 80lb

m_w=0.6*200lb=120lb

T_i = 183\°F\\T_f = 22\°F\\c_w = 1Btu/lbF\\c_v = 0.24Btu/lbF

Replacing at our equation we have,

\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)

\Delta Q = (120)(1)(183-22)+(80)(0.24)(183-22)

\Delta Q = 22411.2Btu

Therefore the heat removed is 22411.2 Btu

6 0
3 years ago
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