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erastova [34]
3 years ago
5

Which of the following best explains the uneven nature of the exposed rock layers

Chemistry
1 answer:
AlexFokin [52]3 years ago
4 0

Answer:

where is the picture?

Explanation:

no picture at all....

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Find the mass of an object on earth if its weight is 100N
Kaylis [27]

Answer:

0.19 kg

8 N

1.964 N

58.86 N

15 kg

Explanation:

w = Weight of an object = 100 N

g = Acceleration due to gravity of Earth = 9.81\ \text{m/s}^2

m = Mass of an object

Weight is given by

w=mg\\\Rightarrow m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{100}{9.81}\\\Rightarrow m=10.19\ \text{kg}

Mass of the object is 0.19 kg

g = 2\ \text{m/s}^2

m = 4 kg

w=mg\\\Rightarrow w=4\times 2\\\Rightarrow w=8\ \text{N}

Weight of the object is 8 N.

m = 200 g

w=mg\\\Rightarrow w=0.2\times 9.81\\\Rightarrow w=1.964\ \text{N}

Weight of the object is 1.964 N.

Weight is the force the Earth exerts on an object which is on the surface of the Earth.

m = 6kg

w=mg\\\Rightarrow w=6\times 9.81\\\Rightarrow w=58.86\ \text{N}

The force of the object is 58.86 N.

w = 150 N

a = g = 10\ \text{m/s}^2

m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{150}{10}\\\Rightarrow m=15\ \text{kg}

Mass of the object is 15 kg.

7 0
3 years ago
If the half-life of technitium-104 is 18.0 min, how much of a 14.2-g sample will remain after 72 minutes?
Debora [2.8K]
The decay of a radioactive isotope can be predicted using the formula: A = Ao[2^(-t/T_0.5)] where A is the amount after time t, Ao is the original amount and T_0.5 is the half-life. Using the equation and the given values, 0.888 g of the sample will remain after 72 minutes. 
3 0
3 years ago
Read 2 more answers
Use the ruler to determine the length of this object. Record your answer to the nearest tenth. The object is ____ long.
VLD [36.1K]
The object is 2.7 cm long
7 0
3 years ago
Which pair of elements are in the same period?Hg & PbLi & NaB & ClO & As
Alexeev081 [22]

Answer:

Hg Pb are in one, Li and B and O are in one, and NaCl are in one, As is alone

Explanation:

Periods are horizontal

Hope this helps!

4 0
2 years ago
A mileage test is conducted for a new car model. Thirty randomly selected cars are driven for a month and the mileage is measure
Leni [432]

Answer:  (27.81,\ 29.39)

Explanation:

Given : Sample size : n= 30 , it means it is a large sample (n≥ 30), so we use z-test .

Significance level : \alpha: 1-0.95=0.05

Critical value: z_{\alpha/2}=1.96

Sample mean : \overline{x}=28.6

Standard deviation : \sigma=2.2

The formula to find the confidence interval is given by :-

\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}

i.e. 28.6\pm (1.96)\dfrac{2.2}{\sqrt{30}}

i.e. 28.6\pm 0.787259889321

\approx28.6\pm 0.79=(28.6-0.79,28.6+0.79)=(27.81,\ 29.39)

Hence, the 95% confidence interval for the mean mpg in the entire population of that car model = (27.81,\ 29.39)

4 0
3 years ago
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