All categories

3 years ago

Please help!!!

Physics

2 answers:

krek1111 [17]3 years ago

6 0

The answer is influenced

swat323 years ago

4 0

The correct answer to your question is B. " Shaped". I just did the assignment and it said C was wrong. Have a nice day

You might be interested in

Which is one use for radioactive isotopes? sanitation architecture meteorology archaeology

e-lub [12.9K]

Answer:

Archaeology

Explanation:

Radioisotopes are radioactive atoms of an element in which their atoms contain excess energy making them unstable. When broken down they become more stable releasing radiations.

Carbon 14 is a radioactive isotope that is used in archaeology to study and estimate the lifespan and age of organic materials such as wood, leather. Carbon 14 can be used to estimate the ages of materials up to 50000 to 60000 years.

7 0

3 years ago

A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t

Anastaziya [24]

Answer:

The gauge pressure is $P_g = 2058 \ P_a$

Explanation:

From the question we are told that

The height of the water contained is $h_w = 30 \ cm = 0.3 \ m$

The height of liquid in the cylinder is $h_t = 40 \ cm = 0.4 \ m$

At the bottom of the cylinder the gauge pressure is mathematically represented as

$P_g = P_w + P_o$

Where $P_w$ is the pressure of water which is mathematically represented as

$P_w = \rho_w * g * h_w$

Now $\rho_w$ is the density of water with a constant values of $\rho_w = 1000 \ kg /m^3$

substituting values

$P_w = 1000 * 9.8 * 0.3$

$P_w = 2940 \ Pa$

While $P_o$ is the pressure of oil which is mathematically represented as

$P_o = \rho_o * g * (h_t -h_w )$

Where $\rho _o$ is the density of oil with a constant value

$\rho _o = 900 \ kg / m^3$

substituting values

$P_o = 900 * 9.8 * (0.4 - 0.3)$

$P_o = 882 \ Pa$

Therefore

$P_g = 2940 - 882$

$P_g = 2058 \ P_a$

6 0

3 years ago

What do acids release in water ?

nata0808 [166]

Acid's release protons, or hydrogen ions.

Hope this helped, have a great day! :D

6 0

3 years ago

If you were to measure the width of your lab manual with a meter stick, and then the length (roughly 50% longer), by what fracti

Leona [35]

Answer:

Explanation:

a ) Let breadth be L , then length will be 1.5 L

Since it is a meter scale , the least count is 1 m.m. So

maximum possible error in a measurement is 1 mm.

In both the measurement of length and breadth , absolute

uncertainty will be 1mm.

Hence there is no difference in absolute uncertainty of their measurement.

b ) percent uncertainty in the measurement of length

$\frac{1 mm}{1.5L\times1000} \times100$

= 1 / 15L%

percent uncertainty in the measurement of breadth

$\frac{1 mm}{L\times1000} \times100$

1 / 10L%

In breadth it will be higher by

( 1/10L - 1/15L)/ 1/10L

= 1/3

c ) Uncertainty in the measurement of R= 1 %

Area A = π R²

ΔA/A X100 = 2 X ΔR/R X 100

Percentage uncertainty in A = 2 X percentage uncertainty in R

= 2 X 1

=2 %

Therefore percentage accuracy in the calculation of area

= 2 %.

7 0

3 years ago

The rate (in liters per minute) at which water drains from a tank is recorded at half-minute intervals. Use the average of the l

lana [24]

Answer:

see explanation

Explanation:

You are missing the chart with the rates and time to do this, however, I wll do it with a similar exercise here, and you only need to replace the procedure with your data:

See the attached table.

From the left we have:

r = 1/2 (50 + 48 + 46 + 44 + 42 + 40) = 135 L/min

From the right we have:

r = 1/2 (48 + 46 +44 + 42 + 40 + 38) = 129 L/min.

And this should be the correct answer. Watch your chart and replace if it's neccesary.

3 0

3 years ago