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3 years ago

Which part of the heating curve corresponds with the heating of liquid water?

1 answer:

8 0

Answer: the part of the heating curve corresponds to showing how the temperature changes as a substance for a constant rate.

Explanation: Let me know if you need any details!

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Convert 0.700 atm of pressure to its equivalent in millimeters of mercury.

inna [77]

Answer:

532 millimeters of mercury

Explanation:

In order to convert the pressure from atm to millimeters of mercury (mm Hg), we should remind the conversion factor between the two units:

1 atm = 760 mm Hg

Therefore, we can solve the problem by setting up the following proportion:

$1 atm : 760 mmHg = 0.700 atm : x$

Solving for x, we find

$x=\frac{(760 mmHg)(0.700 atm)}{1 atm}=532 mmHg$

5 0

4 years ago

You are given an unknown solid substance of mass 500 g and you want to estimate its specific heat. unfortunately, you don’t have

Julli [10]

Answer:

Option C) 2,090 J/(mol K)

Explanation:

Data:

Volume in the beaker = 429 ml

temperature = 20° C

Density = 789 kg/m³

Equilibrium reading = 429

volume change = 29 ml

= 0.029 L

Energy change = mcΔT

U + PΔV

7 0

3 years ago

7. What is another name for the force of

leonid [27]

Weight because both gravity and weight are pulling an object down

3 0

3 years ago

Read 2 more answers

Which of the following statements best describe the frequency of a wave?

Brrunno [24]

Answer:

I believe the answer is B.

6 0

3 years ago

A block is balanced on top of a frictionless sphere of radius R. When the block is given a slight nudge it starts to slide down

Nataly_w [17]

Answer: $\frac{R}{3}$

Explanation:

Given

Sphere of Radius R

Suppose mass of block is m

At any instant \theta Normal reaction(N) and weight(mg) is acting such that

$mg\sin \theta -N=\frac{mv^2}{R}$ , where v is velocity of block at any angle \theta

When block is just about to leave then N=0

therefore

$mg\sin \theta =\frac{mv^2}{R}$

$v^2=gR\sin \theta$ -------------------1

Also by conserving Energy we get

Potential Energy=kinetic Energy of block

$mgh=\frac{mv^2}{2}$

here h=vertical distance traveled by block

From diagram

$h=R-R\sin \theta$

$h=R(1-\sin \theta )$

$mgR(1-\sin \theta )=\frac{mv^2}{2}$

$2gR(1-\sin \theta )=v^2$ -----------------2

From 1 and 2

$2(1-\sin \theta )=\sin \theta$

$3\sin \theta =2$

$\sin \theta =\frac{2}{3}$

Thus from this value of h is

$h=R(1-\sin \theta )$

$h=R(1-\frac{2}{3})$

$h=\frac{R}{3}$

3 0

4 years ago