All categories

3 years ago

Water is most dense at 4 degrees Celsius. Since at this temperature 1 ml of water has a mass of 1 g. What is its density?

Physics

2 answers:

sveticcg [70]3 years ago

7 0

The density is 1. 1/1=1

jolli1 [7]3 years ago

7 0

Answer: 1 g/cm³

Explanation:

The density of a substance is defined as the ratio of its mass to its volume.

At 4 °C, 1 mL water has 1 g mass.

1 mL = 1 cm³

Thus, ρ = m/V

m = 1 g

V = 1 cm³

⇒ ρ = m/V = 1 g/ 1 cm³

So, the density of water at 4° C is 1 g/cm³

You might be interested in

A sound wave has a frequency of 295 Hz and travels the length of a football field, 91.4 m in 0.506 s. What is the period of the

dem82 [27]

Answer:1/295 seconds

Explanation:

Period=1/frequency

Period=1/295 seconds

7 0

3 years ago

All objects emit what radiation

Eduardwww [97]

If an object is not at Absolute Zero, then it is
absorbing and radiating thermal (heat) energy.

5 0

3 years ago

Peering through a telescope could be considered Step 1 of the scientific method. True False

aleksklad [387]

Answer:

True: because when u look thru a telescope you are making an observation

Explanation:

4 0

2 years ago

Read 2 more answers

Help, please. I am not sure what to do.

miss Akunina [59]

Answer:

option D) -3m

Explanation:

if 6m is diplaced by -3m then it would be -3+6=3m

feel free to ask if you are confused

3 0

2 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago