A car is at velocity of 20 km/h. If the car traveled 120 km in 3 hours at constant acceleration, what is its final velocity?

Why is it important to use the correct number of significant digits when

Artemon [7]

Answer:

D

Explanation:

Scientists use significant figures to avoid claiming more accuracy in a calculation than they actually know.

6 0

3 years ago

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Un vas plin cu lichid cântăreşte 175kg. Ceea ce reprezintă de 5 ori masa vasului gol. Ştiind că volumul interior al vasului este

Mariana [72]

a) Density of the liquid: $823.5kg/m^3$

b) Weight of the liquid: 1372 N

Explanation:

Translation of the text:

"A full tank with liquid weighs 175kg. Which is 5 times the mass of the empty vessel. Knowing that the inside volume of the vessel is 0.17kl, calculate:

a) the density of the liquid;

b) the weight of the liquid."

a)

We know that the full tank with liquid has a total mass of M = 175 kg. We can write the total mass as

$M=m_L + m_V$ (1)

where

$m_L$ is the mass of the liquid

$m_V$ is the mass of the vessel

We also know that the total mass M is 5 times the mass of the empty vessel, so we have:

$M=5m_V\\m_V=\frac{M}{5}=\frac{175}{5}=35 kg$

which is the mass of the empty vessel.

Therefore, we can find the mass of the liquid only using (1):

$m_L=M-m_V=175-35=140 kg$

The density of the liquid is given by

$d=\frac{m}{V}$

where

m = 140 kg (mass of the liquid)

V = 0.170 kL = 170 L = $0.170 m^3$ (volume of the liquid, which is equal to the volume of the vessel)

So we get

$d=\frac{140}{0.170}=823.5kg/m^3$

b)

The weight of a body is given by

$F=mg$

where

m is its mass

g is the acceleration due to gravity

For the liquid in this problem, we have

m = 140 kg (mass)

$g=9.8 m/s^2$ (acceleration due to gravity)

Therefore, its weight is

$F=(140)(9.8)=1372 N$

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

6 0

3 years ago

An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver

Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, $E=2.3\times 10^{-10}\ N/C$

Vertical distance covered, $s=1\ \mu m=10^{-6}\ m$

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

$v^2-u^2=2as$

$v^2=2as$ .............(1)

Electric force is $F_e$ and force of gravity is $F_g$ . As both forces are acting in downward direction. So, total force is:

$F=mg+qE$

$F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}$

$F=4.57\times 10^{-29}\ N$

Acceleration of the electron, $a=\dfrac{F}{m}$

$a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}$

$a=50.21\ m/s^2$

Put the value of a in equation (1) as :

$v=\sqrt{2as}$

$v=\sqrt{2\times 50.21\times 10^{-6}}$

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0

3 years ago

1 example of a conductor and 1 example of a insulator in your EVERYDAY world.

ratelena [41]

Answer: Examples of conductors include metals, aqueous solutions of salts (i.e., ionic compounds dissolved in water), graphite, and the human body. Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.

4 0

2 years ago

A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its

dlinn [17]

Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.

The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:

 70 km per second per megaparsec.

We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

 (Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =

 (150 million) x (1 / 3,262,000) x (70 km/sec) =

 3,219 km/sec in the direction away from us (rounded)

4 0

3 years ago