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1 year ago

Two cars approach each other. How does the

Physics

1 answer:

Lemur [1.5K]1 year ago

7 0

Answer:

<u>The car's fast. The ground isn't moving.</u>

Hope this helped! :D

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What is the ultimate energy source for most wind?

Vinvika [58]

The ultimate energy source for wind comes from Earth's uneven heating on Earth's surface which can be caused mainly by the sun.

5 0

3 years ago

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A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in

dexar [7]

Answer:

K = 0.076 J

Explanation:

The height of the target, h = 0.860 m

The mass of the steel ball, m = 0.0120 kg

Distance moved, d = 1.50 m

We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

$h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}$

Put all the values,

$t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s$

The velocity of the ball is :

$v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s$

The kinetic energy of the ball is :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J$

So, the required kinetic energy is 0.076 J.

6 0

2 years ago

A car moves at 20 m s' for 15 minutes. Calculate the distance travelled

Firdavs [7]

Explanation:

Calculating this according to the m/s rate

Now solving the question

If the car goes 20 m/s mathematically we can generate it as

15 × 60 = 600 secs

If the car goes 20 meters per every second it means

The car will go 20 meters for 900 secs

Which is 900 ×20 = 18000m/s

=18000÷1000 = 1km

Therefore the answer is 18km

6 0

2 years ago

Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.

DaniilM [7]

Answer: $2(10)^{-9} mg$

Explanation:

We know the total diameter of the cell (assumed spherical) is:

$d=1.9\mu m=1.9(10)^{-6} m$

Then its total radius $r=\frac{d}{2}=\frac{1.9(10)^{-6} m}{2}=9.5(10)^{-7} m$

On the other hand, we know the thickness of the cell wall is $r_{t}=60 nm= 60(10)^{-9} m$ and its density is the same as water ( $\rho=997 kg/m^{3}$ ).

Since density is the relation between the mass $m$ and the volume $V$ :

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

The mass is: $m=\rho V$ (1)

Now if we are talking about this cell as a thin spherical shell, its volume will be:

$V=\frac{4}{3}\pi R^{3}$ (2)

Where $R=r-r_{w}=9.5(10)^{-7} m - 60(10)^{-9} m$

Then:

$V=\frac{4}{3}\pi (9.5(10)^{-7} m - 60(10)^{-9} m)^{3}$ (3)

$V=2.952(10)^{-18} m^{3}$ (4)

Substituting (4) in (1):

$m=(997 kg/m^{3})(2.952(10)^{-18} m^{3})$ (5)

$m=2.94(10)^{-15} kg$ (6)

Knowing $1 kg=1000 g$ and $1 mg=0.001 g$ :

$m=2.94(10)^{-15} kg=2(10)^{-9} mg$

7 0

3 years ago

Mga halimbawa ng metapora

faust18 [17]

Answer:

UMm If i understood ide answer

Explanation:

4 0

2 years ago

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