What is the average velocity of an arrow that travels 12 m [E] in 0.15 s?

1 answer:

6 0

Velocity is defined as Distance divided by Time.
In other words, V = D/T.

Now that we have our formula, we can solve.
Let's plug in the numbers we have.

We have 12m [East (direction not necessary when solving yet)] for our distance, and 0.15s as our time.

Divide the distance (12 /) by the time (0.15)
12 / 0.15 = 80.

Your velocity is 80 m/s [E]

I hope this helps!

You might be interested in

If the mass of a 1.8 g paperclip was able to be completely converted to energy, how much energy would you obtain?

Anton [14]

Answer:

$E=1.62\times 10^{14}\ J$

Explanation:

Given that,

The mass of the paperclip, m = 1.8 g = 0.0018 kg

We need to find the energy obtained. The relation between mass and energy is given by :

$E=mc^2$

Where

c is the speed of light

So,

$E=0.0018\times (3\times 10^8)^2\\\\E=1.62\times 10^{14}\ J$

So, the energy obtained is $1.62\times 10^{14}\ J$ .

7 0

3 years ago

What happens when sediment builds up over time

Morgarella [4.7K]

When sediment has built up over time layers of rock start to form, starting with sedimentry rocks, then metamorphic rocks

6 0

3 years ago

Read 2 more answers

Answer and explanation please!!

expeople1 [14]

Answer:

Option 3

Explanation:

O Option C is NEGATIVELY CHARGED, meaning it has GAINED ELECTRONS resulting in a GREATER number of ELECTRONS than PROTONS.

8 0

2 years ago

Read 2 more answers

A car with mass 950 kg and a speed of 16 m/s approaches an intersection. A 1300 kg minivan traveling at 21 m/s is heading for th

Alex73 [517]

Answer:

$V_f = 13.8863 \angle 60.89\°$

Explanation:

Our values are,

$m_1 = 950Kg\\v_1 = 16m/s \\m_2 =1300Kg\\v_2 = 21m/s$

We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.

$m_1v_1=(m_1+m_2)v_fcos\theta$