In science, the prefix hemo- refers to

We run a distance of 1000 m at a speed of 4.3 m/s. Calculate the time elapsed to cover this distance

pashok25 [27]

Answer:

So if we need to cover 1000 meters. And we go at a speed of 4.3 m/s. That means that every 4.3 meters we cover is 1 second. So we divide both amd get

1000/4.3 = 232.56 is approx the answer. Also the meters cancel out because

m/(m/s) = m*s/m, cancels out giving s as a unit.

<h2><u>Therefore the answer is 232.56 seconds</u></h2>

6 0

3 years ago

A concert loudspeaker suspended high off the ground emits 28.0 W of sound power. A small microphone with a 0.700 cm2 area is 55.

grandymaker [24]

Explanation:

It is known that wave intensity is the power to area ratio.

Mathematically, I = $\frac{P}{A}$

As it is given that power is 28.0 W and area is $7 \times 10^{-5} m^{2}$ .

Therefore, sound intensity will be calculated as follows.

I = $\frac{P}{A}$

= $\frac{28.0 W}{4 \times 3.14 \times 7 \times 10^{-5} m^{2}}$

= $0.318 \times 10^{5} W/m^{2}$

or, = $3.18 \times 10^{4} W/m^{2}$

Thus, we can conclude that sound intensity at the position of the microphone is $3.18 \times 10^{4} W/m^{2}$ .

7 0

3 years ago

A plane passes over Point A with a velocity of 8000 m/s north. Forty seconds later it passes over Point B at a velocity of 10,00

Airida [17]

Answer:

The planes’ acceleration from A to B is 500m/s^2

Explanation:

Given that the initial velocity u is 8000m/s

and also given the final velocity v=10,000 m/s

the time taken to move from A to B = 40 second

The acceleration is defined as the rate of change of velocity with time

we know that the expression for acceleration is given as

a=(v-u)/t

substituting our given data into the expression for a we have

a=(10000-8000)/40

a=2000/40

a=500m/s^2

The planes’ acceleration from A to B is 500m/s^2

7 0

3 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )

Also at equilibrium the moment about the point where the string holding the bar (where ( $m_1 +m_2$ ) and $m_3$ are hanged ) is attached is zero

So basically

$(m_1 + m_2 ) * 20 = m_3 * 30$

$(0.42 + 0.47) * 20 = 30 * m_3$

Making $m_3$ subject

$m_3 = \frac{(0.42 + 0.47) * 20 }{30 }$

$m_3 = 0.59 kg$

3 0

3 years ago

The force applied when using a simple machine ??

beks73 [17]

Answer : I hope this helps !

The Effort Force is the force applied to a machine. Work input is the work done on a machine. The work input of a machine is equal to the effort force times the distance over which the effort force is exerted.

8 0

3 years ago