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2 years ago

(0.50kg)(6.0mls)+(1.00kg)(-12.00mls)

Physics

1 answer:

Lana71 [14]2 years ago

6 0

Answer:

-9.0 × 10-6 m3 kg

Explanation:

I'm not sure if that's what you're looking for nor do I know how to explain it.

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Which of the following is the best definition of an isotope?

almond37 [142]

Option A looks like the best definition

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3 years ago

Which of the following accurately describes the process of Gas Exchange? *

Marrrta [24]

Answer:

<u><em>Oxygen in your alveoli is diffused in to your blood and Carbon Dioxide is taken out of the blood and into the lungs</em></u>

Explanation:

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8 0

3 years ago

Assuming atmospheric pressure to be 1.01 x 10^5 Pa and the density of sea water to be 1025 kg/m^3, what is the absolute pressure

Rudik [331]

To solve for absolute pressure, you will need this formula:

$P_{total} = P_{atm} + (rgh)$

Where: $P_{total}$ = absolute pressure
$P_{total}$ = atmospheric pressure
r (rho) = density
g = acceleration due to gravity constant $9.8 \frac{m}{ s^{2} }$
h = depth (in this case)

rgh is the formula for pressure of fluids

So with your given, we just need to insert it into the formula:

$P_{total} = P_{atm} + (rgh)$
$P_{total}$ = 1.01 x $10^{5}$ x (1,025 $\frac{kg}{m^3}$ x 9.8 $\frac{m}{{s^2}}$ x 15 m
$P_{total}$ = 1.01 x $10^{5}$ + 150,675
$P_{total}$ = 1.01 x $10^{5}$ + 1.51 x $10^{5}$

$P_{total}$ = 2.52 x $10^{5}$ This is your absolute pressure.

3 0

3 years ago

If a flea can jump straight up to a height of 21.1 cm , what is its initial speed as it leaves the ground, neglecting air resist

Vikentia [17]

The flea jumps at high velocity, reaches zero velocity at the maximum height 21.1 cm before starting to fall. Solve for initial velocity given final velocity is zero.
since time is not given, use the equation:
v^2 = u^2 + 2as
convert gravity or displacement to have same units. 9.8 m/s = 980 cm/s
0 = u^2 + 2(-980)(21.1)
41356 = u^2
sqrt(41356) = u
203.4 cm/s = u

5 0

4 years ago

Read 2 more answers

A block of weight 45.7 N is hanging from a rope. The tension from the rope is pulling upward on the block. The block is accelera

viktelen [127]

<h2>Answer:</h2><h2></h2>

52.555 N

<h2>Explanation:</h2>

Let's use Newton's second law of motion here which states that the resultant force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) due to this force. i.e

∑F = m x a ---------------------(i)

<em>Now, let's get the resultant force;</em>

Two main forces are acting on the rope;

i. the weight (W) of the block acting downwards.

Where;

W = mass of block(m) x gravity(g) = m x g

ii. the tension (T) in the rope acting upwards.

Therefore, the resultant force is the vector sum of these two forces as follows;

∑F = - W + T [upward motion is taken as positive. hence -W and +T]

<em>Substitute ∑F = - W + T into equation (i) as follows;</em>

- W + T = m x a ---------------------(ii)

<em>From the question;</em>

* Weight (W) of the block = 45.7N

=> mass (m) of the block = W / g = 45.7 / 10 [Taking g = 10m/s²]

=> m = 4.57 kg

* acceleration (a) = 1.50m/s²

<em>Substitute these values into equation (ii) as follows;</em>

- 45.7 + T = 4.57 x 1.50

- 45.7 + T = 6.855

<em>Solve for T;</em>

T = 6.855 + 45.7

T = 52.555 N

Therefore, the tension in the rope is 52.555 N

6 0

4 years ago