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3 years ago

Pain in the squadron can be a sign of appemditicis

Physics

2 answers:

alexgriva [62]3 years ago

6 0

No it can also be a sign of bruising or a stomach infection <span />

timofeeve [1]3 years ago

4 0

Yes. Or it can be a sign of intestinal gas, or of constipation. Or it can be a sign of an internal bruise caused by sleeping too long on the same side. Or it can be the result of the new squadron commander cancelling all scheduled leaves.

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A boat travels at 30 mph for a huge, solid cliff that is about 3,000 meters away. When the horn on the boat makes a toot, you ca

AleksandrR [38]

Answer:The frequency of the echo is slightly decreased

Explanation:

Given

speed of boat $=30\ mph$

cliff is $3000\ m$ away

when boat is still , suppose t is the time taken by the echo to reach observer on the boat

But as soon as boat starts moving the distance between cliff and boat decreasing and time for echo to reach observer also decreases

and we know $time \propto \frac{1}{frequency}$

therefore frequency of the echo slightly decreased.

5 0

3 years ago

A student practicing for a cross country meet runs 250 m in 30 s. What is the average speed

Kaylis [27]

Answer:8.3m/sec 30 sec,

Explanation:

A student practicing for a track meet, ran 250 m in 30 sec. a. What was her average speed? 250 m = 8.3 m/sec 30 sec.

5 0

2 years ago

Read 2 more answers

Suppose you run into a wall at 4.5 meters per second (about 10 mph). Let's say the wall brings you to a complete stop in 0.5 sec

STatiana [176]

Answer with Explanation:

We are given that

Initial velocity,u=4.5 m/s

Time=t =0.5 s

Final velocity=v=0m/s

We have to find the deceleration and estimate the force exerted by wall on you.

We know that

Acceleration= $\frac{v-u}{t}$

Using the formula

Acceleration= $a=\frac{0-4.5}{0.5}$

deceleration=a= $-9m/s^2$

We know that

Force =ma

Using the formula and suppose mass of my body=m=40 kg

The force exerted by wall on you

Force= $40\times (9)=360N$

3 0

2 years ago

A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between

elena-14-01-66 [18.8K]

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

W ≅ -589.05 J

4 0

2 years ago

Which is true regarding diffusion and osmosis? Question 3 options:

Scilla [17]

Answer:

Option (A)

Explanation:

Diffusion is defined as the transfer of particles such as ions or molecules from a higher concentration region to a lower concentration region.

Osmosis is a special type of diffusion in which the solvents such as water are allowed to move from a region of higher to a lower concentration through a layer of a semi-permeable membrane. This is an energy-independent process and occurs along the concentration gradient.

Thus, the correct answer is option (A) that is true for both osmosis and diffusion.

4 0

3 years ago