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2 years ago

In a harbor, you can see sea waves traveling around the edges of small stationary boats. Why does this happen?

1 answer:

8 0

Below are the choices that can be found in the other sources:

A. diffraction
B. refraction 
C. reflection 
D. transmission

The answer is diffraction. It means that the process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the wave forms produced.

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A physician orders Humulin R 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many

jekas [21]

The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration

Complete question:

A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?

Answer:

the total units of insulin admistered each morning

= 22 units of qam and humulin

Explanation:

given

44 units and Humnlin N

with concentration 50/100 = 1/2 = 0.5

∴ 44 × 0.5 ≈ 22 units in the morning

regular insulin administered each day

(22 + 35)units of qam and humulin

= 57units

5 0

2 years ago

A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball

liberstina [14]

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$1.45=\frac{u^2\sin 90}{9.8}$

u=3.77 m/s

Conserving Energy

$E_{bottom}=E_{initial\ point\ at\ cliff}$

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

$\frac{mv^2}{2}=mgh+\frac{mu^2}{2}$

$v^2=u^2+2gh$

$v=\sqrt{u^2+2gh}$

$v=\sqrt{3.77^2+2\time 9.8\times 0.64}$

$v=\sqrt{26.75}=5.17 m/s$

5 0

1 year ago

A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

$initial\ momentum=final\ momentum$

$1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v$

$210000-21600=9000\ kg\times v$

$v=20.93\ m/s$

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

8 0

1 year ago

If a 1,300 kg car with no people inside is on the edge of a cliff 1,500 m above the ground, what is its potential energy?

Ghella [55]

Given that

mass (m) = 1300 Kg ,

height (h) = 1500 m

Determine the potential energy ?

P.E = m × g × h

= 1300 × 9.81 × 1500

= 19129500 Joules

= 19129.5 KJ

3 0

2 years ago

When the mass of the bottle is 0.125 kg, the KE is______ kg m2/s2.

tensa zangetsu [6.8K]

Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2

Explanation:

(1) Given mass = 0.125 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2

(2) Given mass = 0.250 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2

(3) Given mass = 0.375 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2

(4) Given mass = 0.500 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2

5 0

2 years ago