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Karo-lina-s [1.5K]

2 years ago

15

Hydrogen (H-1), deuterium (H-2), and tritium (H-3) are the three isotopes of hydrogen. What are the values of a, b, and c in the

table?

2 answers:

Shtirlitz [24]2 years ago

7 0

foe plato the answer is option b: a = 1, b = 2, c = 3

asambeis [7]2 years ago

6 0

I believe it is a=1 b=2 c=3

plato students

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The grooved pulley of mass m is acted on by a constant force F through a cable which is wrapped securely around the exterior of

Sidana [21]

Answer:

Answer; v= 1.2654m/s

T= 110.76N

Explanation:

Apply Momentum Principle

Fdtro - Mgridt = Iow +Mvr

Fdtro - Mgridt = mK2 v/r1 + Mvr1

85 x 3x 0.345 -11 x 9.81 x 0.23 x 3 =30 x 0.25 x 0.25 x v/0.23 + 11 x v x 0.23 =

v = 1.2654m/s

To find the timed average value

Tdt -Mgdt =MV

T x 3 - 11 x 9.81 x 3 = 11 x 0.778

T= 110.76N

3 0

2 years ago

If the angle of elevation of the cannon is decreased from 35 to 30 degrees, the vertical component of the ball's initial velocit

stealth61 [152]

Answer:

Determine the initial horizontal and vertical velocity for a dart launched with a ... an angle of 30 degrees. Vi=4m/s Vix Vicoso Mas cos 30 = = 30°. Viy=visine ... What is the resultant direction (angle) of the cannon ball (relative to a fixed frame)? ... Which of the following launch angles would result in the greatest time of flight iſ the.

Explanation:

I hope it helped

4 0

2 years ago

While practicing S-turns, a consistently smaller half-circle is made on one side of the road than on the other, and this turn is

erica [24]

Answer:

The answer is "4-5-6 because the bank is growing too quickly in the beginning of its turn".

Explanation:

The S-turns is also known as the reference technique, in which the ground track of the aircraft on both sides of a defined ground-based straight-line distance represents 2 different but equivalent circles.

Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

5 0

1 year ago

Match each label to the boundary it describes. convergent boundary new crust forms transform boundary crust submerges into the m

Paraphin [41]

The answers would be:

CONVERGENT boundary - Crust submerges into the mantle

TRANSFORM boundary - neither forms nor submerges

DIVERGENT boundary - new crust forms

If you'd like to know more about the different boundaries, read on:

Convergent boundaries occur when two plates move TOWARDS each other. The event where crust submerges into the mantle is called <em><u>subduction</u></em> and this occurs when an oceanic plate and a continental plate collide. The oceanic plate is more dense and thinner than the continental plate, so it slides under it.

Transform boundaries occur when two plates slide against each other. They move slide side by side, so nothing is formed nor do they go under each other. Although, this type of boundaries create strong earthquakes.

Lastly, divergent boundaries occur when two plates move apart. The separation creates a way for magma to come up. New crust is formed when the magma that seeps out is cooled by its cooler surroundings. This is observed in the mid oceanic ridge.

7 0

2 years ago

Read 2 more answers

A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener

Bingel [31]

The ball was dropped from a height 20 meters

Explanation:

The given is

1. A ball is dropped from the top of a cliff

2. By the time it reaches the ground, all the energy in its gravitational

potential energy store has been transferred into its kinetic energy

store, that mean K.E = P.E

3. The ball is travelling at 20 m/s when it hits the ground

4. The gravitational field strength is 10 N/kg

We need to find the height that the ball dropped from it

The ball dropped from the top of a cliff means the initial speed is 0

→ K.E = $\frac{1}{2}m(v^{2}-v_{0}^{2})$

where v is the final speed, $v_{0}$ in the initial speed and m

is the mass

→ v = 20 m/s and $v_{0}$ = 0 m/s

→ K.E = $\frac{1}{2}m(20^{2}-0^{2})$

→ K.E = $\frac{1}{2}m(400)$

→ K.E = 200 m joules ⇒ when the ball hits the ground

→ P.E = m g h

where g is the gravitational field strength, m is the mass and h is

the height

→ g = 10 N/kg

→ P.E = m(10)(h)

→ P.E = 10 m h joules

→ P.E = K.E

→ 10 m h = 200 m

Divide both sides by 10 m

→ h = 20 meters

The ball was dropped from a height 20 meters

Learn more

You can learn more about gravitational potential energy in brainly.com/question/1198647

#LearnwithBrainly

8 0

2 years ago