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3 years ago

Select the correct answer.

Physics

2 answers:

Lemur [1.5K]3 years ago

8 0

Answer:

B.Temperature and density increase with layer thickness.

Explanation:

Serhud [2]3 years ago

5 0

Answer:

b hope this helps

Explanation:

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First Stan txt (tomorrow by together) and stream freeze on YT Hybe labels !!!!!!!!!!!!!

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3 years ago

I need help with #4 and #5?

algol [13]

Explanation:

just pick a

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2 years ago

A small plane flies 40.0 km in a direction 60° north of east and then flies 30.0 km in a direction 15° north of east. Use the an

lana66690 [7]

Answer:

d= 64.7 km

$\theta = 40.9^{o}$

displacement vector $=r_xi + r_yj = 48.9i + 42.4j$

Explanation:

total distance = 40 + 30 = 70 km

during 1st flight

$r_1 x = 40*cos60$

$r_1 x = 20 km$

$r_1 y = 40*sin60$

$r_1 y = 34.64 km$

during 2nd flight

$r_2 x = 30*cos15$

$r_2 x = 28.9 km$

$r_2 y = 30*sin15$

$r_2 y = 7.76 km$

the two component of r are:

$r_x = r_1x + r_2x = 20 + 28.9 = 48.9 km$

$r_y = r_1y + r_2y = 34.64 + 7.76 = 42.4 km$

Geographical Direction $\theta = tan^{-1}\frac{r_y}{r_x} [tex]\theta = 40.9^{o}$

Displacement d $= \sqrt{r_x^2 + r_y^2}$

$d = \sqrt{48.9^2+42.4^2} = 64.7 km$

d= 64.7 km

displacement vector $=r_xi + r_yj = 48.9i + 42.4j$

8 0

3 years ago

A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it f

Rufina [12.5K]

Answer:

Kf = 470 mJ

Explanation:

According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.
Since the puck is initially at rest, the change is kinetic energy is just the final kinetic energy of the puck.
Assuming that the net force is horizontal, and causes a horizontal displacement also, we can find the net work on the puck as follows:

$W_{net} = F_{net} * \Delta X = 0.56 N * 0.84 m = 0.47 J = 470 mJ (1)$

As we have already said, (1) is equal to the final kinetic energy of the puck:
⇒ Kf = 470 mJ (2)

8 0

2 years ago

An object of unknown mass is hung on the end of an unstretched spring and is released from rest. The acceleration of gravity is

lakkis [162]

Answer:

0.33 s

Explanation:

For this case, as the object is hung on the end of an unstretched spring, we can consider this system as a simple pendulum.

For this system, we can determine the period of the motion using the following formula:

T = 2π√(L/g)

Where: T = period (in sec), L = lenght of the spring, g = acceleration of garvity = 9.8 m/s²

By the exact time the object is 2.75 cm before coming to rest, that will be the lenght of the spring we can consider (2.75 cm = 0.0275 m)

Finally:

T = 2π√(0.00275/9.8)

T = 0.33 sec

4 0

3 years ago