All categories

2 years ago

As your rockets went upwards how would you describe how the energies changed?

1 answer:

4 0

As a rocket increases height and slows down, it gains more and more potential energy and loses more and more kinetic energy. Potential energy is store energy (usually determined by height), and kinetic energy increases as speed increases.

You might be interested in

What are some types of landforms on Earth’s surface? PLS ANSWER QUICK 11 POINTS

Brrunno [24]

Answer:

plateau, mountains, hills, plains

5 0

3 years ago

Read 2 more answers

Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a forc

Maslowich

Answer:

mass of box 1 = 2.20 kg

mass of box 2 = 5.93 kg

Explanation:

Let the mass of box 1 and box 2 is respectively

$m_1$ and $m_2$

so we will have

Force applied on box 1 then acceleration

$a = \frac{F}{m_1 + m_2}$

$1.50 = \frac{12.2}{m_1 + m_2}$

$m_1 + m_2 = 8.13$

Now we know that contact force between them in above case is given as

$F_n = m_2a$

$8.90 = m_2(1.5)$

$m_2 = 5.93 kg$

now we have

$m_1 = 2.20 kg$

8 0

2 years ago

A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu

malfutka [58]

Answer:

Acceleration, $a=-2.48\ m/s^2$

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}$

$a=-2.48\ m/s^2$

So, the acceleration on the rough ice $-2.48\ m/s^2$ and negative sign shows deceleration.

8 0

3 years ago

Describe the relationship between motion and a reference point.

steposvetlana [31]

Answer:

An object is in motion when its distance from another object is changing. ... A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point. You assume that the reference point is stationary, or not moving.

Explanation:

4 0

3 years ago

Read 2 more answers

Please help, my homework is due in a couple of days and I'm stuck. A plane is travelling 10.0 meters per second (this means the

ss7ja [257]

Answer:

< 25 m/s
triangle inequality
between north and east
45° < angle < 60°

Explanation:

(a) Just as one-dimensional numbers add on a number line by putting them end-to-end, so two-dimensional numbers add on a coordinate plane the same way.

Here, we choose to let the positive y-axis represent North, and the positive x-axis, East. This is the way a map is conventionally oriented. The velocity of the plane is represented by a vector pointing north (up). Its length represents the magnitude of the velocity. Likewise, the wind is represented by a vector of length 15 pointing east (right). The sum of these is the hypotenuse of the triangle they form.

The magnitude of the sum can be found here using the Pythagorean theorem, but for the purpose of this question, you're not asked to find that.

Instead, you're asked to estimate whether it is more or less than 25 (m/s).

Your knowledge of the triangle inequality will tell you that the hypotenuse (resultant) must be shorter than the sum of the lengths of the sides of the triangle, hence must be less than 10+15 = 25.

(b) The triangle inequality says the resultant is less than the sum of the other two sides of the triangle.

(c) Since the wind is blowing the plane toward the east, but the plane is traveling toward the north, the resulting direction is somewhere between north and east.

(d) "Somewhere between north and east" can be expressed as the inequality ...

0° < angle < 90°

4 0

3 years ago