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fomenos
3 years ago
14

Find the magnitude of the sum of these two vectors: 101 m 60.0 ° 85.0 m

Physics
1 answer:
attashe74 [19]3 years ago
8 0

Answer: 161.3

I have a acellus too and got this question correct, so I hope this helps y’all out

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Difference between constant speed and average speed
Tresset [83]
If a body is traveling with constant speed , it means that it's distance is constantly increasing with time.

If it goes 5m in 3min , it will go 5m in next 3 min.

Average velocity is the total displacement divided by total time.

When the body travels 5m in 3 min and 25 m in next 3min , average velocity
=(25+5)/(3+3)
=30/6
=5m/min
8 0
3 years ago
Consider a transition at 5000 Å with a width of 1 Å and a cavity 2 cm3 in volume. How many electromagnetic modes exist in this f
Lena [83]

Answer:

total number of modes is 8

Explanation:

attached here is the calculations

3 0
3 years ago
As the speed/velocity of an object increases what happens to the kinetic energy of the object
Elanso [62]
The kinetic energy increases
8 0
3 years ago
Read 2 more answers
In order for work to happen you MUST have?
Nataly_w [17]

Explanation:

it requires energy

hope it helps

7 0
3 years ago
Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea
Setler [38]

Answer:

a)  M = 4,997 10²⁰ kg ,  b)   T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

          v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

         a = (v₀ - v) / t

         a = (15 - (-15)) /9.00 = 30/9

         a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

          F = m a

         G m M / r² = m a

         M = a r² / G

Let's calculate

         M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

         M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

         R = R_{m} + h

         R = 1 10⁵ + 2.00 10⁴

         R = 12 10⁴ m

         F = m a

        G m M / R² = m a

Centripetal acceleration is

         a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

        d = 2π R

        v = d / t

        v = 2π R / T

Let's replace

        G m M / R² = m (2π R / T)² / R

        G M = R³ 4π² / T²

        T² = 4π² R³ / G M

       T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

       T² = 6.82 10¹⁶ / 3.33 10¹⁰

       T = √ (2,048 10⁶)

       T = 1.43 10³ s

3 0
3 years ago
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