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Likurg_2 [28]
3 years ago
12

Please help me with question B.

Physics
1 answer:
Step2247 [10]3 years ago
5 0
It accelerates in the y component (bc of gravity) AND the x-component (b/c of the friction force).
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A boy runs 400m at an average speed of 4.0m/s he runs the first 200m in 40 s how long does he take to run the second 200m?
IRISSAK [1]
If he runs at the same speed he will cover next 200m in 40s
that is at the average of 4.0m
8 0
4 years ago
How much force would a 55kg person experience on a roller coaster if they travel through a loop with a radius of 55kg at a speed
Inessa [10]

Answer:

200 N

Explanation:

For a body moving in uniform circular motion, the force acting on it will be <em>centripetal force</em> and its direction is <em>radially inward</em> , pointing to the center.

The radially inward acceleration, or the centripetal acceleration is given by :

                                          a = v² / r

           where v is the speed at which the body is moving and r is the radius of the circle

Given-

m = 55kg

v = 14.1 m/s

r= 55m

We know that F = ma

⇒   F = m (  v²/ r )

⇒ F = 55 x 14.1 x 14.1 / 55

⇒ F =14.1 x 14.1 = 200 N

∴ <em>The force acting is 200 N</em>.

5 0
3 years ago
what is a point of view of an object used to determine another obejects motion i nedd help asap plsss​
Igoryamba
A believe that’s called a reference point.
9 0
3 years ago
0.
larisa [96]

Answer:21

Explanation:

4 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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