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Bas_tet [7]
3 years ago
15

What is the center of our solar system?

Physics
1 answer:
Yakvenalex [24]3 years ago
8 0
The sun is the centre of our solar system
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Chapter 05, Problem 15 Multiple-Concept Example 7 and Concept Simulation 5.2 review the concepts that play a role in this proble
Llana [10]

Answer:

Explanation:

The question relates to motion on a circular path .

Let the radius of the circular path be R .

The centripetal force for circular motion is provided by frictional force

frictional force is equal to μmg , where μ is coefficient of friction and mg is weight

Equating cenrtipetal force and frictionl force in the case of car A

mv² / R = μmg

R = v² /μg

= 26.8 x 26.8 / .335 x 9.8

= 218.77 m

In case of moton of car B

mv² / R = μmg

v²  = μRg

= .683  x 218.77x 9.8

= 1464.35

v = 38.26 m /s .

3 0
3 years ago
First extinguish a match or candle by blasting it violently. Why?​
Andreas93 [3]

Answer:

The wax vapor on burning candles could reignite the flame.

Explanation:

7 0
3 years ago
It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft
Paul [167]

Answer: 1.23\ m/s^2

Explanation:

Given

At an elevation of y=34.7\ km, spacecraft is dropping vertically at a speed of u=293\ m/s

Final velocity of the spacecraft is v=0

using equation of motion i.e. v^2-u^2=2as

Insert the values

\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2

Therefore, magnitude of acceleration is 1.23\ m/s^2.

8 0
2 years ago
Assume we’re able to travel to your planet and decide to take some fireworks with us to celebrate our journey.
julia-pushkina [17]

Answer:

The horizontal distance covered by the firework will be \frac{1876.8}{g}

Explanation:

Let acceleration due to gravity on the planet be g, initial velocity of the firework be u and angle made with the horizontal be ∅.

writing equation of motion in vertical direction:

v_{y}=u_{y}+(-g) t

u_{y}= u\sin \phi

and v_{y}=0

therefore \frac{u\sin \phi }{g} =t

writing equation of motion in horizontal direction:

s_{x}=u_{x}t

u_{x} = u\cos \phi

therefore the equation becomes s_{x}=\frac{u^{2}   \sin \phi  \cos \phi}{g}

therefore horizontal distance traveled =\frac{u^{2}\sin 2\alpha \phi }{2g}=\frac{1876.8}{g}\frac{m}{s}

5 0
3 years ago
A toy plane is flying in a horizontal circle by being attached to a 0.75 meter string. The plane has a mass of 101.7 grams and m
nadezda [96]

Answer:

F = 2,894 N

Explanation:

For this exercise let's use Newton's second law

      F = m a

The acceleration is centripetal

     a = v² / r

Angular and linear variables are related.

     v = w r

Let's replace

     F = m w² r

The radius r and the length of the rope is related

    cos is = r / L

    r = L cos tea

Let's replace

    F = m w² L cos θ

Let's reduce the magnitudes to the SI system

     m = 101.7 g (1 kg / 1000g) = 0.1017 kg

      θ = 5 rev (2π rad / rev) = 31,416 rad

     w =  θ / t

     w = 31.416 / 5.1

     w = 6.16 rad / s

     F = 0.1017 6.16² 0.75 cos  θ

     F = 2,894 cos  θ

The maximum value of F is for  θ equal to zero

     F = 2,894 N

7 0
3 years ago
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