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Dominik [7]
4 years ago
7

If the total momentum of a system is changing:

Physics
1 answer:
DENIUS [597]4 years ago
3 0

Answer:

(d) a net external force must be acting on the system

Explanation:

Momentum is given as the product of mass and velocity.

P = MV

According to Newton's second law of motion, " Force applied to a body (system) is directly proportional to the rate of change of momentum of the body (system) which takes place in the direction of the applied force (external force).

F ∝ΔMV

Therefore, If the total momentum of a system is changing, a net external force must be acting on the system.

(d) a net external force must be acting on the system

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The internet is most useful to them because they use it to communicate.

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How much time will it take for a bug to travel 5 meters across the floor if it is
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It will take 5 seconds

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To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit
DiKsa [7]

To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

R = 350ft

a_t = 1.1ft/s^2

PART A )

a_c = \frac{V^2}{R}

a_c = \frac{V^2}{350}

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 5.25ft/s^2

a = \sqrt{a_t^2+a_r^2}

5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

27.5625 = 1.21 + \frac{v^4}{122500}

v=42.3877ft/s

Now calculate the angular velocity of the motorcycle

v = r\omega

42.3877 = 350\omega

\omega = 0.1211rad/s

Calculate the angular acceleration of the motorcycle

a_t = r\alpha

1.1 = 350\alpha

\alpha = 3.1428*10^{-3}rad/s^2

Calculate the time needed by the motorcycle to reach an acceleration of

5.25ft/s^2

\omega = \alpha t

0.1211 = 3.1428*10^{-3}t

t = 38.53s

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a_r = \frac{v^2}{R}

a_r = \frac{21.5^2}{350}

a_r =1.3207ft/s^2

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a = \sqrt{a_t^2+a_r^2}

a = \sqrt{(1.1)^2+(1.3207)^2}

a = 1.7187ft/s^2

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

3 0
3 years ago
"determine the resultant internal loadings acting at the cross sections at points f and g of the frame. set θ = 27º and t = 178
Leviafan [203]

Hi you didn't provide any images to solve the question, hence I am going to solve a different question of same concept so you can have an idea how to tackle such types of questions.(please refer to the attachment for question)

Answer:

<u> Please refer to the attachment for answers and explanation</u>

Explanation:

<u> Please refer to the attachment for answers and explanation</u>

5 0
4 years ago
An electron has a charge of 1.602 X 10-19.coulomb. When two electrons are separated by 1.2 X 10-9m, what force will they exert o
amid [387]

Answer:

The force they will exert on each other is 1.6*10⁻¹⁰ N

Explanation:

The electromagnetic force is the interaction that occurs between bodies that have an electric charge. When the charges are at rest, the interaction between them is called the electrostatic force. Depending on the sign of the interacting charges, the electrostatic force can be attractive or repulsive. The electrostatic interaction between charges of the same sign is repulsive, while the interaction between charges of the opposite sign is attractive.

Coulomb's law is used to calculate the electric force acting between two charges at rest. This force depends on the distance "r" between the electrons and the charge of both.

Coulomb's law is represented by:

F=k*\frac{q1*q2}{r^{2} }

where:

  • F = electric force of attraction or repulsion in Newtons (N). Like charges repel and opposite charges attract.
  • k = is the Coulomb constant or electrical constant of proportionality.
  • q = value of the electric charges measured in Coulomb (C).
  • r = distance that separates the charges and that is measured in meters (m).

In this case:

  • k= 9*10⁹ \frac{N*m^{2} }{C^{2} }
  • q1= 1.602*10⁻¹⁹ C
  • q2= 1.602*10⁻¹⁹ C
  • r= 1.2*10⁻⁹ m

Replacing:

F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{1.602*10^{-19} C*1.602*10^{-19} C}{(1.2*10^{-9} )^{2} }

and solving you get:

F=1.6*10⁻¹⁰ N

<u><em>The force they will exert on each other is 1.6*10⁻¹⁰ N</em></u>

3 0
3 years ago
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