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Anastaziya [24]
4 years ago
7

Question 1

Physics
1 answer:
Yuki888 [10]4 years ago
3 0

Joseph is conducting an investigation on how two objects compare in mass, volume, and force with each other. When determining the  gravitational force between the two objects, the following characteristics of the objects should be considered :

<u>MASS AND DISTANCE</u>

Explanation:

  • The gravitational force is a force that attracts any two objects with mass. In fact, every object, including you, is pulling on every other object in the entire universe. This is called Newton's Universal Law of Gravitation
  • The strength of the gravitational force between two objects depends on two factors, mass and distance.
  • The force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.
  • When two objects are gravitational locked, their gravitational force is centered in an area that is not at the center of either object, but at the barycenter of the system. The principle is similar to that of a see-saw.
  • Gravity is often assumed to be the same everywhere on Earth, but it varies because the planet is not perfectly spherical or uniformly dense. In addition, gravity is weaker at the equator due to centrifugal forces produced by the planet's rotation.

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Vector A is in the direction 44.0 degrees clockwise from the y-axis. The x-component of A Ax=-15.0 m. Part A: What is the y-comp
Gnesinka [82]

Answer:

a)  Y component of the vector =15.54 m

b) Vector magnitude = 21.6 m

Explanation:

The given vector makes 44 degree angle with Y axis, as given. This is same as 90 -44 = 46 degrees with the horizontal or X axis.

b) X component of the given vector = A_{x} = A cos 46 =15

⇒ A = 15/cos 16 = 21.6 m = Total vector magnitude.

a) Y component of the vector = 21.6 sin 46 = 15.54 m

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3 0
4 years ago
A car with mass m-1000 kg completes a turn of radius r 540 m at a constant speed of v =25 m/s. As the car goes around the turn,
Nataly [62]

Answer:

friction coefficient on the rough road is 0.117

Explanation:

As we know that car is taking turn on flat rough road

So here we can say that required centripetal force for circular motion of car is due to frictional force

So we have

F_c = \frac{mv^2}{R}

now we have

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so we have

\mu = \frac{v^2}{Rg}

\mu = \frac{25^2}{540(9.81)}

\mu = 0.117

6 0
3 years ago
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