Which of the following describes the net force acting on an object?The sum of all forces acting on an object
The gravitational force minus any contact forces acting on an object
The difference between the normal force and the gravitational force acting on an object
The sum of all the forces acting on an object in the same direction

Answers

Answer 1
Answer:

The sum of all forces acting on an object in the same direction is described for the net force acting on an object.

What is a Net force?

  • When the forces are acting in the same direction of movement of the object it can be said as sum of the two individual forces will be equal to the "Net Force" .
  • The net force is the combined force of all individual forces acting on an object.
  • If the object with the forces in the opposite direction, then the net force will not be equal to the sum of the forces.

Example : If two forces (2 kids pushing in the same direction to move the object big box) act on an object (big box) in the same direction, then the net force is equal to the sum of the two forces. If the kids pushed in the opposite direction, the net force will not occur.

Hence, Option D is the correct answer.

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Answer 2
Answer:

Answer:

The sum of all the forces acting on an object in the same direction.


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A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force

Answers

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

Answer all three parts and show work.

Answers

The distance for both Parts A and B are given in the question.

A balloon drifts 140m toward the west in 45s.

The wind suddenly changes and the balloon flies 90m toward the east in the next 25s.

To find the total distance, we can just add.

140 + 90 = 230m

Best of Luck!

wo balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L = 0.550 m. One is attached at one end of the rod and the other at the middle of the rod. If the rod is held by the open end and rotates in a circular motion with angular speed of 45.6 revolutions per second,

Answers

Answer:

  T_1 =677224.40\ N

Explanation:

given,

mass of the both ball = 5 Kg

length of rod = 2 L            

where L = 0.55 m            

angular speed = 45.6 rev/s

ω = 45.6 x 2 π                      

ω = 286.51 rad/s                

v₁ = r₁ ω₁                        

v₁ =0.55 x 286.51 = 157.58 m/s

v₂ = r₂ ω₂                                

v₂ = 1.10 x 286.51 = 315.161 m/s

finding tension on the first half of the rod

r₁ = 0.55  r₂ = 2 x r₁ = 1.10

  T_1 = m ((v_1^2)/(r_1)+(v_2^2)/(r_2))

  T_1 = 5 ((157.58^2)/(0.55_1)+(315.161^2)/(1.1))

  T_1 =677224.40\ N

A mouse is running across a room with a speed of 2.2 m/s. The mass of the mouse is 1.4 kg. What is the Kinetic Energy of the mouse?

Answers

Answer:

3.39 J

Explanation:

The kinetic energy of an object can be found by using the formula

k =  (1)/(2) m {v}^(2)  \n

m is the mass

v is the velocity

From the question we have

k =  (1)/(2)  * 1.4 *  {2.2}^(2)  \n  = 0.7 * 4.84 \n  = 3.388 \:  \:  \:  \:  \:  \:  \:  \:  \:

We have the final answer as

3.39 J

Hope this helps you

Answer:

im not sure

Explanation:

A point charge that is exactly q =3 mu or micro CC is at the origin. In this problem, assume that the Coulomb constant k = 8.99 times 109 N m2/C2 exactly.(a) Find the potential V on the x axis at x = 3.00 m and at x = 3.01 m. In this part, enter your answers to exactly 6 signfificant figures.

Answers

Answer:

a) 8,990.00 V   b) 8,960.13 V

Explanation:

a) The potential due to a point charge, can be found from the expression of  Coulomb's Law, as follows:

V = (k*q)/(r)

where k = 8.99*10⁹ N*m²/C², q = 3.00*10⁻⁶ C, and r = 3.00 m.

Replacing by this values, we can find the potential V as follows:

V = (8.99e9 N*m2/C2*3.00e-6C)/(3.00m) = 8,990.00 V

b) Repeating the process for r = 3.01m:

V = (8.99e9 N*m2/C2*3.00e-6C)/(3.01m) = 8,960.13 V

Final answer:

The potential V at x=3.00 m and x=3.01 m from a point charge at the origin is 8.99 * 10^3 V and 8.97 * 10^3 V, respectively. This calculation is based on Coulomb's Law.

Explanation:

The potential V at a distance r from a point charge q is given by Coulomb's Law:

V = k*(q/r)

Here, k = 8.99 * 10^9 N*m^2/C^2 is the Coulomb constant, q = 3 µC is the charge, and r is the distance from the origin along the x-axis. For x = 3.00 m and x = 3.01 m, we can substitute these values into the equation to find V:

  • V1 = ((8.99 * 10^9 N*m^2/C^2)* (3 * 10^-6 C)) / 3.00 m = 8.99 * 10^3 V (to 6 significant figures)
  • V2 = ((8.99 * 10^9 N*m^2/C^2)* (3 * 10^-6 C)) / 3.01 m = 8.97 * 10^3 V (to 6 significant figures)

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Each mass in the figure is 3 kg. Find the magnitude and direction of the net gravitational force on mass A due to the other masses.A. 2.45 × 10–7 N toward B
B. 3.75 × 10–7 N toward C
C. 2.00 × 10–7 N toward D
D. 1.15 × 10–7 N toward D

Answers

The magnitude and direction of the net gravitationalforce on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this case, all of the masses are equal to 3 kg, and the distance between mass A and mass D is 3 m.

The gravitational force between mass A and mass D is therefore:

F = G * m_A * m_D / r²

= 6.674 × 10⁻¹¹ N m² / kg² * 3 kg * 3 kg / 3 m²

= 1.15 × 10⁻⁷ N

The direction of the gravitational force is towards mass D.

Therefore, the net gravitational force on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

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Answer:

THE ANSER IS B

Explanation: