If an object is being pulled by two forces, one 4 N to the left and the other 2 N to the right, what is the net force acting on the object?

Answers

Answer 1
Answer:

Since this object is being pulled by two forces, the net  force acting on the object is 2 Newton, to the left.

  • Let the first force be A.
  • Let the second force be B.

Given the following data:

  • Force A = 4 Newton, to the left.
  • Force A = 2 Newton, to the right.

To find the net  force acting on the object:

A net force can be defined as the vector sum of all the forces that are acting on a physical object or body.

This ultimately implies that, a net force is a single (one) force that substitutes the effect of all the forces acting on a physical object or body.

Mathematically, net force is given by the following:

Net \;force = F_1 + F_2 + F_3 + F_4 + .....F_n

Where:

  • F represents each force.

In this scenario, the forces are acting in opposite directions (left and right) and as such, we would subtract the two forces:

Net\;force = F_A - F_B\n\nNet\;force = 4 - 2

Net force = 2 Newton, to the left.

Note: The direction is to the left because Force A has more magnitude than Force B, so the object would be pulled to the left.

Read more: brainly.com/question/22210180

Answer 2
Answer:

Answer:

2N

Explanation:

subtract  rthe two forces to see which is greater

4-2=2


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How does mass effect my room?

Answers

It is by how much weight there is in your room and what can make your room the double sized of your whole room.

near the earth surface the amount of mass an object has does not affect its acceleration due to gravity

Answers

Answer:

True

Explanation:

As per the gravitational law, the gravitational force (F) can be expressed as

F = (GMm)/(r^(2) ) -------------  (1)

where,

F= Gravitational force

G= Gravitational constant

M= Mass of the Earth

m= Mass of the object

r= Distance between Earth and object

Also, the force due to gravity,

F=mg -------------  (2)

where,

F= Gravitational force

g= Acceleration due to gravity

m= Mass of the object

Deducing both the equations

mg = (GMm)/(r^(2) )\n g = (GM)/(r^(2) )\n

From the above equation it is clear that the acceleration due to gravity is independent of the mass of the object.

2. A force of 24x10² N exists between a positive charge of 8x10-5C and a positive charge of 3x10C.What distance separates the charges?

Answers

Answer: 94.9 metres

Explanation:

Given that,

Force between the two charges (F) = 24x10² N

Negative charge (q1) = 8x10-5C

Positive charge (q2) = 3x10C

Distance between charges (r) = ?

Apply the formula for

F= (kq1q2) / r²

Recall that k is Coulomb's constant with a value of 9 × 10^9 Nm2C−2

Then,

24x10² N = (9 × 10^9 Nm2C−2 x 8x10-5C x 3x10C) / r²

24x10² N = (2.16 x 10^7) / r²

r² = (2.16 x 10^7 / 24x10² N)

r² = 9000

To get the value of r, find the square root

r = √9000

r = 94.868m (rounded to the nearest tenth as 94.9m)

Thus, a distance of 94.9 metres separates the charges.

Does the bowling ball have more potential energy or kinetic energy just before it hits the ground? Why?Answer:
EXPLAIN YOUR ANSWER OR YOU WILL NOT GET POINTS

Answers

Answer:

Explanation:

kenetic energy because it already hit the ground

Why are frames of reference important for motion

Answers

Because frame of reference decides are your forces positive or negative. It is important for the direction of motion.

Listen →A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared east for 30 seconds. What is the speed
of the cart at the end of this 3.0 second interval?

Answers

Answer:

V = 20.5 m/s

Explanation:

Given,

The mass of the cart, m = 6 Kg

The initial speed of the cart, u = 4 m/s

The acceleration of the cart, a = 0.5 m/s²

The time interval of the cart, t = 30 s

The final velocity of the cart is given by the first equation of motion

                              v = u + at

                                  = 4 + (0.5 x 30)

                                = 19 m/s

Hence the final velocity of cart at 30 seconds is, v = 19 m/s

The speed of the cart at the end of  3 seconds

                                    V = 19 + (0.5 x 3)

                                       = 20.5 m/s

Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s