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N76 [4]
3 years ago
5

To understand Newton's 1st law. Newton's Principia states this first law of motion: An object subject to no net force maintains

its state of motion, either at rest or at constant speed in a right line. This law may be stated as follows: If the sum of all forces acting on an object is zero, then the acceleration of that object is zero. Mathematically this is just a special case of the 2nd law of motion, F =ma , when F =0. When studying Newtonian mechanics, it is best to remember the 1st law in two ways: If the net force (i.e., sum of all forces) acting on an object is zero, the object will keep moving with constant velocity (which may be zero). If an object is moving with constant velocity, that is, with zero acceleration, then the net force acting on that object must be zero.
Part A
If a car is moving to the left with constant velocity, one can conclude that ?
(a) there must be no forces applied to the car.
(b) the net force applied to the car is directed to the left.
(c) the net force applied to the car is zero.
(d) there is exactly one force applied to the car.
Part B
An object cannot remain at rest unless

(a) there are no forces at all acting on it.
(b) the net force acting on it is zero.
(c) the net force acting on it is constant.
(d) there is only one force acting on it.
Part C

An object will have constant acceleration if?
(a)there are no forces at all acting on it.
(b)the net force acting on it is zero.
(c)the net force acting on it is constant in magnitude and direction.
(d)there is only one force acting on it.
Physics
1 answer:
Doss [256]3 years ago
7 0

Explanation:

Part A,

If a car is moving to the left with constant velocity, the change in velocity of the car is equal to zero. So the acceleration of the car is zero. As a result, the the net force applied to the car is zero.

So, the correct option is (c) "the net force applied to the car is zero".

Part B,

An object cannot remain at rest unless the net force acting on it is zero. This is Newton's first law of motion.

So, the correct option is (b).

Part C,

An object will have constant acceleration, the net force acting on it is constant in magnitude and direction. The net force acting on the object is directly proportional to its acceleration.

So, the correct option is (c).

Hence, this is the required solution.

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natka813 [3]
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3 0
2 years ago
A 25N force is acting on a body moving on a straight line with Initial momentum 20 kam's. Find the final momentum after 4 second
Nezavi [6.7K]

The final momentum of the body is equal to 120 Kg.m/s.

<h3>What is momentum?</h3>

Momentum can be described as the multiplication of the mass and velocity of an object. Momentum is a vector quantity as it carries magnitude and direction.

If m is an object's mass and v is its velocity then the object's momentum p is: {\displaystyle \mathbf {p} =m\mathbf {v} . The S.I. unit of measurement of momentum is kg⋅m/s, which is equivalent to the N.s.

Given the initial momentum of the body = Pi = 20 Kg.m/s

The force acting on the body, Pf = 25 N

The time, Δt = 4-0 = 4s

The Force is equal to the change in momentum: F ×Δt = ΔP

25 × 4 = P - 20

100 = P - 20

P = 100 + 20 = 120  Kg.m/s

Therefore, the final momentum of a body is 120 Kg.m/s.

Learn more about momentum, here:

brainly.com/question/4956182

#SPJ1

5 0
8 months ago
A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he
zmey [24]

Answer:

851.33 N

Explanation:

Using newton;s equation of motion,

v² = u² + 2gh ......... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height

Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

v² = 196

v = √196

v = 14 m/s.

Note: As the Diver touches the water,  u = 14 m/s and v = 0 m/s( stopped)

Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

Given: v = 0 m/s, u = 14 m/s t = 2.10 s

Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest

making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

F = 851.33 N

Hence the upward force the water exert on her = 851.33 N

8 0
2 years ago
What type of plate boundary decreases the amount of the Earth's crust?
rosijanka [135]

Answer:

Convergent.

Explanation:

Just as oceanic crust is formed at mid-ocean ridges, it is destroyed in subduction zones. Subduction is the important geologic process in which a tectonic plate made of dense lithospheric material melts or falls below a plate made of less-dense lithosphere at a convergent plate boundary.

3 0
2 years ago
A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ
valentinak56 [21]

Explanation:

The given data is as follows.

                    mass = 0.20 kg

              displacement = 2.6 cm

              Kinetic energy = 1.4 J

       Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

         Total energy = Kinetic energy + spring potential energy

                           = 1.4 J + 2.2 J

                            = 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So,      K.E = Total energy

                = 3.6 J

Also, we know that

                  K.E = \frac{1}{2}mv^{2}_{m}

or,                   v = \sqrt{\frac{2K.E}{m}}

                        = \sqrt{2 \times 3.6 J}{0.2 kg}

                        = \sqrt{36}

                        = 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0
3 years ago
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