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Korolek [52]
3 years ago
14

according to newton's second law of motion of the net force acting on the object increases while the mass of the object remains

constant, what happens to the acceleration?
Physics
1 answer:
Licemer1 [7]3 years ago
4 0

Answer:

The Acceleration will increase

Explanation:

Newton's Second Law of motion: It states that the rate of change of momentum is directly proportional to the applied force and takes places along the direction of the force.

It can be expressed mathematically as,

F ∝ m(v-u)/t

Where (v-u)/t = a

F  = kma.

F = force, m = mass of the body, a = acceleration, k = constant of proportionality which tend to unity for a unit force, a unit mass, and a unit acceleration.

Therefore,

F = ma.

From the equation above,

If the net force acting on a body increase, while the mass of the body remains constant, the acceleration will also increase.

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The illustration shows a rollercoaster and indicates four different positions the car might be at as it moves along the track. A
Vanyuwa [196]

Answer:

Point a

Explanation:

The potential energy of an object is given by :

P = mgh

m is mass, g is acceleration due to gravity, h is height above ground level.

Potential energy is directly proportional to the position of an object.

In the attached figure, the maximum height is shown at point (a). It means it will have maximum potential energy at a as compared to b,c and d.

5 0
2 years ago
A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge
Digiron [165]

Answer:

F = 2.49 x 10⁻⁹ N

Explanation:

The electrostatic force between two charged bodies is given by Colomb's Law:

F = \frac{kq_1q_2}{r^2}\\

where,

F = Electrostatic Force = ?

k = colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

q₂ = second charge = 1.4 C

r = distace between charges = 0.9 m

Therefore,

F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.4\ C)}{(0.9\ m)^2}

<u>F = 2.49 x 10⁻⁹ N</u>

8 0
3 years ago
Read 2 more answers
A basketball player can leap upward 0.43 m. how long does he remain in the air? use an acceleration due to gravity of 9.80 m/s2
MArishka [77]
From the equations of linear motion,
v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,
Thus, v² = u² -2gs, but v=0
hence, u² = 2gs
                = 2×9.81×0.43
                = 8.4366
            u = √8.4366
               =2.905 m/s
Hence the initial velocity is 2.905 m/s
 Then using the equation v= u +gt .
Therefore, v = u -gt. (-g because the player is jumping against the gravity)
but, v = 0
Thus, u= gt
Hence, t = u/g
              = 2.905/9.81
              = 0.296 seconds


3 0
3 years ago
A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it
makkiz [27]

Answer:D

Explanation:

Given

mass of object m=5 kg

Distance traveled h=10 m

velocity acquired v=12 m/s

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy=mgh=5\times 9.8\times 10=490 J

Final Energy=\frac{1}{2}mv^2+W_{f}

=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}

where W_{f} is friction work if any

490=360+W_{f}

W_{f}=130 J

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0
3 years ago
As an object falls to the ground, its potential energy is being converted to kinetic energy.
alekssr [168]

Answer:

<em>The statement is true</em>

Explanation:

<u>Energy Conversion </u>

When an object starts to fall in free air, it speeds up as it falls. The force of gravity acting on the object causes energy to be transferred from its gravitational potential energy to its kinetic energy. We can safely say the height converts to speed and vice-versa. If no external forces act on the system, we can easily calculate heights and speeds by knowing the total mechanical energy (gravitational potential plus kinetic) is conserved.

Answer:

\boxed{\text{The statement is true}}

5 0
3 years ago
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