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tankabanditka [31]
3 years ago
15

What is the effect on acceleration if the velocity is increased by 4 times?

Physics
1 answer:
Liula [17]3 years ago
4 0

Answer:

If velocity is higher, then traversed distance during any time increment dt is higher. This tells you that the amount of energy delivered to the object is proportional to the velocity, which is proportional to elapsed time t.

Explanation:

If velocity is higher, then traversed distance during any time increment dt is higher. This tells you that the amount of energy delivered to the object is proportional to the velocity, which is proportional to elapsed time t.

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When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
pantera1 [17]

Answer:

The period is T =  0.700 \ s

Explanation:

From the question we are told that  

    The mass is m =  0.350  \ kg

     The extension of the spring is  x =  12.0 \ cm = 0.12 \ m

       

The spring constant for this is mathematically represented as

       k  = \frac{F}{x}

Where F is the force on the spring which is mathematically evaluated as

       F  =  mg  =  0.350 * 9.8

       F  =3.43 \ N

So  

    k  = \frac{3.43 }{ 0.12}

    k  = 28.583 \ N/m

The period of oscillation is mathematically evaluated as

      T =  2 \pi \sqrt{\frac{m}{k} }

substituting values

     T =  2  *  3.142*  \sqrt{\frac{0.35 }{28.583} }

     T =  0.700 \ s

   

7 0
3 years ago
Define moment of momentum. at which condition is it's magnitude zero?​
ololo11 [35]

Let's start with the concept of momentum. What is it? Linear momentum in physics is mathematically written as a product of mass and velocity of an object. Now let us suppose a body of mass m is moving in an inertial frame of reference with velocity v. Consider the fact that no external force is acting on the system. The momentum of this body is given by mv, where m is the mass and v is its velocity. In case of simple real world problems not delving into the realms of relativity, mass is a conserved quantity and it cannot be zero. Hence the velocity of the body must be zero and hence the momentum.

However, photons are considered to have a rest mass zero.

However note the point carefully "rest mass". A body in motion cannot have mass to be zero.

<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em>

7 0
3 years ago
Which has more friction? wax paper, aluminum foil, wood
Nadusha1986 [10]
The answer to this  is  aluminum foil.
5 0
3 years ago
Read 2 more answers
A boy throws a ball of mass 0.22 kg straight upward with an initial speed of 29 m/s. When the ball returns to the boy, its speed
maksim [4K]

Answer:

The work is -67.76 J

Explanation:

The law of conservation of energy is considered one of one of the fundamental laws of physics and states that the total energy of an isolated system remains constant. except when it is transformed into other types of energy.

This is summed up in the principle that energy can neither be created nor destroyed in the universe, only transformed into other forms of energy.

In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.

So, the loss of kinetic energy is \frac{1}{2} *m*(vf^{2} -vi^{2} )

You know:

  • mass=m=0.22 kg
  • Initial velocity of the ball: vi= 29 \frac{m}{s}

Final velocity of the ball: vf= 15 \frac{m}{s}

Replacing:

\frac{1}{2} *0.22 kg*(15^{2} -29^{2} )= -67.76 J

Friction work is always negative because friction is always against displacement.

<u><em>The work is -67.76 J</em></u>

5 0
3 years ago
n an experiment of a simple pendulum, measurements show that the pendulum has length m, mass kg, and period s. Take m/s2 . i. Us
barxatty [35]

Answer:

The answer is "(1.265 \pm 0.010) \ s \ and \ 0.709 \%"

Explanation:

In point i:

T_{theo}= 2\pi \sqrt{\frac{l}{g}}

        =2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s

If  error in the theoretical time period :

\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2}  \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}

           = 0.010 \ s

 T_{theo} = (1.265 \pm 0.010) \ s

In point ii:

\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100

<h3>                     = \frac{1.274 -1.265}{\frac{1.274+1.265}{2}} \times 100\\\\=0.709 \%</h3>
5 0
3 years ago
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