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Rudiy27
3 years ago
5

Can someone help pleaseeee

Physics
1 answer:
Eduardwww [97]3 years ago
4 0

Answer:

Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .

Explanation:

The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.

You might be interested in
Skater 1 has a mass of 45 kg and is at rest. Skater 2 has a mass of 50 kg and is moving slowly at a constant velocity of 3.2 m/s
algol13

Answer:

None

Explanation:

Force, F is given by ma where m is the mass of an object and a is acceleration

Acceleration is the rate of change in velocity per unit time. Since skaters with mass of 75 kg and 50 kg are moving at a constant speed, there is no acceleration hence F=50*0=0 and F=75*0=0

For skater of 45 kg, he is at rest to mean the initial and final velocitu of the skater is zero hence no acceleration, the force will be 45*0=0

Therefore, none of the skaters will experience a greater net force.

3 0
3 years ago
Why is it impossible to create a perpetual motion machine?
enyata [817]

Answer:

The first law of thermodynamics is the law of conservation of energy. It states that energy is always conserved. ... To keep a machine moving, the energy applied should stay with the machine without any losses. Because of this fact alone, it is impossible to build perpetual motion machines.

Explanation:

8 0
3 years ago
Suppose you take a short piece of wire that is not attached to anything and move it up and down in a magnetic field. Explain whe
Otrada [13]

Answer:

No.

Explanation:

  • According to Faraday's law, the induced emf in the circuit is given by :

         e=\dfrac{d\phi}{dt}, it is proportional to the rate of change of magnetic flux.

  • In this case, a short piece of wire that is not attached to anything and move it up and down in a magnetic field. It means that the circuit is not completed here. It is an open circuit. For the induction of current, a circuit must be completed.
  • Hence, no current will induce.
4 0
4 years ago
Read 2 more answers
A ____ is the time required for one half of the nuclei in a radio- ____ isotope to decay.
sukhopar [10]

Answer:

A half-life is the time required for one half of the nuclei in a radio- active isotope to decay.

Explanation:

A radio-active isotope is an isotope which undergoes radioactive decay.

Radioactive decay is a spontaneous process in which the nucleus of an atom changes its state (turning into a different nucleus, or de-exciting), emitting radiation, which can be of three different types: alpha, beta or gamma.

The half-life of a radio-active isotope is the time required for half of the nuclei of the initial sample to decay.

The law of radio-active decay can be expressed as follows:

N(t) = N_0 (\frac{1}{2})^{t/t_{1/2}}

where

N(t) is the number of undecayed nuclei left at time t

N0 is the initial number of nuclei

t is the time

t_{1/2} is the half-life

We see that when t=t_{1/2} (that means, when 1 half-life has passed), the number of undecayed nuclei left is

N(t) = N_0 (\frac{1}{2})^{t_{1/2}/t_{1/2}}=N_0 (\frac{1}{2})^1=\frac{N_0}{2}

So, half of the initial nuclei.

5 0
3 years ago
a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f
34kurt

Answer:

    \frac{h_{liquid} }{ h_{body} } = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

         B = ρ_liquid  g V_liquid

let's write the translational equilibrium condition

         B - W = 0

let's use the definition of density

        ρ_body = m / V_body

        m = ρ_body  V_body

        W = ρ_body  V_body  g

we substitute

          ρ_liquid  g  V_liquid = ρ_body  g  V_body

          \frac{\rho_{body}   }{\rho_{liquid} } } =  \frac{V_{liquid}   }{V_{body} } }

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

          V = A h_bogy

Thus

          \frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }

we substitute

           5/9 = \frac{h_{liquid} }{ h_{body} }

8 0
3 years ago
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