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slavikrds [6]
3 years ago
5

a toboggan loaded with vacationing students (total weight 1300 N) slides down a slope at 30 degrees and there is no friction. wh

at is the acceleration of the toboggan?
Physics
1 answer:
sasho [114]3 years ago
4 0

Answer:

The acceleration of the sliding toboggan is, a = 4.9 m/s²

Explanation:

Given data,

The total weight of the toboggan, W = 1300 N

The slope is, Ф = 30°

The acceleration of a body under the influence of the gravitational field does not depend on its mass, size and shape in the absence of the air resistance.

Therefore,

The acceleration of the toboggan is given by the formula,

                           a = g Sin Ф

Substituting the given values in the above equation,

                           a = 9.8 x Sin 30°

                              = 4.9 m/s²

Hence, the acceleration of the sliding toboggan is, a = 4.9 m/s²

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If two particles have equal kinetic energies, are their momenta necessarily equal? explain.
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Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

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

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For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective kinetic energies is given by

K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}

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Similarly For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective momenta are given by

p_{1}=m_{1}\times v_{1}

p_{2}=m_{2}\times v_{2}

Now since it is given that the two kinetic energies are equal thus we have

\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0
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When making maps of the large-scale universe, astronomers estimate distances to the vast majority of galaxies by using:
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Answer:

<em>The comoving distance and the proper distance scale</em>

<em></em>

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The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster.  Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.

4 0
3 years ago
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Answer:

Time, t = 0.015 seconds.

Explanation:

Given the following data;

Mass, m = 0.2kg

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To find the time;

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Substituting into the equation, we have;

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Time, t = 0.015 seconds.

Note: We ignored the negative sign because time can't be negative.

8 0
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