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3 years ago

Which situation describes the best example of rotational motion?

Physics

2 answers:

noname [10]3 years ago

5 0

Answer: C

Explanation:

Serggg [28]3 years ago

3 0

The answer is C

! I took the test lol :)

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A 28 kg student stand on the surface of the Earth. What is the magnitude of the

Contact [7]

Answer:

F = 274.68[N]

Explanation:

The gravitational force is equal to the weight of a body, or this case that of a person. Weight can be calculated by means of the product of mass by gravitational acceleration. In this way we have the following equation:

$F=m*g$

where:

F = force or weight [N]

m = mass = 28 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

$F=28*9.81\\F=274.68[N]$

4 0

2 years ago

a bus takes to reach from station A to station B and then 3 hour to return from station B to station A find the average velocity

Anton [14]

Answer:

you have probably missed some details in the question.

3 0

2 years ago

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$

So, the speed at point B is $3.78\times 10^5\ m/s$ .

7 0

3 years ago

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

2 years ago

Two objects, A and B, each of mass o.22 kg, are moving at 0.34 m/s directly toward each other. What is the speed of Object A aft

ki77a [65]

Theoretically, if the objects have the same mass and are moving towards each other at a speed of $0.34m s^{-1}$ , after a perfectly elastic collision, the object A is supposed to move with the same velocity in the opposite direction.

6 0

2 years ago