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Free_Kalibri [48]

2 years ago

11

On a position vs. time graph for an object, a horizontal line segment indicates that the object is(1 point)

1 answer:

babunello [35]2 years ago

7 0

Answer:

moving at a constant rate. (could be zero m/s)

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A planet has a period of revolution about the sun equal to T and a mean distance from the sun equal to R. T2 varies directly as

Tamiku [17]

Answer:

T² ∝ R³

Explanation:

Given data,

The period of revolution of the planet around the sun, T

The mean distance of the planet from the sun, R

According to the III law of Kepler, " Law of Periods' states that the square of the orbital period to go around the sun once is directly proportional to the cube of the mean distance between the sun and the planet.

T² ∝ R³

$\frac{T^{2}}{R^{3}} = Constant$

From the above equation it is clear that T² varies directly as the R³.

7 0

3 years ago

Identical charges A, B, C, and D, each with a charge of magnitude 1.5 micro Coulomb, are situated at the corners of a square who

olga nikolaevna [1]

This the correct answer who will know this answer

8 0

2 years ago

17. The four points built into the camera to ensure that each exposure can be oriented

3241004551 [841]

The diameter of the circle is 18 m. Eugene incorrectly says that the circumference of the circle is about 113.04 m. What mistake did Eugene make? Use 3.14 for pi.

7 0

3 years ago

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

2 years ago

Find the resultant of an easterly force of 100 N and a southeast force of 80 N acting at 65 degrees to the 100 N force

saveliy_v [14]

Answer:

Resultant is 152 N at 28.5 degrees south to the 100 N force

Explanation:

7 0

3 years ago