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3 years ago

Question 1

1 answer:

3 0

Joseph is conducting an investigation on how two objects compare in mass, volume, and force with each other. When determining the gravitational force between the two objects, the following characteristics of the objects should be considered :

<u>MASS AND DISTANCE</u>

Explanation:

The gravitational force is a force that attracts any two objects with mass. In fact, every object, including you, is pulling on every other object in the entire universe. This is called Newton's Universal Law of Gravitation
The strength of the gravitational force between two objects depends on two factors, mass and distance.
The force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.
When two objects are gravitational locked, their gravitational force is centered in an area that is not at the center of either object, but at the barycenter of the system. The principle is similar to that of a see-saw.
Gravity is often assumed to be the same everywhere on Earth, but it varies because the planet is not perfectly spherical or uniformly dense. In addition, gravity is weaker at the equator due to centrifugal forces produced by the planet's rotation.

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Two hypothetical discoveries in Part A deal with moons that, like Earth's moon, are relatively large compared to their planets.

arlik [135]

Answer:

Unusually large moons form in giant impacts, which are relatively rare events

Explanation:

Solution:

- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.

- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible. The transition from an undifferentiated cloud to a star system complete with planets and moons takes about 100 million years.

8 0

3 years ago

Consider a step-down transformer with 15 turns in the primary and 6 turns in the secondary windings. Calculate the input impedan

AURORKA [14]

Answer:

Input impedance of this transformer is 50 ohms.

Explanation:

Given that,

Number of turns in the primary coil, $N_p=15$

Number of turns in the secondary coil, $N_s=6$

Output impedance of the transformer, $V_o=8\ \Omega$

The number of turns and the impedance ratio in the step down transformer is given by :

$\dfrac{15}{6}=\sqrt{\dfrac{Z_p}{Z_s}}\\\\\dfrac{15}{6}=\sqrt{\dfrac{Z_p}{8}}\\\\Z_p=50\ \Omega$

So, the input impedance of this transformer is 50 ohms. Hence, this is the required solution.

8 0

3 years ago

Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel

castortr0y [4]

Answer:

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

$\vec{V_{AB}} =\vec{V_A} - \vec{V_B}$

$\vec{V_A} = 22 km/hr$

$\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}$

$= 31.52\hat{i} + 24.62 \hat{j}$

putting respective value to get velocity of A with respect to B

$\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})$

$\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}$

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

6 0

4 years ago

A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse

NISA [10]

Answer:

The time interval is $t = 5.48 *10^{-3} \ s$

Explanation:

From the question we are told that

The length of the string is $l = 3.00 \ m$

The mass of the string is $m = 5.00 \ g = 5.0 *10^{-3}\ kg$

The tension on the string is $T = 500 \ N$

The velocity of the pulse is mathematically represented as

$v = \sqrt{ \frac{T}{\mu } }$

Where $\mu$ is the linear density which is mathematically evaluated as

$\mu = \frac{m}{l}$

substituting values

$\mu = \frac{5.0 *10^{-3}}{3}$

$\mu = 1.67 *10^{-3} \ kg /m$

Thus

$v = \sqrt{\frac{500}{1.67 *10^{-3}} }$

$v = 547.7 m/s$

The time taken is evaluated as

$t = \frac{d}{v}$

substituting values

$t = \frac{3}{547.7}$

$t = 5.48 *10^{-3} \ s$

5 0

4 years ago

The stoplight had just changed and a 2100 kg Cadillac had entered the intersection, heading north at 3.2 m/s , when it was struc

elena55 [62]

Explanation:

It is given that,

Mass of the Cadillac, $m_1=2100\ kg$

Speed of Cadillac, $v_1=3.2\ m/s$ (towards north)

Mass of Volkswagen, $m_2=1100\ kg$

The cars stuck together and slid to a halt, leaving skid marks angled 35 degrees north of east.

According to the law of conservation of momentum,

The momentum along x axis, $m_1v_1=(m_1+m_2)\ usin35$ ............(1)

The momentum along y axis, $m_2v_2=(m_1+m_2)\ ucos35$ ...........(2)

From equation (1) and (2) it is clear that,

$\dfrac{m_1v_1}{m_2v_2}=tan35$

$v_2=\dfrac{m_1v_1}{m_2\ tan35}$

$v_2=\dfrac{2100\times 3.2}{1100\times \ tan35}$

$v_2=8.72\ m/s$

So, the Volkswagen going just before the impact is 8.72 m/s. Hence, this is the required solution.

6 0

3 years ago