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4 years ago

Help fast please!!!!!

Physics

2 answers:

Svet_ta [14]4 years ago

8 0

Answer:

the answer is A

Hope tis helps:)

Explanation:

inysia [295]4 years ago

6 0

We're told that the planets have EQUAL MASS.

If that's true, then the strength of the gravitational forces between
each planet and the star depends only on the distance between
them ... the farther a planet is from the star, the smaller the
gravitational forces are IF we're talking about planets with
equal masses.

Planet-X is closer to the star, and Planet-Y is farther from it.
From this we know that the gravitational forces between the
star and Planet-X are greater, and the forces between the star
and Planet-Y are smaller.

'A' says this.

'B' is totally absurd, because it talks about gravity repelling things.

'C' says exactly the opposite for the two planets.

'D' says that distance doesn't matter. We know this is absurd,
simply because we're never pulled toward Jupiter in our daily life.

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A graph here shows the displacement of two walkers over time. Waliper 1 is graphed in red while walker 2 is graphed

natita [175]

894984+894390548-098548948+8-734897389696563418686528687

thats your answe thank oyu for asking

5 0

3 years ago

Explain the process of a nuclear fusion reaction using hydrogen. Include the particles that are used to start and maintain the c

Zielflug [23.3K]

Answer it might be this

Explanation:

8 0

3 years ago

11. A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds.

aleksklad [387]

Answer:

A) a = 7.89 m/s²

B) a = 3.51 m/s²

C) 4.38 m/s²

D) Frictional force

E) F_f = 5552.83 N

Explanation:

A) Formula for Force is;

F = ma

Where;

m is mass

a is acceleration

We are given;

F = 10,000 N

m = 1267 kg

Thus;

10000 = 1267a

a = 10000/1267

a = 7.89 m/s²

B) We are told the car covers 394.6 m in 15 seconds.

Using Newton's third equation equation of motion, we can find the actual acceleration.

s = ut + ½at²

u is zero since the object began from rest.

Thus;

S = ½at²

394.6 = ½ × a × 15²

a = 394.6 × 2/225

a = 3.51 m/s²

C) difference in accelerations = 7.89 - 3.51 = 4.38 m/s²

D) The force that caused the difference in acceleration is frictional force

E) To find the magnitude of the force that caused the difference in acceleration, we will use the formula;

F - F_f = ma

F_f = F - ma

Where F_f is the frictional force

Thus;

F_f = 10000 - 1267(3.51)

F_f = 5552.83 N

7 0

3 years ago

The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (Th

Bezzdna [24]

Answer:

hello the diagram relating to this question is attached below

a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec

b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

a) Determine the angular accelerations of B1 and B2 when couple is applied

at point P ; Co = I* ∝B2'

150 = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

at point 0 ; Co = Inet * ∝B1

150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

hence angular acceleration of B1 = 180 rad/sec

b) Determine the force exerted on B2 at P

T2 = mB1g + T1 -------- ( 1 )

where ; T1 = mB2g ( at point p )

= 2 * 9.81 = 19.6 N

back to equation 1

T2 = (2 * 9.8 ) + 19.6 = 39.2 N

3 0

3 years ago

An 80g meter stick is supported at its 30 cm mark by a string attached to the ceiling. A 20 g mass is hung from the 80 cm mark w

BARSIC [14]

Answer:

$104\; {\rm g}$ , assuming that the meter stick is uniform with the center of mass precisely at the $50\; {\rm cm}$ mark.

Explanation:

Refer to the diagram attached. The meter stick could be considered as a lever. The string at the $30\; {\rm cm}$ mark would then act as the fulcrum of this lever.

The $m_{A} = 20\; {\rm g}$ mass at the $80\; {\rm cm}$ mark is at a distance of $r_{A} = 50\; {\rm cm}$ to the right of the fulcrum at $30\; {\rm cm}$ .

The weight of the $80\; {\rm g}$ meter stick acts like a weight of $m_{B} = 80\; {\rm g}$ attached to the center of mass of this meter stick. Under the assumptions, this center of mass of this meter stick would be at the $50\; {\rm cm}$ mark, which is $r_{B} = 20\; {\rm cm}$ to the right of the fulcrum at $30\; {\rm cm}$ .

Let $m_{C}$ be the mass attached to the meter stick at the $5\; {\rm cm}$ mark. This mass would be at a distance of $r_{C} = 25\; {\rm cm}$ to the left of the fulcrum at $30\; {\rm cm}$ .

At equilibrium:

$\begin{aligned} & m_{C}\, r_{C} && (\text{mass on the left of fulcrum})\\ &= m_{A}\, r_{A} + m_{B} \, r_{B} && (\text{mass on the right of fulcrum})\end{aligned}$ .

Solve for $m_{C}$ , the unknown mass attached to the meter stick at the $5\; {\rm cm}$ mark:

$\begin{aligned}m_{C} &= \frac{m_{A}\, r_{A} + m_{B}\, r_{B}}{r_{C}} \\ &= \frac{20\; {\rm g} \times 50\; {\rm cm} + 80\; {\rm g} \times 20\; {\rm cm}}{25\; {\rm cm}} \\ &= 104\; {\rm g}\end{aligned}$ .

5 0

3 years ago