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3 years ago

What would the answer be

Physics

1 answer:

wariber [46]3 years ago

6 0

Answer:

Explanation:

The density has a formula of d = m / V.

The volumes are given as the same.

The density will be

d_lead = 113/ V

d_aluminum = 27/V

It does not matter for this question what the volume actually is. Just note that the value of the volume is the same for each ball.

A: incorrect. If the volumes are the same, the density is not the same because the masses are different.

B: incorrect. If the density of the two balls was less than water, the two balls would float. Never going to happen unless the balls are thin shells.

C: incorrect. The exact opposite is the answer.

D: Correct. This is the answer.

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Sharon the ant (Aaron’s sister) sits at the edge of a turntable of radius R that is spinning with period T. As she makes one-hal

Dmitry_Shevchenko [17]

Answer:

$a = \dfrac{4\pi^2R}{T^2}$

Explanation:

The acceleration of a circular motion is given by

$a = \omega^2 R$

where $\omega$ is the angular velocity and $R$ is the radius.

Angular velocity is related to the period, T, by

$\omega=\dfrac{2\pi}{T}$

Substitute into the previous formula.

$a = (\dfrac{2\pi}{T})^2 R$

$a = \dfrac{4\pi^2R}{T^2}$

This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.

6 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

What is noble gases

Marysya12 [62]

Answer:

helium, neon, argon,krypton, xenon, and radonoccupying Group 0 (18) of the periodic table. They were long believed to be totally unreactive but compounds of xenon, krypton, and radon are now known.

6 0

3 years ago

Please help don't for get to show work

emmasim [6.3K]

-- Bob covered a distance of (32m + 45m) = 77 meters.

-- His displacement is the straight-line distance and direction
from his starting point to his ending point.

The straight-line distance is

D = √(32² + 45²)
D = √(1,024 + 2,025)
D = √3,049 = 55.22 meters

The direction is the angle whose tangent is (32/45) south of east.

tan⁻¹(32/45) = tan⁻¹(0.7111...) = 35.42° south of east.

3 0

3 years ago

45. A particle moves in a straight line with an initial velocity of 30 m/s and constant acceleration 30 m/s2 . (a) What is its d

omeli [17]

Answer:

Displacement after 5 seconds is 155/2 meters

Explanation:

Let X (t) represent the equation of the position, then you have to d2x / dt2 = 5.

Applying the fundamental theorem of the calculation dx/dt = 5t + vo. The speed equation is V (t) = 5t + vo. Since the initial velocity is 30m/s, V (0) = 5 (0) + vo = 30. Therefore, V (t) = dx/dt = 5t + 30. Applying again the fundamental theorem of the calculation X (t) = 5t^2 / 2 + 30t + xo.

Displacement in 5 seconds is given by X (5) - X (0).

X (5) - X (0) = 5 (5)^2/2 +3 (5) + Xo - 5 (0)^2/2 -3 (0) -Xo = 155/2

Displacement after 5 seconds is 155/2 meters

7 0

3 years ago

Read 2 more answers

What would the answer be​

What would the answer be