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AleksAgata [21]
2 years ago
5

Two spheres of mass M and 2M float in space in the absence of external gravitational forces, as shown in the figure. Which of th

e following predictions is correct about the motion of the center of mass of the two-sphere system?

Physics
1 answer:
anygoal [31]2 years ago
4 0

-- If the two spheres are both at rest when we arrive and start watching them, then their total momentum is zero.  

-- If their center of mass were to move, then that would mean that some momentum had somehow magically appeared.

-- But we know that momentum is conserved so this can't happen.

-- The center of mass must remain at rest.<em> (A)</em>

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If it were possible to remove gravity and friction, think about what would happen to a football if it were tossed into the air.
elena-14-01-66 [18.8K]
Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>
6 0
2 years ago
Read 2 more answers
A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,
Annette [7]

Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

d_x = 10 + 2= 12 blocks

Total displacement towards South

d_y = 5 block

now to find the net displacement we can use vector addition

d = \sqrt{d_x^2 + d_y^2}

d = \sqrt{12^2 + 5^2}

d = 13 blocks

<em>so magnitude of net displacement will be equal to 13 blocks</em>

6 0
1 year ago
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
1 year ago
If a 10 meter ramp helps you move a 500 kg object up 1 meter. What was the mechanical advantage of the ramp?
sattari [20]
The mechanical advantage of an inclined plane is

(Length of the incline) / (its height)

= (10m) / (1m)

= 10 .

It's the same for any load, and doesn't depend on the mass that you're trying to move up or down the ramp.
6 0
1 year ago
Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

6 0
2 years ago
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