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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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An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
2 years ago
How does giving this award help the BLM more
Kazeer [188]

Answer:

Number 3

Explanation:

6 0
2 years ago
Read 2 more answers
An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v
Aneli [31]

We are given: Final velocity (v_f)=20 m/s .

Time t= 2.51 s and

distance s = 82.9 m.

We know, equation of motion

v_f = v_i + at.

Let us plug values of final velocity, and time in above equation.

20=v_i+a(2.51)

20=v_i+2.51a

Subtracting 2.51a from both sides, we get

20-2.51a=v_i  -----------equation(1)

Using another equation of motion

v_f-v_i=2as

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.

We get,

20-(20-2.51a)=2*a(82.90)

Now, we need to solve it for a.

20-20+2.51a=165.8a.

-163.29a=0

a=0.

So, the acceleration would be 0 m/s^2.


5 0
2 years ago
Select the correct answer from each drop-down menu. A baking tray is made of metal because it’s of heat. An oven mitt is used to
Margarita [4]

An oven mitt is used to take the tray out of the oven because it’s an insulator.

5 0
2 years ago
Read 2 more answers
The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 600°C is 1 × 10-6. Calculate the number of vacancies
algol [13]

Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

= 5.778 × 10^22/m^3.

6 0
2 years ago
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