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Leya [2.2K]
2 years ago
7

Find the moments Mx and My and the center of mass of the system, assuming that the particles have equal mass m.

Physics
1 answer:
kogti [31]2 years ago
5 0
<span> If we plot the points on the coordinate system we can see that for x, equal masses are occupying x = 1 through 4. with an equal density value. because they are all uniform we can intuit the moment to be the center of that distribution so Mx = 2.5.
Looking at y we have two equal masses at y = 1 and another set of masses at y = 3. This means that we have a moment in between those distributions and our My = 2.
The center of mass is the combination of the two, and it is at pt (2.5, 2). Because the masses considered are in general and also equal to each other.</span>
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A cyclist moving with a constant velocity of 6.0 m/s forward passes a car that is just starting. If the car has a constant accel
sergiy2304 [10]

After 6 seconds, the car will surpass the cyclist.

<h3><u>Explanation:</u></h3>

The speed of the cyclist = 6 m/s.

Let after time t sec, the car will overtake the cyclist.

So, distance covered by the cyclist in t sec = 6t m

Initial velocity of the car is 0 m/s, because the car is just starting.

Acceleration of the car =2 m/s^2.

Final velocity of the car =6 m/s.

So to cover the distance 6t, the time required by the car = \frac{1}{2} \times a \times t^2 = \frac{1}{2} \times 2 \times t^2

6t = \frac{1}{2} \times 2 \times t^2\\6t = t^2

t =6 sec

So, after 6 seconds, the car will surpass the cycle.  

6 0
2 years ago
A spring stores 10. joules of elastic potential
Mekhanik [1.2K]
The answer would be . Since we are looking for the spring constant you would need to use the formula
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. Then you'd substitute, for PEs and x.
10j=1/2k(.20m^2).
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6 0
2 years ago
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A 3.0-kilogram object is acted upon by an impulse having a magnitude of 15 newton•seconds. What is the magnitude of the object’s
Ilia_Sergeevich [38]
The mass of the object doesn't matter. The change in its momentum is equal to the impulse that changed it ... 15 N-sec.
3 0
2 years ago
11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0
Greeley [361]

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or  it is filled with water , in both the cases pressure centre will lie at the centre of the car .

7 0
2 years ago
When a 75.0-kg man slowly adds his weight to a vertical spring attached to the ceiling, he reaches equilibrium when the spring i
Sergio039 [100]

Answer:

1)k=11.319kN/m

2)displacement=13.02cm

3)k_{eq}=5.65kN/m

Explanation:

At equilibrium position the weight of the man should be balanced by force in the spring

thus we have at equilibrium

kx=mg\\\\k=\frac{mg}{x}

Applying values we get

k=\frac{75\times 9.81}{0.065}\\\\k=11.319kN/m

2)

When we add another identical spring we get an equivalent spring with spring constant as  

\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}

Applying values we get

\frac{1}{k_{eq}}=\frac{1}{11.319}+\frac{1}{11.319}\\\\k_{eq}=5.65kN/m

Thus at equilibrium we have

x_{2}k_{eq}=mg\\\\x_{2}=\frac{mg}{k_{eq}}\\\\x_{2}=\frac{75\times 9.81}{5.65}\times 10^{-3}=13.02cm

3) Equivalent spring constant will be as calculated earlier k_{eq}=5.65kN/m

3 0
2 years ago
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