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sergeinik [125]
3 years ago
13

Help please! Thanks so much!

Physics
1 answer:
qaws [65]3 years ago
4 0
You pretty much just have to convert and be mindful of significant figures.

5 km = 3 miles
0.3 cm = 0.12 in
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If you push any floating object down from equilibrium and release it, it bobs up and down. That looks like an oscillation, so le
GarryVolchara [31]

Answer:

  F_{y} = ( ρ_fluid g A) y

Explanation:

This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force

for the first part, let's write Newton's equilibrium equation

        B₀ - W = 0

        B₀ = W

        ρ_fluid g V_fluid = W

the volume of the fluid is the area of ​​the cube times the height it is submerged

      V_fluid = A y  

For the second part, the body introduces a quantity and below this equilibrium point, the equation is

        B - W = m a

        ρ_fluid g A (y₀ + y) - W = m a

        ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a

       ρ_fluid g A y + (B₀-W) = ma

the part in parentheses is zero since it is the force when it is in equilibrium

      ρ_fluid g A y = m a

      this equation the net force is

      F_{y} = ( ρ_fluid g A) y

we can see that this force varies linearly the distance and measured from the equilibrium position

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3 years ago
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3 years ago
Over 4 seconds, a car's momentum decreases by 1000 kg m/s how much force did it take to make this happen?
kotykmax [81]

Answer:

250N

Explanation:

Given parameters:

Time  = 4s

Momentum  = 1000kgm/s

Unknown:

Force  = ?

Solution:

To solve this problem, we use Newton's second law of motion;

      Ft  = Momentum

F is the force

t is the time

So;

          F x 4 = 1000kgm/s

          F  = 250N

6 0
3 years ago
the temperature of a 2.0-kg increases by 5*c when 2,000 J of thermal energy are added to the block. What is the specific heat of
nata0808 [166]
To calculate the specific heat capacity of an object or substance, we can use the formula

c = E / m△T

Where
c as the specific heat capacity,
E as the energy applied (assume no heat loss to surroundings),
m as mass and
△T as the energy change.

Now just substitute the numbers given into the equation.

c = 2000 / 2 x 5
c = 2000/ 10
c = 200

Therefore we can conclude that the specific heat capacity of the block is 200 Jkg^-1°C^-1
3 0
3 years ago
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