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Alexxx [7]
3 years ago
9

What is the force if the mass is 75kg and the acceleration is 24.5m/s^2

Physics
1 answer:
Snowcat [4.5K]3 years ago
4 0

<u>Given;</u>

mass m = 75 kg

acceleration a = 24.5 ms²

<em>F = ma </em>

F  =  75 kg * 24.5 ms²

    =  1837.5 kg ms².

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Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled wi
Korvikt [17]

Complete Question:

Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled with a liquid. Both objects float in equilibrium. Less of object T is submerged than of object B, which floats, fully submerged, closer to the bottom of the container. Which of the following statements is true?

  • Object T has a greater density than object B.
  • Object B has a greater density than object T.
  • Both objects have the same density.  

Answer:

Object B has a greater density than object T

Explanation:

Any object partially or completely submerged in a liquid, experiments an upward force, equal to the weight of  the volume displaced by the liquid. This force is called the buoyant force, and can be expressed as follows:

Fb = ρl * Vs*g

where ρl is the density of the liquid, and Vs is the submerged volume.

This force must be compared with the weight of the object, which is always downward, and can be expressed as follows:

Fg = ρb* Vb * g

where ρb, is the density of the object, and Vb is the total volume of the object, regardless which portion is submerged.

For object B, as it floats fully submerged, this means that both forces are equal in magnitude:

Fg = Fb⇒ ρb* Vb * g = ρl * Vs*g

As Vb = Vs (the object is fully submerged) this means that ρb =ρl.

For object T, as it floats partially submerged, this means that Fg < Fb:

Fg= ρt* Vt * g < Fb = ρl * Vs*g.

Now, we know that ρb =ρl, so we can replace in the equation above:

ρT* Vt * g < ρb*Vs*g

Simplifying common terms, and replacing Vs by KVt (where K is the fraction of the total volume which is submerged, i.e. K<1), we have:

ρt*Vt < ρb*K*Vt ⇒ ρt / ρb < K < 1 ⇒ ρt < ρb ⇒ ρb > ρt

3 0
3 years ago
Newly discovered planet has twice the mass and three times the radius of the earth. What is the free-fall acceleration at its su
skad [1K]

Answer:

g_n=\dfrac{2}{9}g

Explanation:

M = Mass of Earth

G = Gravitational constant

R = Radius of Earth

The acceleration due to gravity on Earth is

g=\dfrac{GM}{R^2}

On new planet

g_n=\dfrac{G2M}{(3R)^2}\\\Rightarrow g_n=\dfrac{2GM}{9R^2}

Dividing the two equations we get

\dfrac{g_n}{g}=\dfrac{\dfrac{2GM}{9R^2}}{\dfrac{GM}{R^2}}\\\Rightarrow \dfrac{g_n}{g}=\dfrac{2}{9}\\\Rightarrow g_n=\dfrac{2}{9}g

The acceleration due to gravity on the other planet is g_n=\dfrac{2}{9}g

4 0
3 years ago
Read 2 more answers
An object that is magnetic all of the time is called a.
Novosadov [1.4K]

Answer:

Bar magnets are permanent magnets. This means that their magnetism is there all the time and cannot be turned on or off as it can with electromagnets .

Explanation:

5 0
2 years ago
Two football players are attempting to tackle each other. If one football player has a mass of 100 kg and pushes with a force of
Vaselesa [24]

Answer:

is it 20kg. Two opposing forces pushing onto each other

8 0
3 years ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
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