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Ganezh [65]
3 years ago
13

The SAF operates the M113 Ultra APC.

Physics
1 answer:
HACTEHA [7]3 years ago
8 0

The volume of the object must be no larger than 11.15 m^3.

Explanation:

In order for an object to be able to float in water, its density must be equal or smaller than the water density.

The density of water is:

\rho = 1000 kg/m^3

This means that the density of the object must be no larger than this value.

We also know that the density of an object is given by

\rho = \frac{m}{V}

where

m is the mass of the object

V is its volume

For the object in this problem, the mass is

m=1.115\cdot 10^4 kg

Therefore, we can re-arrange the equation to find its volume:

V=\frac{m}{\rho}=\frac{1.115\cdot 10^4}{1000}=11.15 m^3

So, the volume of the object must be no larger than 11.15 m^3.

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

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Gulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells.
kumpel [21]

Answer:

v = 17.71 m / s

Explanation:

We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity

                v² = v₀² - 2 g (y -y₀)

                v² = 0 - 2g (y -y₀)

when it hits the stone the height is zero and part of the height of the seagull I

              v² = 2g y₀

              v = Ra (2g i)

let's calculate

              v =√ (2 9.8 16)

              v = 17.71 m / s

8 0
3 years ago
It takes 160 kj of work to accelerate a car from 24.0 m/s to 27.5 m/s. what is the car's mass?
ankoles [38]
Work-Energy :W = 1/2 m ( Vf^2 -Vo^2 )
Vo = 24.0 m/s Initial speed 
 Vf = 27.5 m/s  Final speed 

W = 1/2 m ( Vf^2 -Vo^2 )
160 kj = 1/ 2 m ( 27.5^2  -24.0 ^2)
160kj =  4680 x m
convert kilo joules to jeoules                     160000 j = 4689 xm
m = 160000 j/4689
m = 34.18 kg
4 0
4 years ago
Apply a force of 50N to the left describe the motion of the box
GenaCL600 [577]
When a force is applied to the box , this will cause an acceleration to the box.
(force =mass×acceleration)

So the box has a constant acceleration and a changing velocity.
4 0
3 years ago
A graph of the net force F exerted on an object as a function of x position is shown for the object of mass M as it travels a ho
saul85 [17]

The change in kinetic energy is \Delta K = 3Fd

Explanation:

According to the work-energy theorem, the work done on an object is equal to the change in kinetic energy of the object. Mathematically:

W=K_f -K_i= \Delta K

where :

W is the work done on the object

K_f is the final kinetic energy of the object

K_i is the initial kinetic energy

Also, the work done on an object is (assuming that the force is applied parallel to the motion of the object):

W=F\Delta x

where

F is the magnitude of the force

\Delta x is the displacement of the object

In this problem, the force acting on the object is

F

While the displacement is the horizontal distance travelled, so

\Delta x = 3d

Therefore, the work done is

W=(F)(3d)=3Fd

And so the change in kinetic energy is

\Delta K = 3Fd

Learn more about work and kinetic energy:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/6536722

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5 0
3 years ago
Help please!
Talja [164]

The correct expression for the maximum speed of the object during its motion is \frac{FT}{m}.

<h3>Maximum speed of the object</h3>

The maximum speed of the object is determined using the following formulas;

v(max) = Aω

where;

  • A is the amplitude of the motion
  • ω is angular speed

The maximum speed of the object can also be obtained from the maximum net force on the object,

F = ma

where;

  • F is the maximum net force
  • a is the acceleration
  • m is mass of the object

F = m(v/t)

mv = Ft

v = Ft/m

Thus, the correct expression for the maximum speed of the object during its motion is \frac{FT}{m}.

Learn more about maximum speed here: brainly.com/question/4931057

3 0
2 years ago
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