Answer:Final volume after pressure is applied=4,292cm3
Explanation:
Using the bulk modulus formulae
We have that The bulk modulus of waTer is given as
K =-V dP/dV
Where K, the bulk modulus of water = 2.15 x 10^9N/m^2
2.15 x 10^9N/m^2= - 4,300 x 4 × 106N/m2 / dV
dV = - 4,300 x 4 × 10^6N/m^2/ 2.15 x 10^9N/m^2
dV (change in volume)= -8.000cm^3
Final volume after pressure is applied,
V= V+ dV
V= 4300cm3 + (-8.000cm3)
=4300cm3 - 8.000cm3
Final Volume, V =4,292cm3
A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?
Answer:
The displacement of the car after 6s is 43.2 m
Explanation:
Given;
velocity of the car, v = 12 m/s
acceleration of the car, a = -1.6 m/s² (backward acceleration)
time of motion, t = 6 s
The displacement of the car after 6s is given by the following kinematic equation;
d = ut + ¹/₂at²
d = (12 x 6) + ¹/₂(-1.6)(6)²
d = 72 - 28.8
d = 43.2 m
Therefore, the displacement of the car after 6s is 43.2 m
In the winter sport of curling, players give a 20 kg stone a push across a sheet of ice. The Slone moves approximately 40 m before coming to rest. The final position of the stone, in principle, onlyndepends on the initial speed at which it is launched and the force of friction between the ice and the stone, but team members can use brooms to sweep the ice in front of the stone to adjust its speed and trajectory a bit; they must do this without touching the stone. Judicious sweeping can lengthen the travel of the stone by 3 m.1. A curler pushes a stone to a speed of 3.0 m/s over a time of 2.0 s. Ignoring the force of friction, how much force must the curler apply to the stone to bring it op to speed?A. 3.0 NB. 15 NC. 30 N
D. 150 N2The sweepers in a curling competition adjust the trajectory of the slope byA. Decreasing the coefficient of friction between the stone and the ice.
B. Increasing the coefficient of friction between the stone and the ice.C. Changing friction from kinetic to static.D. Changing friction from static to kinetic.3. Suppose the stone is launched with a speed of 3 m/s and travel s 40 m before coming to rest. What is the approximate magnitude of the friction force on the stone?A. 0 NB. 2 NC. 20 ND. 200 N4. Suppose the stone's mass is increased to 40 kg, but it is launched at the same 3 m/s. Which one of the following is true?A. The stone would now travel a longer distance before coming to rest.B. The stone would now travel a shorter distance before coming to rest.C. The coefficient of friction would now be greater.D. The force of friction would now be greater.
Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.
The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2
Fd = (1/2)mv^2
F = (1/2)mv^2/d.
Plug in m = 20 kg, v = 3 m/sec, d = 40 m.
83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.
Explanation:Hope I helped :)
Acceleration is sometimes expressed in multiples of g, where g = 9.8 m/s^2 is the magnitude of the acceleration due to the earth's gravity. In a test crash, a car's velocity goes from 26 m/s to 0 m/s in 0.15 s. How many g's would be experienced by a driver under the same conditions?
Answer:
Acceleration = 18g
Explanation:
Given the following data;
Initial velocity, u = 26m/s
Final velocity, v = 0
Time = 0.15 secs
To find the acceleration;
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
Substituting into the equation, we have;
[tex]a = \frac{0 - 26}{0.15}[/tex]
[tex]a = \frac{26}{0.15}[/tex]
Acceleration = 173.33m/s2
To express it in magnitude of g;
Acceleration = 173.33/9.8
Acceleration = 17.7 ≈ 18g
Acceleration = 18g
A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.
Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.
The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:
n + (-w) = 0
n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N
The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be
f = 0.67 (51.94 N) ≈ 35 N
so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.
The block starts moving as soon as x = 35 N, at which point f = 35 N.
At any point after the block starts moving, we have
f = 0.48 (51.94 N) ≈ 25 N
so that x = 25 N is the required force to keep the block moving at a constant speed.
As x is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force
Magnitude of normal force ( object at rest ); n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x after object start moving x = 25 NGiven data :
mass of block at rest ( m ) = 5.3 kg
Coefficient of static friction ( μ_s ) =0.67
Coefficient of kinetic friction is ( μ_k ) = 0.48
Horizontal force applied to block = x
First step : magnitude of normal force ( n ) when object is at rest
n = w where w = m*g
n - w = 0
n - ( 5.3 * 9.81 ) = 0 ∴ n = 51.94 N
Second step : Required magnitude of x before the movement of object
F = μ_s * n
F = 0.67 * 51.94 = 34.79 N ≈ 35 N
∴ The object will start moving once F and x = 35 N
Final step : Magnitude of x after object start moving
F = μ_k * n
= 0.48 * 51.94 = 24.93 N ≈ 25 N
∴ object will continue to accelerate at a constant speed once F and x = 25N
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8. A rectangle is measured to be 6.4 +0.2 cm by 8.3 $0.2 cm.
a) Calculate its perimeter in cm
b) Calculate the uncertainty in its perimeter.
Answer:
a) The perimeter of the rectangle is 29.4 centimeters.
b) The uncertainty in its perimeter is 0.8 centimeters.
Explanation:
a) From Geometry we remember that the perimeter of the rectangle ([tex]p[/tex]), measured in centimeters, is represented by the following formula:
[tex]p = 2\cdot (w+l)[/tex] (1)
Where:
[tex]w[/tex] - Width, measured in centimeters.
[tex]l[/tex] - Length, measured in centimeters.
If we know that [tex]w = 6.4\,cm[/tex] and [tex]l = 8.3\,cm[/tex], then the perimeter of the rectangle is:
[tex]p = 2\cdot (6.4\,cm+8.3\,cm)[/tex]
[tex]p = 29.4\,cm[/tex]
The perimeter of the rectangle is 29.4 centimeters.
b) The uncertainty of the perimeter ([tex]\Delta p[/tex]), measured in centimeters, is estimated by differences. That is:
[tex]\Delta p = 2\cdot (\Delta w + \Delta l)[/tex] (2)
Where:
[tex]\Delta w[/tex] - Uncertainty in width, measured in centimeters.
[tex]\Delta l[/tex] - Uncertainty in length, measured in centimeters.
If we know that [tex]\Delta w = 0.2\,cm[/tex] and [tex]\Delta l = 0.2\,cm[/tex], then the uncertainty in perimeter is:
[tex]\Delta p = 2\cdot (0.2\,cm+0.2\,cm)[/tex]
[tex]\Delta p = 0.8\,cm[/tex]
The uncertainty in its perimeter is 0.8 centimeters.
If the particles were moving with a speed much less than c, the magnitude of the momentum of the second particle would be twice that of the first. However, what is the ratio of the magnitudes of momentum for these relativistic particles?
Answer:
p₂ / p₁ = 2 (v₁ / v₂)
Explanation:
The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,
p = m v / √[1+ (v/c)² ]
for the case of speeds much lower than the speed of light this expression is close to
p = m v
In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small
p₂ = 2 p₁
p₂/p₁ = 2
in consecuense
m v₂ = 2 m v₁
v₂ = 2 v₁
consider particles of equal mass.
By the time their speeds increase they enter the relativistic regime
p₂ = mv₂ /√(1 + v₂² /c²)
p₁ = m v₁ /√(1 + v₁² / c²)
let's look for the relationship between these two moments
p₂ / p₁ = mv₂ / mv₁ [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)
from the initial statement
p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)
we take c from the root
p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]
this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1
p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]
p₂ / p₁ = 2 (v₁ / v₂)
we see the value of the moment depends on the speed of the particles