The thrust F of a screw propeller is known to depend upon the diameter d, Speed of advance v, fluid density e, revolution per second N, and the coefficient of viscosity M, of the fluid. Find the expression for F, in terms of the quantities
Answer:
[tex]{ \bf{F = { \tt{ \frac{4}{3} \pi {r}^{3}v gM}}}}[/tex]
A stopped organ pipe of length L has a fundamental frequency of 220 Hz. For which of the following organ pipes will there be a resonance if a tuning fork of frequency 660 Hz is sounded next to the pipe?
a. a stopped organ pipe of length L
b. a stopped organ pipe of length 2L
c. an open organ pipe of length L;
d. an open organ pipe of length 2L.
Answer:
Option (a), (d) are correct.
Explanation:
Frequency, f = 220 Hz
Resonant frequency = 660 Hz
The next frequency of stopped organ pipe is
2f, 3 f, 4 f ....
= 2 x 220 , 3 x 220 , 4 x 220 ....
= 440 Hz, 660 Hz, 880 Hz
So, the option (a) is correct.
The next resonant frequency of open organ pipe is
3 f, 5 f,...
= 3 x 220, 5 x 220 , ..
= 660 Hz, 1100 Hz,...
So, option (d) is correct.
Choose the force diagram that best represents a ball thrown upward by Peter, at the
top of its path.
Diagram A
Diagram B
Diagram C
Diagram D
Answer:Diagram A
Explanation:
Since the air resistance is to be neglected, only the gravitational force acts on the ball ( and has acted all the way from the throw upward). Diagram A reflects this fact correctly indicating the gravity acting on the ball downward.
A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x-axis.
Answer:
-1.0778×10⁻¹⁰ N/C
Explanation:
Applying,
E = kq/r²................ equation 1
Where E = elctric field, q = charge, r = distance, k = coulomb's law
From the question,
Given: q = -3.0×10 C, r = 5.0 m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values in equation 1
E = (-3.0×10)(8.98×10⁹)/5²
E = -1.0778×10⁻¹⁰ N/C
Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C
if 145kl of energy is added to water, what mass of water can be heated from 35C to 100C then vaporized at 100C
Answer:
m = 0.057 kg = 57 g
Explanation:
Energy Added to Water = Heat added to raise the temperature of water + Heat used to vaporize water
[tex]E = mC\Delta T + mH\\E = m(C\Delta T + H)[/tex]
where,
E = Energy added to water = 145 KJ
m = mass of water = ?
C = specific heat capacity of water = 4.2 KJ/kg.°C
ΔT = change in temperature = 100°C - 35°C = 65°C
H = Latent heat of vaporization of water = 2260 KJ/kg
Therefore,
[tex]145\ KJ = m[(4.2\ KJ/kg.^oC)(65^oC)+2260\ KJ/kg]\\\\145\ KJ = m(2533\ KJ/kg)\\\\m = \frac{145\ KJ}{2533\ KJ/kg}[/tex]
m = 0.057 kg = 57 g
The mass of water that can be heated is equal to 0.527 kilograms.
Given the following data:
Quantity of energy = 145 kJ = 145,000 Joules.Initial temperature = 35.0°CFinal temperature = 100.0°CScientific data:
Specific heat capacity of water = 4200 J/kg°CLatent heat of vaporization of water = 2260 KJ/kgTo calculate the mass of water that can be heated:
The quantity of energy and heat.Note: The quantity of energy added to water is equal to the quantity of heat used to vaporize water and the quantity of heat that is added to raise the temperature of water.
Mathematically, this is given by this expression:
[tex]E=mc\theta + mH\\\\E= m(c\theta + H)[/tex]
Making m the subject of formula, we have:
[tex]m=\frac{E}{c\theta + H}[/tex]
Substituting the parameters into the formula, we have;
[tex]m=\frac{145000}{[42000\times (100-35)] + 2260}\\\\m=\frac{145000}{(4200\times 65) + 2260}\\\\m=\frac{145000}{273000 + 2260}\\\\m=\frac{145000}{275260}[/tex]
Mass, m = 0.527 kilograms.
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In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-called cathode rays. One remarkable series of experiments with cathode rays, conducted by J. J. Thomson around 1897, led to the discovery of the electron.
With the idea that cathode rays were charged particles, Thomson used a cathode-ray tube to measure the ratio of charge to mass, q/m, of these particles, repeating the measurements with different cathode materials and different residual gases in the tube.
Part A
What is the most significant conclusion that Thomson was able to draw from his measurements?
He found a different value of q/m for different cathode materials.
He found the same value of q/m for different cathode materials.
From measurements of q/m he was able to calculate the charge of an electron.
From measurements of q/m he was able to calculate the mass of an electron.
Part B
What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
Express your answer in terms of e, m, d, v0, L, and E0.
Part C
Now imagine that you place your entire apparatus inside a region of magnetic field of magnitude B0 (Figure 2) . The magnetic field is perpendicular to E⃗ 0 and directed straight into the plane of the figure. You adjust the value of B0 so that no deflection is observed on the screen.
What is the speed v0 of the electrons in this case?
Express your answer in terms of E0 and B0.
Answer:
a) He found the same value of q/m for different cathode materials.
b) y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex] , c) v = [tex]\frac{E_o}{B_o}[/tex]
Explanation:
In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.
b) In the part of the plates the electrons are accelerated by the electric field,
F = ma
- e E = m a
a = - (e/m) E₀
the distance traveled is
X axis
x = v₀ t
the separation of the plates is x = d
t = vo / d
Y axis
y = v_{oy} t + ½ to t²
y = ½ a t²
y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex]
c) In this case there is a magnetic field B₀ and the electrons have no deflection
F = - e E + e v x B
if there is no deviation F = 0
e E = e v B
v = [tex]\frac{E_o}{B_o}[/tex]
It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.
1. What is the passenger's apparent weight before the elevator starts moving?
2. What is the passenger's apparent weight whilethe elevator is speeding up?
3. What is the passenger's apparent weight afterthe elevator reaches its cruising speed?
Answer:
1. 588 N
2. 738 N
3. 588 N
Explanation:
time, t = 4 s
initial velocity, u = 0
final velocity, v = 10 m/s
mass, m= 60 kg
1.
Weight of passenger before starts
W =m g = 60 x 9.8 = 588 N
2.
When the elevator is speeding up
v = u + a t
10 = 0 + a x 4
a = 2.5 m/s2
Now the weight is
W' = m (a + g) = 60 (9.8 + 2.5) = 738 N
3.
When he reaches the cruising speed, the weight is
W = 588 N
Suppose the pucks start spinning after the collision, whereas they were not before. Will this affect your momentum conservation results
Answer:
No, it will not affect the results.
Explanation:
For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.
What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.
If an object of a constant mass experiences a constant net force, it will have a constant what?
Explanation:
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If an object of a constant mass experiences a constant net force, it will have a constant acceleration.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.
According to Newton's second law of motion:
Applied force = mass × acceleration.
Hence, if an object of a constant mass experiences a constant net force, it will have a constant acceleration.
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Astronauts in space move a toolbox from its initial position ????????→=<15,14,−8>m to its final position ????????→=<17,14,−1>m. The two astronauts each push on the box with a constant force. Astronaut 1 exerts a force ????1→=<18,7,−12>???? and astronaut 2 exerts a force ????2→=<16,−10,16>????.
Required:
What is the total work performed on the toolbox?
If both forces are measured in Newtons, then the net force is
F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N
The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector
d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m
The total work done by the astronauts on the toolbox is then
F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J
The work done by the two astronauts is equal to 96 J.
What is work done?work done?Work done is defined as the product of force applied and the distance moved by the force.
Work done = Force × DistanceThe forces applied = 18+16 N, 7+ -10 N, and -12 + 16N
Forces = 34 N, -3 N, and 4N
Distances = (17 - 15, 14 - 14, -1 - - 8) m
Distances = 2, 0, 7
Work done = 34 × 2 + -3 × 0 + 4 × 7
Work done = 96 J
Therefore, the work done by the two astronauts is equal to 96 J.
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How do you find the product of gamma decay?
Answer:
The mass and atomic numbers don't change
Explanation:
An excited atom relaxes to the ground state emitting a photon...called a gamma ray.
The answer is that the mass and atomic numbers don't change.
In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.
To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.
During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.
The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).
For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.
It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).
Thus, the product nucleus remains unchanged in terms of atomic number and mass number.
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Which phase of matter makes up stars?
O liquid
O gas
O plasma
Answer:
The answer to this question is plasma
Answer:
Plasma
Explanation:
* A ball is projected horizontally from the top of
a building 19.6m high.
a, How long when the ball take to hit the ground?
b, If the line joining the point of projection to
the point where it hits the ground is 45
with the horizontal. What must be the
initial velocity of the ball?
c,with what vertical verocity does the ball strike
the grounds? (9= 9.8 M152)
Explanation:
Given
Ball is projected horizontally from a building of height [tex]h=19.6\ m[/tex]
time taken to reach ground is given by
[tex]\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s[/tex]
(b) Line joining the point of projection and the point where it hits the ground makes an angle of [tex]45^{\circ}[/tex]
From the figure, it can be written
[tex]\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6[/tex]
Considering horizontal motion
[tex]\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s[/tex]
(c) The vertical velocity with which it strikes the ground is given by
[tex]\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s[/tex]
Thus, the ball strikes with a vertical velocity of [tex]19.6\ m/s[/tex]
Explanation:
Given
Ball is projected horizontally from a building of height
time taken to reach ground is given by
(b) Line joining the point of projection and the point where it hits the ground makes an angle of
From the figure, it can be written
Considering horizontal motion
(c) The vertical velocity with which it strikes the ground is given by
Thus, the ball strikes with a vertical velocity of
Click Stop Using the slider set the following: coeff of restitution to 1.00 A velocity (m/s) to 6.0 A mass (kg) to 6.0 B velocity (m/s) to 0.0 Calculate what range can the mass of B be to cause mass A to bounce off after the collision. Calculate what range can the mass of B be to cause mass A to continue forward after the collision. Check your calculations with the simulation. What are the ranges of B mass (kg)
Answer:
[tex]M_b=6kg[/tex]
Explanation:
From the question we are told that:
Coefficient of restitution [tex]\mu=1.00[/tex]
Mass A [tex]M_a=6kg[/tex]
Initial Velocity of A [tex]U_a=6m/s[/tex]
Initial Velocity of B [tex]U_b=0m/s[/tex]
Generally the equation for Coefficient of restitution is mathematically given by
[tex]\mu=\frac{V_b-V_a}{U_a-U_b}[/tex]
[tex]1=\frac{v_B}{6}[/tex]
[tex]V_b=6*1[/tex]
[tex]V_b=6m/s[/tex]
Generally the equation for conservation of linear momentum is mathematically given by
[tex]M_aU_a+M_bU_b=M_aV_a+M_bV_b[/tex]
[tex]6*6+=M_b*6[/tex]
[tex]M_b=6kg[/tex]
If 5kg Stone and 1kg stone throw the from the building which will land more fa ster and why?
Answer:
Both stones will land at the same time because both stones will fall with the same acceleration through the same height.
Explanation:
We are given that
Mass of stone ,m1=5 Kg
Mass of stone, m2=1 kg
We have to find which stone more faster will land and why.
[tex]h=u+\frac{1}{2}gt^2[/tex]
Initial velocity of both stones=0
[tex]h=\frac{1}{2}gt^2[/tex]
[tex]t^2=\frac{h}{g}[/tex]
[tex]t=\sqrt{\frac{h}{g}}[/tex]
[tex]t_1=t_2=\sqrt{\frac{h}{g}}[/tex]
Because both stones are thrown from the same height.
Both stones will land at the same time because both stones will fall with the same acceleration through the same height and the acceleration does not depend of its mass.
vector A has a magnitude of 8 unit make an angle of 45° with posetive x axis vector B also has the same magnitude of 8 unit along negative x axis find the magnitude of A+B?
Answer:
45 × 8 units = A + B as formular
Two projectiles A and B are fired simultaneously from a level, horizontal surface. The projectiles are initially 62.2 m apart. Projectile A is
fired with a speed of 19.5 m/s at a launch angle 30° of while projectile B is fired with a speed of 19.5 m/s at a launch angle of 60°. How long
it takes one projectile to be directly above the other?
Let the point where A is launched act as the origin, so that the horizontal positions at time t of the respective projectiles are
• A : x = (19.5 m/s) cos(30°) t
• B : x = 62.2 m + (19.5 m/s) cos(60°) t
These positions are the same at the moment one projectile is directly above the other, which happens for time t such that
(19.5 m/s) cos(30°) t = 62.2 m + (19.5 m/s) cos(60°) t
Solve for t :
(19.5 m/s) (cos(30°) - cos(60°)) t = 62.2 m
t = (62.2 m) / ((19.5 m/s) (cos(30°) - cos(60°))
t ≈ 8.71 s
What is true when an object floats in water? A. When an object floats, it exceeds the volume of water available. B. When an object floats, it displaces a volume of water equal to its own volume. C. When an object floats, it does not displace its entire volume.
Answer:
C. When an object floats, it does not displace its entire volume.
Explanation:
Buoyancy can be defined as an upward force which is created by the water displaced by an object.
According to Archimede's principle, it is directly proportional to the amount (weight) of water that is being displaced by an object.
Basically, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up. The buoyancy of an object is given by the formula;
[tex] Fb = pgV [/tex]
[tex] But, \; V = Ah [/tex]
[tex] Hence, \; Fb = pgAh [/tex]
Where;
Fb = buoyant force of a liquid acting on an object.
g = acceleration due to gravity.
p = density of the liquid.
v = volume of the liquid displaced.
h = height of liquid (water) displaced by an object.
A = surface area of the floating object.
The unit of measurement for buoyancy is Newton (N).
Additionally, the density of a fluid is directly proportional to the buoyant force acting on it i.e as the density of a liquid decreases, buoyancy decreases and vice-versa.
Furthermore, an object such as a boat, ship, ferry, canoe, etc, are able to float because the volume of water they displace weigh more than their own weight. Thus, if a boat or any physical object weighs more than the volume of water it displaces, it would sink; otherwise, it floats.
In conclusion, the true statement is that when an object floats, it does not displace its entire volume.
15.1.2 Exam: Semester Exam
ŽA
This graph shows the energy of a reaction over time. Which statement is
true?
Potential energy
N
Reaction progress
O A. G represents the activation energy
OB. H represents the energy of the products,
O C. G represents the energy of the products.
< PREVIOUS
Answer: D. F represents the activation energy
Explanation:
The activation energy is the energy required to get the reactants to begin reacting with one another such that products are created. This energy ranges from the minimum to the maximum energy required.
F is therefore the activation energy because it shows the range between the minimum energy it took for the reaction to start and the maximum energy that was required to continue the reaction.
what is conservation energy?
Explanation:
Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant
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When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum
Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;
[tex]E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J[/tex]
The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;
[tex]\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm[/tex]
A body of mass 2kg is released from from a point 100m above the ground level. calculate kinetic energy 80m from the point of released.
Answer:
1568J
Explanation:
Since the problem states 80 m from the point of drop, the height relative to the ground will be 100-80=20m.
Use conservation of Energy
ΔUg+ΔKE=0
ΔUg= mgΔh=2*9.8*(20-100)=-1568J
ΔKE-1568J=0
ΔKE=1568J
since KEi= 0 since the object is at rest 100m up, the kinetic energy 20meters above the ground is 1568J
What is the total number of moles of products involved in the following reaction?
CaCO3 (s) + 2HCl (aq) - CaCl2 (aq) + CO2 (g) + H20 (g)
O 6
2.
3
5
Answer:
3
Explanation:
You must first make sure the equation is balanced. This one is. Then, you simply add up the coefficients of each compound on the products side of the equation. When the coefficient is not specified, you can assume it is 1 mole. So, in this equation, there is 1 mole of CaCl₂, 1 mole of CO₂, and 1 mole of H₂O = 3 moles.
The reactant side of the equation also has three moles:
1 mole of CaCO₃ and 2 moles of HCl.
. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force
Answer:
elative magnitude of the two forces is the same and they are applied in a constant direction.
Explanation:
Newton's second law states that the sum of the forces is equal to the mass times the acceleration
∑ F = m a
in this case there are two forces on the x axis
F_applied - fr = 0
since they indicate that the velocity is constant, consequently
F_applied = fr
the relative magnitude of the two forces is the same and they are applied in a constant direction.
Which indicates the first law of thermodynamics
Answer:
(d)
Explanation:
because dU = Q -W so ,that the option d(D) is correct
A boy throws a ball upward with a velocity of 4.50 m/s at 60.0o. What is the maximum height reached by the ball?
Answer:
3.1m
Explanation:
Since we only care about the y direction we only need to find vy. Once u draw your vector you will realize that vy= 4.5sin60=3.897m/s.
use vf²=v²+2a(y)
At the maximum height the velocity is 0 and since the object is in freefall, a=-g
Plug in all values
0=15.1875-2*9.8(y)
solve for y
-15.1875*2/-9.8=y
y=3.1m
Answer:
0.774m
Explanation:
The formula for maximum height is given by:
hmax = ∨₀² ₓ Sin (α)² / 2 × g
where;
∨₀ = initial velocity
Sin (α) = angle of launch
g = gravitational acceleration which is equal to 9.8m/s²
Plugging in our values, we will have:
hmax = (4.50m/s)² × (Sin 60.0)² / 2 × 9.8m/s²
hmax= 20.25m/s × 0.75 / 19.8m/s²
hmax = 15.1875 / 19.8
hmax = 0.774m
A uniform ladder of length 24 m and weight w is supported by horizontal floor at A and by a vertical wall at B. It makes an angle 45 degree with the horizontal. The coefficient of friction between ground and ladder is 1/2 and coefficient of friction between ladder and wall is 1/3. If a man whose weight is one-half than the ladder, ascends the ladder, how much length x of the ladder he shall climb before the ladder slips
Answer:
I could not find the answer or do it myself if I did find it I would defenetly share
If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additional 4.0 cm
What happens to the acceleration if you triple the force that you apply to the painting with your hand? (Use the values from the example given in the previous part of the lecture.) Submit All Answers Answer: Not yet correct, tries 1/5 3. A driver slams on the car brakes, and the car skids to a halt. Which of the free body diagrams below best matches the braking force on the car. (Note: The car is moving in the forward direction to the right.] (A) (B) (C) (D) No more tries. Hint: (Explanation) The answer is A. The car is moving to the right and slowing down, so the acceleration points to the left. The only significant force acting on the car is the braking force, so this must be pointing left because the net force always shares the same direction as the object's acceleration. 4. Suppose that the car comes to a stop from a speed of 40 mi/hr in 24 seconds. What was the car's acceleration rate (assuming it is constant). Answer: Submit Al Answers Last Answer: 55 N Only a number required, Computer reads units of N, tries 0/5. 5. What is the magnitude (or strength) of the braking force acting on the car? [The car's mass is 1200 kg.) Answer: Submit Al Answers Last Answer: 55N Not yet correct, tries 0/5
Answer:
2) when acceleration triples force triples, 3) a diagram with dynamic friction force in the opposite direction of movement of the car
4) a = 2.44 ft / s², 5) fr = 894.3 N
Explanation:
In this exercise you are asked to answer some short questions
2) Newton's second law is
F = m a
when acceleration triples force triples
3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.
The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car
4) let's use the scientific expressions
v = v₀ - a t
as the car stops v = 0
a = v₀ / t
let's reduce the magnitudes
v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s
a = 58.667 / 24
a = 2.44 ft / s²
5) let's use Newton's second law
fr = m a
We must be careful not to mix the units, we will reduce the acceleration to the system Yes
a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²
fr = 1200 0.745
fr = 894.3 N
water contracts on freezing is it incorrect or conrrect
Answer:
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