Answer: 2:1
Explanation:
Given
Mother has four times the mass of young son
If they possess same kinetic energy then [tex]K_m=K_s[/tex]
[tex]\Rightarrow K_m=\dfrac{1}{2}\times m_m\times v^2_m[/tex]
[tex]\Rightarrow K_s=\dfrac{1}{2}\times m_s\times v^2_s[/tex]
Equate the energy
[tex]\Rightarrow \dfrac{1}{2}\times m_m\times v^2_m=\dfrac{1}{2}\times m_s\times v^2_s\\\\\Rightarrow \dfrac{1}{2}\times 4m_s\times v^2_m=\dfrac{1}{2}\times m_s\times v^2_s\\\\\Rightarrow 4 v^2_m=v^2_s\\\\\Rightarrow \dfrac{v_s}{v_m}=\sqrt{4}\\\\\Rightarrow \dfrac{v_s}{v_m}=2[/tex]
Thus, the ratio of speed is 2:1.
An ideal heat engine operates between two
temperatures 367.7 K and 865.1 K. What is
the efficiency of the engine?
0.787
0.42504
None
0.213
0.5750
Answer:
As it is the most efficient heat engine, it's efficiency is [T1 - T2]/T1. It can be measured for every Carnot cycle. From the formula and diagram, we can understand that the efficiency of an ideal heat engine also depends on the difference between the hot and cold reservoirs.
Explanation:
an object that has momentum must have
Answer:
Mechanical energy
Explanation:
If an object does not have any momentum, then it doesn't have mechicanl energy.
Select the correct answer.
What are the directions of an object's velocity and acceleration vectors when the object moves in a circular path with a constant speed?
OA. The question is meanimgless, since the acceleration is zero.
ов.
The vectors point in opposite directions.
Oc.
Both vectors point in the same direction.
OD
The vectors are perpendicular,
Answer:
A
Explanation:
If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.
Answer: C.
Explanation: plato users
Find the tension in cord A in the figure if the weight of the suspended object is 250 N. the cord is at a 30 degree angle
366 N
183 N
276 N
224 N
Answer:
276 your welcome give me brainliest
A bike and rider of total mass 75.0 kg moving at a velocity of 30 m/s to the right
collides with a skater of mass 60,0 kg moking at a velocity of 6.0 m/s to the left.
After they collide, the skater has a velocity of O m/s. What is the final velocity of the
bike and rider?
O A. O m/s
O B. 1.8 m/s to the left
O C. 3.0 m/s to left
O D. 1.8 m/s to the right
Answer:
[tex](75.0 \times 30) + (60.0 \times 6.0) = (75.0 \times V) + (60.0 \times 0) \\ 2250 + 360 = 75V \\ 75V = 2610 \\ V = 34.8 \: m {s}^{ - 1} [/tex]
What does it mean when work is positive?
O Velocity is greater than kinetic energy.
O Kinetic energy is greater than velocity.
O The environment did work on an object.
O An object did work on the environment.
Answer:
O The environment did work on an object
Explanation:
Each year 500 runners run up the stairs to the 86th floor of the Empire State Building in New York City. There are 1576 steps and each step is 0.241 m high. In 2003, Australian Paul Crake (20-29 age group) set the overall record by reaching the 86th floor in 9:33. His mass was 70.0 kg. Question 2 HomeworkUnanswered What was Paul Crake's power output during this climb
Answer:
The power is 465.44 W.
Explanation:
mass, m = 70 kg
number of steps, n = 1576
height of each step, h = 0.241 m
time taken, t = 9.33 min= 9.33 x 60 s
The power is given by the rate of doing work.
W = n m g h
W = 1576 x 70 x 9.8 x 0.241
W = 260553.776 J
The power is given by
[tex]P = \frac{W}{t}\\\\P = \frac{260553.776}{9.33\times 60}\\\\P = 465.44 W[/tex]
A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J at point A. The particle subsequently passes through point B which has an electric potential of -2.5 kV relative to point A. Determine the kinetic energy of the particle as it moves through point B.
Answer:
16.9000000000000001 J
Explanation:
From the given information:
Let the initial kinetic energy from point A be [tex]K_A[/tex] = 1.9000000000000001 J
and the final kinetic energy from point B be [tex]K_B[/tex] = ???
The charge particle Q = 6 mC = 6 × 10⁻³ C
The change in the electric potential from point B to A;
i.e. V_B - V_A = -2.5 × 10³ V
According to the work-energy theorem:
-Q × ΔV = ΔK
[tex]-Q \times ( V_B - V_A) = (K_B - K_A)[/tex]
[tex]-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)[/tex]
[tex]15 = (K_B - 1.9000000000000001 \ J)[/tex]
[tex]K_B = 15+ 1.9000000000000001 \ J[/tex]
[tex]\mathbf{K_B =1 6.9000000000000001 \ J}[/tex]
A mass is suspended on a spring. The spring is compressed so that the mass is located 5 cm above its rest position. The mass is released at time t= 0 and allowed to oscillate. It is observed that the mass reaches its lowest point 1/2s after it is released.
Required:
Find an equation that describes the motion of the mass.
Answer:
y = 5 cos 2πt
Explanation:
We will use the formula for simple harmonic motion curve where;
y = a cos ωt
Where;
a is amplitude
t is period
ω is angular frequency with the formula; ω = 2π/t
We are told that when the spring is compressed, the mass is located 5 cm above its rest position.
Thus;
a = 5 cm
it's highest point is 5 cm, but we are told that after 1/2 second of being released, it reaches its lowest point.
Since highest point is 5, then lowest point will be -5.
The difference in time between the highest and lowest point is ½ s. Which is half of the period.
Thus;
t/2 = ½
Thus, t = 1 s
Now, we know that;
t = 1/f = 2π/ω
Since t = 1, then 1 = 1/f
f = 1
Thus;
2π/ω = 1
Thus, ω = 2π
Thus, the equation is;
y = 5 cos 2πt
The equation that describes the motion of the mass is y = 5 cos 2πt.
The given parameters;
maximum displacement of the spring, A = 5 cmtime taken for the mass to reach the lowest point (half period), t = 0.5 sThe general equation of the wave is given as;
[tex]y = A\ cos\ \omega t[/tex]
where;
A is the amplitude of the vibrationω is the angular speed of massThe angular speed of the mass is calculated as;
[tex]\omega = 2\pi f\\\\[/tex]
The period of the oscillation is calculated as;
[tex]T = 2t \\\\T = 2(0.5 s) = 1 \ s[/tex]
The frequency of the wave is calculated as;
[tex]f = \frac{1}{T} \\\\f = \frac{1}{1} \\\\f = 1\ Hz[/tex]
The equation that describes the motion of the mass is calculated as;
[tex]y = A\ cos \ \omega t\\\\y = A\ cos \ 2\pi ft\\\\y = 5\ cos \ 2\pi (1) t\\\\y = 5 \ cos \ 2\pi t[/tex]
Thus, the equation that describes the motion of the mass is y = 5 cos 2πt.
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Which describes farsightedness? O Distant objects are blurry. O Concave lenses can correct it. O Objects appear larger when wearing corrective glasses. O Corrective glasses do not change apparent the size of objects.
Answer:
O Distant objects are blurry. describes farsightedness.
Explanation:
Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.
An arrow is shot from a height of 1.5 m toward a cliff of height H. It is shot with a velocity of 30 m/s at an angle of 60 degrees above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff
Answer:
The height of the cliff is 27.02 m
Explanation:
Given;
height above the ground from which the arrow is shot, h₁ = 1.5 m
initial velocity of the projectile, v = 30 m/s
angle of projection, θ = 60⁰
time taken to reach top of the cliff, t = 4.0 s
The vertical component of the velocity is calculated as;
[tex]v_y = v \times sin(\theta)\\\\v_y = 30 \times sin(60)\\\\v_y = 25.98 \ m/s[/tex]
The height attained by the projectile at the given time is calculated as;
[tex]h_2= v_y t - \frac{1}{2}gt^2\\\\h_2 = 25.98\times 4 \ - \ \frac{1}{2} \times 9.8\times 4^2\\\\h_2= 103.92 \ - \ 78.4 \\\\h_2 = 25.52 \ m[/tex]
The height of the cliff is calculated as;
H = h₁ + h₂
H = 1.5 m + 25.52 m
H = 27.02 m
I really don’t know the answer for this
In an experiment to measure the temperature of a Bunsen burner flame, a 250 g piece of iron is held in the flame for several minutes until it reaches the same temperature as the flame . The hot metal is then quickly transferred to 285 g of water contained in a 40.0 g copper calorimeter at 15.0 oC. The final temperature of the copper and water is 80.0 oC.
Using your answer from determine the temperature of the Bunsen flame.
Answer:
wait
Explanation:
Ethyl alcohol is :
a. None of the above
b. Semi polar solvent
c. Polar solvent
d. Non-Polar solvant
Answer:
D. Non- polar solvant
Explanation:
l think that's it
Answer:
I think the answer is D polar solvent
a.
70. All the following statements are true. Which one can be explained by Kepler's Second Law?
The Sun is not in the precise center of Saturn's orbit.
b. Earth is sightly closer to the Sun in January than in July
All the planets orbit the Sun in nearly the same plane.
d. Mars moves faster in its orbit when it is closer to the Sun than when it is farther from the Sun.
e. Venus orbits the Sun at a faster orbital speed than Earth.
The statement that can be explained using Kepler's Second Law is "Mars moves faster in its orbit when it is closer to the Sun than when it is farther from the Sun."
According to Encyclopedia Britannica, Kepler's second law states that; "a planet moves in its ellipse so that the line between it and the Sun placed at a focus sweeps out equal areas in equal times."
This law is sometimes referred to as the law of equal areas. It describes the fact that a planet moves faster when it is closer to the sun and slower when it farther from the sun.
An implication of this law is that mars moves faster in its orbit when it is closer to the Sun than when it is farther from the Sun.
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Help help help HELP AAAAA
Una pelota de basket es soltada desde 2.5 m de altura y rebota con una velocidad igual a 3/4 partes de la velocidad que llego. ¿ a qué altura alcanza la bola en el rebote ? ¿ cuánto tiempo transcurre desde que rebota ?
Answer:
Tenemos dos problemas a resolver acá:
Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.
Acá podemos usar la conservación de la energía.
E = U + K
U = energía potencial = m*g*H
m = masa
g = aceleración gravitatoria = 9.8m/s^2
H = altura
K = energía cinética = (m/2)*V^2
donde V es la velocidad.
Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:
Ei = U = m*(9.8m/s^2)*2.5m
Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:
Ef = (m/2)*V^2
Y como la energía se conserva, la energía final es igual a la inicial, entonces:
m*(9.8m/s^2)*2.5m = (m/2)*V^2
Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.
V = √(2*(9.8m/s^2)*2.5m) = 7m/s
Ahora respondamos la segunda parte.
Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:
A(t) = -9.8m/s^2
Para obtener su velocidad integramos:
V(t) = (-9.8m/s^2)*t + V0
donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s
V0 = (3/4)*7m/s = (21/4) m/s
Así, la ecuación de la velocidad es:
V(t) = (-9.8m/s^2)*t + (21/4) m/s
Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:
V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s
t = (21/4) m/s/9.8m/s^2 = 0.54 s
Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.
Para ello integramos de vuelta:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0
donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t
La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s
P(0.54s) = (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m
La altura máxima es 1.81 metros.
Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.
Assume you are going to race the three objects (hollow sphere, disk and ring used in Experiment 8.2) by releasing them from rest at the top of an inclined plane.Which object do you expect to reach the bottom of the inclined plan first? Why?
Answer:
a. The disk
b. Because it has the smallest rotational inertia
Explanation:
a. Which object do you expect to reach the bottom of the inclined plan first?
I would expect the disk to reach the bottom first.
b. Why?
This is because the disk has the smallest rotational inertia.
The rotational inertial of the hollow sphere, disk and ring are 2/3MR², 1/2MR² and MR² respectively.
Since the three objects are rolling from the same height, they have the same mechanical energy.
But, since the disk has the smallest rotational inertia, it would have the smallest rotational kinetic energy and largest translational kinetic energy. The disk's smaller rotational kinetic energy will cause to rotate less but translate more than the other objects and thus reach the bottom first.
The object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.
What is a moment of inertia?Moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.
There are three objects, hollow sphere, disk and ring.
The moment of inertia of the hollow sphere object is given as,[tex]I=\dfrac{2}{3}mr^2[/tex]
The moment of inertia of the ring is,[tex]I=mr^2[/tex]
The moment of inertia of the disk is,[tex]I=\dfrac{1}{2}mr^2[/tex]Here, (m) is the mass and (r) is the radius of the object.
These three objects are going to race by releasing from rest at the top of an inclined plane to the bottom of the plane.
As moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.
Thus the less the value of inertia will result in less the time required to reach at the bottom of the inclined plane.
Hence, the object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.
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Determine the Mutual Inductance per unit length between two long solenoids, one inside the other, whose radii are r1 and r2 (r2 < r1) and whose turns per unit length are n1 and n2.
Answer:
M' = μ₀n₁n₂πr₂²
Explanation:
Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.
So, M = N₂Ф₂₁/i₁
substituting the values of the variables into the equation, we have
M = N₂Ф₂₁/i₁
M = N₂B₁A₂/i₁
M = n₂lμ₀n₁i₁πr₂²/i₁
M = lμ₀n₁n₂πr₂²
So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²
M' = μ₀n₁n₂πr₂²
A jogger travels a route that has two parts. The first is a displacement ->A of 2.20 km due south, and the second involves a displacement ->B that points due east.
(a) The resultant displacement ->A + ->B has a magnitude of 3.81 km.
What is the magnitude of B?
______ km
What is the direction of A + B relative to due south?
_____° west of south or east of south?
(b) Suppose that A - B had a magnitude of 3.81 km. What then would be the magnitude of B and what is the direction of A - B relative to due south?
Answer:
a) B = 3.11 km. θ= 54.7º E of S
b) B = 3.11 km θ= 54.7º W of S
Explanation:
a)
Since we know the value of the total displacement, and the value of the displacement A and its direction, we can find the magnitude of B just applying the Pythagorean Theorem, as follows:[tex]C=\sqrt{(2.2km)^{2} + B^{2} } = 3.81 Km (1)[/tex]
Solving for B, the only unknown, we get:[tex]B=\sqrt{(3.81km)^{2} -(2.2km)^{2} } = 3.11 Km (2)[/tex]
Now, applying some simple trig, we can find the angle that (A+B) makes with the S axis, as follows:[tex]\theta = arc tg (\frac{B}{A} )= arc tg ( \frac{3.11}{2.2} )= arctg (1.414) = 54.7 deg (3)[/tex]
Since it's a positive number, applying the convention that the positive angles are measured counterclockwise, this means that this angle is measured East of South.b)
If the magnitude of A-B is the same than the one for A+B, this means that the magnitude of B remains the same, i.e. 3.11 Km.But if we do graphically A-B, as it is the same as adding A + (-B), we find that the angle of A-B is different to the one in A+B, even the magnitudes of both displacements are the same.In this case, B is a negative number, because it's a displacement due west.So, applying the same trig that for a) we can find the angle that (A-B) makes with the S Axis, as follows:[tex]\theta = arc tg (\frac{-B}{A} )= arc tg ( \frac{-3.11}{2.2} )= arctg (-1.414) = -54.7 deg (4)[/tex]
So, since it is negative, it's measured clockwise from the S axis, so it's 54.7º W of S.A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. We assume the upward direction to be positive, and the downward direction to be negative.
(a) How long are her feet in the air?(b) What is her highest point above the board?(c) What is her velocity when her feet hit the water?
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
[tex]2gh = v_f^2 - v_i^2[/tex]
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
[tex](2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\[/tex]
h = 0.82 m
Now, for the time in air during upward motion we use first equation of motion:
[tex]v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s[/tex]
(c)
Now we will consider the downward motion and use the third equation of motion:
[tex]2gh = v_f^2-v_i^2[/tex]
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
[tex]2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\[/tex]
vf = 7.17 m/s
Now, for the time in air during downward motion we use the first equation of motion:
[tex]v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s[/tex]
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
t = 1.14 s
Ahmed is pushing a 4 Kg box to the right and Rashid is Pushing it to the right as well with a force of 12 N , the box accelerates by 5 m/s^2. What is the Force that is applied by Ahmed
Answer:
8 N
Explanation:
Applying,
(F'+F) = ma............... Equation 1
Where F' = Amhed's force, F = Rashid's force, m = mass of the box, a = acceleration of the box.
From the question,
Given: F = 12 N, m = 4 kg, a = 5 m/s²
Substitute these values into equation 1
(F'+12) = 4×5
(F'+12) = 20
F' = 20-12
F' = 8 N.
Hence Ahmed's force is 8 N
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
The question is incomplete, the complete question is;
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
A) The atom moves to a state of lower energy
B) The atom is ionized
C) One of the electrons leaves the atom
D) The atom can be excited to a higher energy state
Answer:
The atom can be excited to a higher energy state
Explanation:
According to the Bohr model of the atom, electrons in an atom can be excited from a lower to a higher energy level when energy is absorbed by the atom.
If electrons having an energy of 10.5ev are incident on a hydrogen atom, this energy is transferred to the atom by collision. Since the energy transferred is less than the ionization energy of hydrogen atom in its ground state(13.6ev), the atom is not ionized.
Rather, the atom is excited from ground state to a higher energy level.
A 5.0 Ω resistor is hooked up in series with a 10.0 Ω resistor followed by a 20.0 Ω resistor. The circuit is powered by a 9.0 V battery. Draw a labeled circuit diagram for the circuit described using correct symbols. Calculate the equivalent resistance. Calculate the voltage drop across each resistor in the circuit.
(a) Attached to the response as Figure 1.
(b) 35.0Ω
(c) Across 5.0Ω = 1.3V
Across 10.0Ω = 2.6Ω
Across 20.0Ω = 5.2Ω
Explanation:(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.
(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e
R = R₁ + R₂ + R₃ -------------------(i)
Where;
R₁ = 5.0 Ω
R₂ = 10.0 Ω
R₃ = 20.0 Ω
Substitute these values into equation (i) as follows;
∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω
∴ R = 35.0 Ω
Therefore, the equivalent resistance is ∴ R = 35.0Ω
(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;
i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.
ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;
V = I R ------------------(ii)
Where;
V = Voltage supplied to the circuit = 9.0V
I = Current through the circuit
R = Resistance of the equivalent resistor = 35.0Ω
Substitute these values into equation (ii)
9.0 = I x 35.0
I = [tex]\frac{9.0}{35.0}[/tex]
I = 0.26A
This is also the current flowing through each of the resistors separately.
iii. Calculate the voltage drop across
1. 5.0 Ω resistor
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 5.0Ω resistor
I = current through the 5.0Ω resistor = 0.26A
R = resistance of the 5.0Ω resistor = 5.0Ω
=> V = 0.26 x 5.0
=> V = 1.3V
2. 10.0 Ω resistor
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 10.0Ω resistor
I = current through the 10.0Ω resistor = 0.26A
R = resistance of the 10.0Ω resistor = 10.0Ω
=> V = 0.26 x 10.0
=> V = 2.6V
3. 20.0 Ω resistor
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 20.0Ω resistor
I = current through the 20.0Ω resistor = 0.26A
R = resistance of the 20.0Ω resistor = 10.0Ω
=> V = 0.26 x 20.0
=> V = 5.2V
A 1540-kg truck has a wheel base of 3.13 m (this is the distance between the front and rear axles). The center of mass of the truck is 1.09 m behind the front axle.a) What is the force exerted by the ground on each of the front wheels?b) What is the force exerted by the ground on each of the back wheels? Hint: Remember that the truck has four wheels.
a. The force exerted by the ground on each of the front wheels is 4918.16 Newton.
b. The force exerted by the ground on each of the back wheels is 2627.84 Newton.
Given the following data:
Mass of truck = 1540 kgDistance between the front and rear axles = 3.13 meters.Center of mass of the truck = 1.09 meters.a. To determine the force exerted by the ground on each of the front wheels:
First of all, we would take moment about the rear wheels.
[tex]F(3.13) - 1540(9.8) \times (3.13 - 1.09) = 0\\\\3.13F - 15092 \times 2.04 =0\\\\3.13F -30787.68=0\\\\F=\frac{30787.68}{3.13}[/tex]
Force, F = 9836.32 Newton
For each front wheel:
[tex]Force = \frac{9836.32}{2}[/tex]
Force = 4918.16 Newton.
b. To determine the force exerted by the ground on each of the back wheels:
We would determine the sum of the vertical forces acting on the wheels.
[tex]9836.32 + B - 1540(9.8) = 0\\\\9836.32 + B - 15092 = 0\\\\B=15092-9836.32[/tex]
B = 5255.68 Newton.
For each back wheel:
[tex]Force = \frac{5255.68}{2}[/tex]
Force = 2627.84 Newton.
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The time-average power carried by a UPEMW propagating in vacuum is 0.05 W/m2. i) What is the amplitude value of the electric field and the amplitude value of the magnetic field in the wave
Answer:
The correct solution is "11.51 mA".
Explanation:
Given:
Time average power,
[tex]P_{avg}=0.05 \ W/m^2[/tex]
n = 377
As we now,
⇒ [tex]P_{avg}=\frac{E_0^2}{n}[/tex]
or,
⇒ [tex]E_0^2=0.05\times 377[/tex]
⇒ [tex]=4.341 \ V[/tex]
hence,
⇒ [tex]H_0=\frac{E_0}{n}[/tex]
By putting the values, we get
[tex]=\frac{4.341}{377}[/tex]
[tex]=11.51 \ mA[/tex]
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff? ( EASY QUESTION.. PLZZ HELPPP MEEE I WILL MARK YOU THE BRAINLIEST PLZZ)
Answer:
144 meters
Explanation:
the ball is thrown with a speed of 24 meters per second right so if the ball reaches the ground in 6 seconds. the hight of the cliff must be S=v.t
S (height cliff)=24m/s×6s=144
It's nighttime, and you've dropped your goggles into a 3m deep swimming pool. If you hold a laser pointer 1.0m abovve the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0m from the edge. How far are the goggles from the edge of the pool?
Answer:
x_total = 3.07 m
Explanation:
Let's analyze the situation presented, in this case we have the data to find the angle of incidence and with the law of refraction we can find the angle of refraction.
Let's start looking for the angle with which the laser pointer reaches the water, let's use trigonometry
tan θ₁ = 1.0 /2.0
θ₁ = tan⁻¹ 0.50
θ₁ = 26.57º
now let's use the law of refraction to find the angle of refraction (in water)
n₁ sin θ₁ = n₂ sin θ₂
the refractive index of air is n₁ = 1 and that of water n₂ = 1.33
sin θ₂ = [tex]\frac{n_1}{n_2} \ sin \theta_1[/tex]
sin θ₂ = [tex]\frac{1}{1.33} \ sin \ 26.57[/tex]
θ₂ = sin⁻¹ 0.3363
θ₂ = 19.65º
now we can find the distance from the entry point to the water to the lenses
tan θ₂ = x₂ / 3
x₂ = 3 tan θ₂
x₂ = 3 tan 19.65
x₂ = 1.07 m
the total distance from the edge of the pool is
x_total = 2 + x₂
x_total = 2 + 1.07
x_total = 3.07 m
When it comes to the movement of air, friction
A. increases with altitude.
B. is greater near the ground surface.
C. diminishes turbulence.
D. is responsible for weaker winds aloft.
Answer: When it comes to the movement of air, friction is greater near the ground surface.
Explanation:
A resistance in motion observed by an object while on another object is called friction.
For example, a vehicle moving on road will have friction between its tires and the road.
Friction is more near the ground surface rather than away from the ground surface.
Thus, we can conclude that when it comes to the movement of air, friction is greater near the ground surface.
what is electricity ?
Answer:
it is energy resulting in charged particles
Electricity a form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current
