drawing shows a force vector that has a magnitude of 475 newtons.
Find the
(a) X,
(b) y, and
(c) z components of the vector.
X, Y, and Z components of the vector are 398, 384 and 279 resp.
Vector is a physical quantity which has both magnitude and direction. Vector A can be written as A = a₁i + a₂j + a₃k where a₁, a₂, a₃ are components along X, Y, Z axis resp. and i,j,k, are the unit vectors along X,Y,Z axis resp.
In this figure
vector F is at angle 36° from y axis, hence
x = Fcos33 = 475cos33 = 398 N
y = Fcos36 = 475cos36 = 384 N
z = Fsin36 = 475sin36 = 279 N
The vector can be written as
F = 398i + 384j + 279k
Hence x, y and z components of this force is 398, 384 and 279 resp.
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Which of the following is NOT an interaction among body systems?
A. The circulatory system works with the digestive system to deliver digested nutrients to cells.
B. The nervous system send signals to the musculoskeletal system to cause movement.
C. The excretory system works with the circulatory system to remove toxic carbon dioxide from cells.
D. The immune system gathers information for the respiratory system to aid in white blood cell production.
Option C, "The excretory system works with the circulatory system to remove toxic carbon dioxide from cells," is NOT an interaction among body systems.
Carbon dioxide is a waste product of cellular respiration, and it is removed from the body by the respiratory system, not the excretory system. The circulatory system plays a role in transporting carbon dioxide from the body's tissues to the lungs, where it can be exhaled.
Answer:
C: The excretory system works with the circulatory system to remove toxic carbon dioxide from cells.
Explanation:
In the figure particle 1 (of charge +8.45 mC), particle 2 (of charge +8.45 mC), and particle 3 (of charge Q) form an equilateral triangle of edge length a. For what value of Q (both sign and magnitude) does the net electric field produced by the particles at the center of the triangle vanish?
The value of Q for which the net electric field produced by the particles at the center of the triangle vanishes is, Q = −5.63 × 10⁻³ C, with a negative sign indicating that the charge is negative.
Let's choose a coordinate system where the center of the equilateral triangle is at the origin, and the particles are located at the vertices of an equilateral triangle of edge length a. Then, the electric field produced by each particle at the center is:
E₁ = k * q₁ / r₁², where q₁ = +8.45 mC, r₁ = a / √3
E₂ = k * q₂ / r₂², where q₂ = +8.45 mC, r₂ = a / √3
E₃ = k * Q / r₃², where r₃ = a
Here, k is Coulomb's constant, which is approximately equal to 9 × 10⁹ N⋅m²/C².
Since the three particles are equally distant from the center of the triangle, the magnitude of the net electric field at the center is:
|E_net| = |E₁ + E₂ + E₃|
Using the above equations for E₁, E₂, and E₃, we can substitute the values and simplify the expression:
|E_net| = k * (q₁ / r₁² + q₂ / r₂² + Q / r₃²)
= k * [8.45 × 10⁻³ C / (a² / 3) + 8.45 × 10⁻³ C / (a² / 3) + Q / a²]
= k * [(16.9 × 10⁻³ C) / (a² / 3) + Q / a²]
For the net electric field to be zero, we need:
|E_net| = k * [(16.9 × 10⁻³ C) / (a² / 3) + Q / a²] = 0
This implies:
(16.9 × 10⁻³ C) / (a² / 3) + Q / a² = 0
Solving for Q, we get:
Q / a² = −(16.9 × 10⁻³ C) / (a² / 3)
Q = −16.9 × 10⁻³ C / 3
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1. Calculate the Energy of skater at all the positions shown. Position C is the highest point the skater reaches
The energy of the skater at each position is:
A: 1920 JB: 1764 JC: 3528 JHow to calculate conservation of energy?At position A, the skater is at the lowest point, so the PE is zero. The KE can be calculated using the formula KE = (1/2)mv², where m is the mass of the skater and v is the velocity:
KE = (1/2)(60 kg)(8 m/s)²
KE = 1920 J
Therefore, at position A, the skater has 1920 J of kinetic energy and 0 J of potential energy.
At position B, the skater has gained some height, so there is some potential energy. The KE can be calculated as before, and the PE can be calculated using the formula PE = mgh, where m is the mass of the skater, g is the acceleration due to gravity (9.81 m/s²), and h is the height:
KE = (1/2)(60 kg)(8 m/s)²
KE = 1920 J
PE = (60 kg)(9.81 m/s²)(3 m)
PE = 1764 J
Therefore, at position B, the skater has 1920 J of kinetic energy and 1764 J of potential energy.
At position C, the skater has reached the highest point, so the KE is zero. The PE can be calculated as before:
PE = (60 kg)(9.81 m/s²)(6 m)
PE = 3528 J
Therefore, at position C, the skater has 0 J of kinetic energy and 3528 J of potential energy.
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A rectangle loop with a length of 3 mm and width of 6 mm is positioned in a uniform magnetic field of magnitude 0.5 N/C so that the plane of the loop makes an angle of 75° with the magnetic field. Find the flux passing through the rectangle loop.
The Gift of the Magi
by O Henry
After Della counted her money she flopped down on the couch and began to scream and cry. Sobs, sniffles and smiles seem to be a progression from sadness to satisfaction.
Read the passage closely and answer the following question:
On reflection, what did Della (Mrs. James Dillingham Young) decide that life was made up of?
On reflection, what did Della (Mrs. James Dillingham Young) decide that life was made up of happiness.
In the context of mental or emotional states, happiness refers to good or pleasant emotions ranging from satisfaction to profound delight. Life satisfaction, well-being, subjective well-being, flourishing, and eudaimonia are some of the other types.
Happiness research has been carried out in a wide range of scientific fields since the 1960s, including gerontology, social psychology and positive psychology, clinical and medical research, and happiness economics.
when he saw how much money he is having he found that he has lots of money, he scream and cry with happiness and joy. Then he decided that the life is made up of happiness.
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A portable power source is available for travelers who need electricity for
appliances. The power scurce provides 54 W of power to operate an air
compressor for inflating tires. This compressor draws 4.5 A of current when
connected to the power supply. What is the voltage across the compressor?
The voltage across the compressor is 12 V.
Power of the power source, P = 54 W
Current utilized, I = 4.5 A
The equation for power of a circuit is given by,
P = VI
Therefore, voltage across the compressor,
V = P/I
V = 54/4.5
V = 12 V
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1. A neutron has a neutral charge because:
a. it is composed of positive quarks and negative electrons
b. it is composed of an equal number of protons and electrons
c. it is composed of an equal number of positive and negative electrons
d. it contains a specific combination of quarks
A neutron has a neutral charge because it is composed of an equal number of protons and electrons. Hence option C is correct.
The neutron is a subatomic particle with a neutral (neither positive nor negative) charge and a slightly larger mass than a proton. Atomic nuclei are made up of protons and neutrons. Protons and neutrons are both referred to as nucleons because they function similarly within the nucleus and each have a mass of around one atomic mass unit. Nuclear physics describes their characteristics and interactions. Protons and neutrons are not elementary particles; they are made up of three quarks apiece.
Hence option C is correct.
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A radioactive source has decayed to 1/10 of 1% of its initial activity in 100 days. What is its half life period?
The half-life period of the radioactive source is approximately 693.15 days.
The activity of a radioactive source is known to follow an exponential decay law given by:
A(t) = A(0) × (1/2)[tex]^{t/T}[/tex]
where,
A(t) = activity at time t
A(0) = initial activity
T = half-life period and (1/2)[tex]^{t/T}[/tex] is the fraction of the original activity remaining after time t.
We are given that the activity of the source has decayed to 1/10 of 1% of its initial activity, which is equivalent to 0.001 times the initial activity. This means that:
A(t) = 0.001 ) × A(0)
We are also given that this has occurred in 100 days, so:
t = 100
Substituting these values in the equation, we get:
0.001 × A(0) = A(0) × (1/2)¹⁰⁰/[tex]^T[/tex]
Simplifying and solving for T, we get:
T = -100 / In(1/2) × log(0.001))
T ≈ 693.15 days
Therefore, the half-life period of the radioactive source is approximately 693.15 days.
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The half-life of a radioactive source that decayed to 1/10 of 1% of its initial activity in 100 days is approximately 14.61 days.
Explanation:The given problem can be solved using the formula for radioactive decay, which is N = N0 * (1/2)^(t/h), where N is the final quantity, N0 is the initial quantity, t is time passed, and h is the half-life time. Here, the radioactive source has decayed to 1/10 of 1% of its initial activity, meaning N = 0.001 * N0. The time passed is 100 days. Plugging these values into the formula we have: 0.001 = (1/2)^(100/h). Solving for h, the half-life time, gives us a half-life of approximately 14.61 days.
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