Answer:
Explanation:
a) Energy absorbed by cancerous tissue E = R x M
R is the radiation dosage
M is the mass of the lymphatic
Energy absorbed by cancerous tissue
[tex]=45\times18\times10^-^3J\\\\=810\times10^-^3J[/tex]
b) If only 1% of the total energy Et is absorbed over 20min session for 5 weeks = E
And the time of the period of the course is t = 20 x 5 = 100 min
E = 0.01Et
Total energy of gamma ray beam
Et = E x 100
= 810 x 10⁻³ x 100
= 8100J
Power of gamma ray beam is P
[tex]P=\frac{Et}{t} \\\\=\frac{8100J}{100\times60} \\\\=0.0135W[/tex]
c) The total activity
[tex]A=\frac{-dN}{dt} = dN decays/unit\ time[/tex]
The Number of decays dN = Es Total energy emitted at sourse per seconds / energy emitted per day
[tex]dN=\frac{ Es}{ 0.03Mev}[/tex]
The Energy source Es per seconds = 0.05Ps
P= 0,0135W
The Power emitted at source
[tex]Ps = \frac{0.0135}{0.05} \\\\=0.27W[/tex]
The Energy emitted per decay = 0.03MeV
[tex]=0.03\times10^6\times1.602\times10^-^1^9J[/tex]
[tex]A=\frac{-dN}{dt} \\\\=\frac{0.27}{0.03\times10^6\times1.602\times10^-^1^9} \\\\=\frac{0.27}{0.04806\times10^-^1^3} \\\\=5.618\times10^1^3/sec[/tex]
A toy cannon uses a spring to project a 5.31-g soft rubber ball. The spring is originally compressed by 4.90 cm and has a force constant of 8.09 N/m. When the cannon is fired, the ball moves 14.1 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.031 7 N on the ball.
Required:
a. With what speed does the projectile leave the barrel of the cannon?
b. At what point does the ball have maximum speed?
c. What is this maximum speed?
A 3.10-mm-long, 430 kgkg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 69.0 kgkg construction worker stands at the far end of the beam.What is the magnitude of the gravitational torque about the point where the beam is bolted into place?
Answer:
Explanation:
Given that,
The length of the beam L = 3.10m
The mass of the steam beam [tex]m_1[/tex] = 430kg
The mass of worker [tex]m_2[/tex] = 69.0kg
The distance from the fixed point to centre of gravity of beam = [tex]\frac{L}{2}[/tex]
and our length of beam is 3.10m
so the distance from the fixed point to centre of gravity of beam is
[tex]\frac{3.10}{2}=1.55m[/tex]
Then the net torque is
[tex]=-W_sL'-W_wL\\\\=-(W_sL'+W_wL)[/tex]
[tex]W_s[/tex] is the weight of steel rod
[tex]=430\times9.8=4214N[/tex]
[tex]W_w[/tex] is the weight of the worker
[tex]=69\times9.8\\\\=676.2N[/tex]
Torque can now be calculated
[tex]-(4214\times1.55+676.2\times3.9)Nm\\\\-(6531.7+2637.18)Nm\\\\-(9168.88)Nm[/tex]
≅ 9169Nm
Therefore,the magnitude of the torque is 9169Nmwhen a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. when the paper is turned 25 degree with respect to the field the flux through it is:
Answer:
22.66Nm²/C
Explanation:
Flux through an electric field is expressed as ϕ = EAcosθ
When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. If the paper is turned 25 degree with respect to the field the flux through it can be calculated using the formula.
From the formula above where:
EA = 25N.m^2/C
θ = 25°
ϕ = 25cos 25°
ϕ = 22.66Nm²/C
A mass of 0.450 kg rotates at constant speed with a period of 1.45 s at a radius R of 0.140 m in the apparatus used in this laboratory. What is the rotation period for a mass of 0.550 kg at the same radius
Answer:
1.603 s
Explanation:
Given that
Initial mass, = 0.45 kg
Initial period, = 1.45 s
Initial radius, = 0.14 m
Final mass, = 0.55 kg
Final period, = ?
Final radios, = 0.14 m
Since we are finding the rotation period of two masses of same radius, we can assume that the outward force is the same in both cases. This means that
m₁r₁ω₁² = m₂r₂ω2²
Where, ω = 2π/T, on substituting, we have
0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²
0.45 / 1.45² = 0.550 / T₂²
T₂² = 0.550 * 1.45² / 0.45
T₂² = 2.56972
T₂ = √2.56972
T₂ = 1.603 sec
The rotation period for a mass of 0.550 kg at the same radius is 1.603 s
Calculation of the radius:Since
Initial mass = 0.45 kg
Initial period = 1.45 s
Initial radius = 0.14 m
Final mass = 0.55 kg
Final radios = 0.14 m
Now the following formulas should be used.
m₁r₁ω₁² = m₂r₂ω2²
here ω = 2π/T
So,
0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²
0.45 / 1.45² = 0.550 / T₂²
T₂² = 0.550 * 1.45² / 0.45
T₂² = 2.56972
T₂ = √2.56972
T₂ = 1.603 sec
hence, The rotation period for a mass of 0.550 kg at the same radius is 1.603 s
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Which one of the following is closely related to the law of conservation of
energy, which states that energy can be transformed in different ways but can
never be created or destroyed?
O A. Charles's Law
B. Boyle's Law
C. Second law of thermodynamics
O D. First law of thermodynamics
Answer:
D
Explanation:
Answer:
It is D
Explanation: No cap
Holding force constant, what will be the effect of increasing the Moment arm?
(a) Depends on the direction of the force.
(b) Torque will increase.
(c) Torque is constant.
(d) Torque will decrease.
(e) The direction of rotation will change.
Answer:
(b) Torque will increase.
Explanation:
Torque is given as the product of force and moment arm (radius).
τ = F x r
F = τ / r
where;
F is force
τ is torque
r is radius (moment arm)
Keeping force constant, we will have the following;
τ ∝ r
This shows that torque is directly proportional moment arm (radius), thus increase in moment arm, will cause increase in torque.
For instance;
let the constant force = 5 N
let the initial moment arm, r = 2m
Torque, τ = 5 N x 2m = 10 Nm
When the moment arm is increased to 4 m
Torque, τ = 5 N x 4m = 20 Nm
Therefore, at a constant force, increasing in the Moment arm, will cause increase in torque.
Coorect option is "(b) Torque will increase."
The uniform dresser has a weight of 91 lb and rests on a tile floor for which μs = 0.25. If the man pushes on it in the horizontal direction θ = 0∘, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight of 151 lb , determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip.
Answer:
F = 22.75 lb
μ₁ = 0.15
Explanation:
The smallest force required to move the dresser must be equal to the force of friction between the man and the dresser. Therefore,
F = μR
F = μW
where,
F = Smallest force needed to move dresser = ?
μ = coefficient of static friction = 0.25
W = Weight of dresser = 91 lb
Therefore,
F = (0.25)(91 lb)
F = 22.75 lb
Now, for the coefficient of static friction between shoes and floor, we use the same formula but with the mas of the man:
F = μ₁W₁
where,
μ₁ = coefficient of static friction between shoes and floor
W₁ = Weight of man = 151 lb
Therefore,
22.75 lb = μ₁ (151 lb)
μ₁ = 22.75 lb/151 lb
μ₁ = 0.15
when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.2525 N. What were the initial charges on the spheres
Answer:
q1 = 7.6uC , -2.3 uC
q2 = 7.6uC , -2.3 uC
( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )
Explanation:
Solution:-
- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.
- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:
[tex]F = k\frac{|q_1|.|q_2|}{r^2}[/tex]
Where,
k: The coulomb's constant = 8.99*10^9
- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.
- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.
- Therefore, the force of attraction between the spheres would be:
[tex]\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1[/tex] ... Eq 1
- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).
- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,
[tex]q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}[/tex]
- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:
[tex]\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6[/tex] .. Eq2
- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:
[tex]-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123[/tex]
[tex]q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\[/tex]
A curtain hangs straight down in front of an open window. A sudden gust of wind blows past the window; and the curtain is pulled out of the window. Which law, principle, or equation can be used to explain this movement of the curtain?
a. Poiseuille's law
b. Bernoulli's equation
c. the equation of continuity
d. Archimedes' principle
e. Pascal's principle
Answer:
Bernoulli's equation.Option B is the correct option.
Explanation:
The phenomenon of the curtain to pull out of the window can be explained using Bernoulli's equation.
According to Bernoulli's principle,when the speed of moving fluid increases the pressure within the fluid decrease.When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more.So the curtain moves outside because of low pressure.
Hope this helps...
Good luck on your assignment...
Ocean waves of wavelength 30m are moving directly toward a concrete barrier wall at 4.8m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.A) How far from the wall is she?B) What is the period of her up and down motion?
Answer:
a)15m
b)6.25s
Explanation:
A) She is ½ a wavelength away, or
d = λ/2 = 30/2 = 15 m
B)Speed of the wave:
V=fλ = λ/T
so,
T=λ/V= 30/4.8
T=6.25s
a) The distance from the wall is 15m
b) The period of her up and down motion is 6.25s
Calculation of the distance and period is:a.
Since Ocean waves of wavelength 30m are moving directly toward a concrete barrier wall at 4.8m/s .
Also,
She is ½ a wavelength away, or
d = λ/2
= 30/2
= 15 m
b)
Here the speed of wave should be used
T=λ/V
= 30/4.8
T=6.25s
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A rock falls from a vertical cliff that is 4.0 m tall and experiences no significant air resistance as it falls. At what speed will its gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy
Answer:
About 6.26m/s
Explanation:
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
Divide both sides by mass:
[tex]gh=\dfrac{1}{2}v^2[/tex]
Since the point of equality of kinetic and potential energy will be halfway down the cliff, height will be 4/2=2 meters.
[tex](9.8)(2)=\dfrac{1}{2}v^2 \\\\v^2=39.4 \\\\v\approx 6.26m/s[/tex]
Hope this helps!
The gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.
Given data:
The height of vertical cliff is, h = 4.0 m.
Since, we are asked for speed by giving the condition for gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy. Then we can apply the conservation of energy as,
Kinetic energy = Gravitational potential energy
[tex]\dfrac{1}{2}mv^{2}=mgh[/tex]
Here,
m is the mass of rock.
v is the speed of rock.
g is the gravitational acceleration.
Solving as,
[tex]v=\sqrt{2gh}\\\\v=\sqrt{2 \times 9.8 \times 4.0}\\\\v =8.85 \;\rm m/s[/tex]
Thus, we can conclude that the gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.
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A 50-kg block is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 150 N. Fp is parallel to the displacement of the block. The coefficient of kinetic friction is 0.25.
a) What is the total work done on the block?
b) If the box started from rest, what is the final speed of the block?
Answer:
a) WT = 137.5 J
b) v2 = 2.34 m/s
Explanation:
a) The total work done on the block is given by the following formula:
[tex]W_T=F_pd-F_fd=(F_p-F_f)d[/tex] (1)
Fp: force parallel to the displacement of the block = 150N
Ff: friction force
d: distance = 5.0 m
Then, you first calculate the friction force by using the following relation:
[tex]F_f=\mu_k N=\mu_k Mg[/tex] (2)
μk: coefficient of kinetic friction = 0.25
M: mass of the block = 50kg
g: gravitational constant = 9.8 m/s^2
Next, you replace the equation (2) into the equation (1) and solve for WT:
[tex]W_T=(F_p-\mu_kMg)d=(150N-(0.25)(50kg)(9.8m/s^2))(5.0m)\\\\W_T=137.5J[/tex]
The work done over the block is 137.5 J
b) If the block started from rest, you can use the following equation to calculate the final speed of the block:
[tex]W_T=\Delta K=\frac{1}{2}M(v_2^2-v_1^2)[/tex] (3)
WT: total work = 137.5 J
v2: final speed = ?
v1: initial speed of the block = 0m/s
You solve the equation (3) for v2:
[tex]v_2=\sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(137.5J)}{50kg}}=2.34\frac{m}{s}[/tex]
The final speed of the block is 2.34 m/s
A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the elastic limit is not reached
Answer:
Final Length = 30 cm
Explanation:
The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:
F = kΔx
where,
F = Force applied
k = spring constant
Δx = change in length of spring
First, we find the spring constant of the spring. For this purpose, we have the following data:
F = 50 N
Δx = change in length = 25 cm - 20 cm = 5 cm = 0.05 m
Therefore,
50 N = k(0.05 m)
k = 50 N/0.05 m
k = 1000 N/m
Now, we find the change in its length for F = 100 N:
100 N = (1000 N/m)Δx
Δx = (100 N)/(1000 N/m)
Δx = 0.1 m = 10 cm
but,
Δx = Final Length - Initial Length
10 cm = Final Length - 20 cm
Final Length = 10 cm + 20 cm
Final Length = 30 cm
Near the surface of Earth, an electric field points radially downward and has a magnitude of approximately 100 N/C. 1) What charge (magnitude and sign) would have to be placed on a penny that has a mass of 3.11 g to cause it to rise into the air with an upward acceleration of 0.190 m/s2
Answer:
Negative charge with a magnitude of 5.9*10^-6C
Explanation:
In order to calculate the required charge, for the penny is able to rise into the air with the given acceleration, you take into account the following formula for the electric force and also the Newton second law:
[tex]F_E=qE=ma[/tex] (1)
FE: electric force:
q: charge = ?
E: electric field = 100N/C
m: mass of the penny = 3.11g = 3.11*10^-3kg
a: acceleration of the charge = 0.190m/s^2
The positive charges experiences a force in the same direction of the electric field, and the negative charge, a force in a opposite direction.
The electric field points downward, then, the charge is negative (it is acceleration upwards).
Next, you solve the equation (1) for a, and replace the values of the other parameters:
[tex]q=\frac{ma}{E}=\frac{(3.11*10^{-3}kg)(0.190m/s^2)}{100N/C}=5.9*10^{-6}C[/tex]
The magnitude of the charge is 5.9*10^-6C and its sign is negative
The drag on a 2-m-diameter satellite dish due to an 81-km/hr wind is to be determined through a wind tunnel test using a geometrically similar 0.4-m-diameter model dish. Assume standard air for both model and prototype. (a) At what air speed should the model test be run
Answer:
The correct answer is - 405 km/hr
Explanation:
given:
2-m- = diameter satellite dish
81-km/hr = Y
0.4-m- = diameter model dish.
Reynolds number is essential for this, thus,
Vm Dm/vm = VD/v
(D is the diameter of the dish)
Vm = (vm/v) (D/Dm) Y
and if vm/v = 1, then
Vm = (2/0.4) (81)
= 405 km/hr
Thus, the correct answer is - 405 km/hr
disadvantage of vb language
Answer:
visual basics
Explanation:
not suited for programming, slower than the other languages. hard to translate to other operating systems
Using only the trainiris dataset, for each feature, perform a simple search to find the cutoff that produces the highest accuracy, predicting virginica if greater than the cutoff and versicolor otherwise. Use the seqfunction over the range of each feature by intervals of 0.1 for this search. Which feature produces the highest accuracy?
A. Sepal. Length
B. Sepal. Width
C. Petal. Length
D. Petal. Width
Answer: C. Petal. Length
Explanation: Petal are unit of Corolla which are usually brightly colored. This part of a plant or flower, helps attracts insects to the plant for pollination. And also provide protection to the reproductive parts of the plant or flower.
Examples of flowers with petals is the Sun Flower, which coincidentally is the flower plant with most petals.
A large motor in a factory causes the
floor to vibrate at a frequency of 10 Hz.
The amplitude of the floor's motion
near the motor is about 3.0 mm.
Estimate the maximum acceleration of
the floor near the motor.
Answer:
12 m/s²
Explanation:
Maximum acceleration is a = Aω², where A is the amplitude and ω is the frequency.
a = (0.0030 m) (10 rev/s × 2π rad/rev)²
a = 12 m/s²
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the floor. A man weighing 700 N could climb up to 7.0 m before slipping. What is the coefficient of static friction between the floor and the ladder
Answer:
μ = 0.37
Explanation:
For this exercise we must use the translational and rotational equilibrium equations.
We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive
let's write the rotational equilibrium
W₁ x/2 + W₂ x₂ - fr y = 0
where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances
cos 60 = x / L
where L is the length of the ladder
x = L cos 60
sin 60 = y / L
y = L sin60
the horizontal distance of man is
cos 60 = x2 / 7.0
x2 = 7 cos 60
we substitute
m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0
fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60
let's calculate
fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)
fr = (735 + 2450) / 8.66
fr = 367.78 N
the friction force has the expression
fr = μ N
write the translational equilibrium equation
N - W₁ -W₂ = 0
N = m₁ g + W₂
N = 30 9.8 + 700
N = 994 N
we clear the friction force from the eucacion
μ = fr / N
μ = 367.78 / 994
μ = 0.37
An ideal, or Carnot, heat pump is used to heat a house to a temperature of 294 K (21 oC). How much work must the pump do to deliver 3000 J of heat into the house (a) on a day when the outdoor temperature is 273 K (0 oC) and (b) on another day when the outdoor temperature is 252 K (-21 oC)
Answer:
a) [tex]W_{in} = 214.286\,J[/tex], b) [tex]W_{in} = 428.571\,J[/tex]
Explanation:
a) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.
[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.
Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 273\,K[/tex]. The Coefficient of Performance is:
[tex]COP_{HP} = \frac{294\,K}{294\,K-273\,K}[/tex]
[tex]COP_{HP} = 14[/tex]
Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.
[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]
The input work to deliver a determined amount of heat to the house:
[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]
If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 14[/tex], the input work that is needed is:
[tex]W_{in} = \frac{3000\,J}{14}[/tex]
[tex]W_{in} = 214.286\,J[/tex]
b) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.
[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.
Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 252\,K[/tex]. The Coefficient of Performance is:
[tex]COP_{HP} = \frac{294\,K}{294\,K-252\,K}[/tex]
[tex]COP_{HP} = 7[/tex]
Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.
[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]
The input work to deliver a determined amount of heat to the house:
[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]
If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 7[/tex], the input work that is needed is:
[tex]W_{in} = \frac{3000\,J}{7}[/tex]
[tex]W_{in} = 428.571\,J[/tex]
A woman stands on a bathroom scale in an elevator that is not moving. The scale reads 588 N. The elevator then moves down at an acceleration of - 2.8 m/s2. What does the scale read while the elevator moves down
Answer:
756 N.
Explanation:
So, we are given the following data or parameters or information which is going to assist us in solving this particular Question or problem effectively;
=> The bathroom scale in an elevator is not moving.
=> " The scale reads 588 N. "
=> "The elevator then moves down at an acceleration of - 2.8 m/s2."
For the second point above: 588 N = mass × gravity.
Thus, the scale that will read while the elevator moves down is given below;
(Mass × gravity) - (mass × acceleration).
(The value of scale when it is not moving) × (1 + acceleration/gravity).
=> 588 × { 1 + 2.8/9.8} = 756 N
Determine the type of reaction (S, D, SD, or DD) and balance the chemical equation. (Hint: Please place a coefficient of 1 in front of any formula
that does not need its atoms to increase.)
Type of reaction
НІ
--
H2
+
12
Blank 1: 1
Blank 2:
Blank 3:
Blank 4:
Answer and Explanation: The types of reaction represented in the question are the following: S is Synthesis; D is Decomposition; SD is Single Displacement; and DD is Double Displacement.
The reaction showed is: HI ⇄ H₂ + I₂
This reaction's type is Decomposition, because it is a reaction in which a compund breaks down into 2 or more simpler compounds. Most of them requires energy that could be in form of heat, light or electricity.
To balance a chemical equation, remember that, according to the Conservation of Mass Law, matter can not be created or destroyed just transformed. Then, the amount of reagents that enters the reaction has to be same amount that reacts to become the products, i.e.:
For the decomposition reaction, there are 2 atoms of hydrogen and 2 atoms of iodine being produced while only 1 atom of each element enters the reaction.
The balanced reaction will be: 2HI ⇄ H₂ + I₂
Explain whether or not there is any difference between a light ray emitted by a candle flame and one reflected off the cover of a book. b. Determine whether the reflection off the cover of a book is specular or diffuse and explain your answer.
Answer:
the difference between the two is that the candle forms an emission spectrum and the book an absorption spectrum.
the book it is observed in all directions so that its reflection has to be diffused
Explanation:
The ray of light emitted by a candle is the light generated by the temperature of the flame, which is made up of the emissions of a black body at this temperature plus the emissions of the chemical elements that make up the candle.
The Light reflected from the cover of a book is the same incident light spectrum minus the wavelengths that create transitions in the elements of the cover, these wavelengths will be seen as dark areas.
As a consequence of the above, the difference between the two is that the candle forms an emission spectrum and the book an absorption spectrum.
For the cover of the book form a specular reflection the incident rays are reflected in one direction and the rest would be dark, but in the book it is observed in all directions so that its reflection has to be diffused
Answer:
There is no difference between a light ray emitted by a candle flame and one reflected off the cover of a book. The candle flame is a source of light rays, but those rays could travel to the book cover and reflect off the book cover diffusely so that the same ray is actually emitted by the candle flame and reflected by the book. The reflection off the book is diffuse reflection because the book is visible from any angle.
Explanation:
A traveling electromagnetic wave in a vacuum has an electric field amplitude of 62.5 V/m. Calculate the intensity S of this wave. Then, determine the amount of energy ???? that flows through area of 0.0231 m2 over an interval of 14.9 s, assuming that the area is perpendicular to the direction of wave propagation.
Answer:
a) 5.19 W/m²
b) 1.79 J
Explanation:
For the calculation of intensity, I. We have
I = E(rms)² / (cμ), where
c = speed of light
μ = permeability of free space
I = (62.5 / √2)² / [(2.99 x 10^8) (1.26 x 10^-6)]
I = 1954 / 376.74
I = 5.19 W/m²
Therefore, the intensity, I = 5.19 W/m²
t = 14.9 s
A = 0.0231 m²
Amount if energy flowing, U = IAt
U = (5.19) (0.0231) (14.9) J
U = 1.79 J
A car traveling with velocity v is decelerated by a constant acceleration of magnitude a. It takes a time t to come to rest. If its initial velocity were doubled, the time required to stop would
Answer:
If the initial speed is doubled the time is also doubled
Explanation:
You have that a car with velocity v is decelerated by a constant acceleration a in a time t.
You use the following equation to establish the previous situation:
[tex]v'=v-at[/tex] (1)
v': final speed of the car = 0m/s
v: initial speed of the car
From the equation (1) you solve for t and obtain:
[tex]t=\frac{v-v'}{a}=\frac{v}{a}[/tex] (2)
To find the new time that car takesto stop with the new initial velocity you use again the equation (1), as follow:
[tex]v'=v_1-at'[/tex] (3)
v' = 0m/s
v1: new initial speed = 2v
t': new time
You solve the equation (3) for t':
[tex]0=2v-at'\\\\t'=\frac{2v}{a}=2t[/tex]
If the initial speed is doubled the time is also doubled
A 20-kg object sitting at rest is struck elastically in a head-on collision with a 10-kg object initially moving at 3.0 m/s. Find the final velocity of the 10-kg object after the collision.
Answer:1m/s
Explanation: As the stationary ball is hit by the moving ball ,the two moves together after collision, with a single velocity. The attached photo further explains how the answer is calculated
When stationary ball is hit by the moving ball, both the balls moves together after collision. The final velocity of the object after collision is 1 m/s.
When stationary ball is hit by the moving ball, both the balls moves together after collision
The conservation of momentum,
[tex]\bold {m_1 u_1 + m_2u_2 = (m_1+m_2) V}\\[/tex]
Where,
m1 - initial mass = 20 kg
m2 - final mass =10 kg
u1 - initial velocity = 0 m/s (object at rest)
u2 - final velocity = 3 m/s
V- velocity after collision = ?
Put the values int he formula and calculate for V2,
[tex]\bold { 10 \times 0 + 20 \times 3 = (10+20) V}\\\\\bold {V = \dfrac {30}{30}}\\\\\bold {V = 1\ m/s}[/tex]
Therefore, final velocity of the object after collision is 1 m/s.
To know more about velocity,
https://brainly.com/question/13639113
A child of mass 46.2 kg sits on the edge of a merry-go-round with radius 1.9 m and moment of inertia 130.09 kg m2 . The merrygo-round rotates with an angular velocity of 2.4 rad/s. The child then walks towards the center of the merry-go-round and stops at a distance 0.779 m from the center. Now what is the angular velocity of the merry-go-round
Answer:
The angular velocity is [tex]w_f = 4.503 \ rad/s[/tex]
Explanation:
From the question we are told that
The mass of the child is [tex]m_c = 46.2 \ kg[/tex]
The radius of the merry go round is [tex]r = 1.9 \ m[/tex]
The moment of inertia of the merry go round is [tex]I_m = 130.09 \ kg \cdot m^2[/tex]
The angular velocity of the merry-go round is [tex]w = 2.4 \ rad/s[/tex]
The position of the child from the center of the merry-go-round is [tex]x = 0.779 \ m[/tex]
According to the law of angular momentum conservation
The initial angular momentum = final angular momentum
So
[tex]L_i = L_f[/tex]
=> [tex]I_i w_i = I_fw_f[/tex]
Now [tex]I_i[/tex] is the initial moment of inertia of the system which is mathematically represented as
[tex]I_i = I_m + I_{b_1}[/tex]
Where [tex]I_{b_i}[/tex] is the initial moment of inertia of the boy which is mathematically evaluated as
[tex]I_{b_i} = m_c * r[/tex]
substituting values
[tex]I_{b_i} = 46.2 * 1.9^2[/tex]
[tex]I_{b_i} = 166.8 \ kg \cdot m^2[/tex]
Thus
[tex]I_i =130.09 + 166.8[/tex]
[tex]I_i = 296.9 \ kg \cdot m^2[/tex]
Thus
[tex]I_i * w_i =L_i= 296.9 * 2.4[/tex]
[tex]L_i = 712.5 \ kg \cdot m^2/s[/tex]
Now
[tex]I_f = I_m + I_{b_f }[/tex]
Where [tex]I_{b_f}[/tex] is the final moment of inertia of the boy which is mathematically evaluated as
[tex]I_{b_f} = m_c * x[/tex]
substituting values
[tex]I_{b_f} = 46.2 * 0.779^2[/tex]
[tex]I_{b_f} = 28.03 kg \cdot m^2[/tex]
Thus
[tex]I_f = 130.09 + 28.03[/tex]
[tex]I_f = 158.12 \ kg \ m^2[/tex]
Thus
[tex]L_f = 158.12 * w_f[/tex]
Hence
[tex]712.5 = 158.12 * w_f[/tex]
[tex]w_f = 4.503 \ rad/s[/tex]
A girl and her bicycle have a total mass of 40 kg. At the top of the hill her speed is 5.0 m/s. The hill is 10 m high and 100 m long. If the force of friction as she rides down the hill is 20 N, what is her speed at the bottom
Answer:
v = 11 m/s is her final speed
Explanation:
work done by gravity = m g Δh = 40×9.8×10 = 3920 Joules
Work done by friction = - force×distance = - 20×100 = - 2000 Joules
[minus sign because friction force is opposite to the direction of motion]
Initial K.E. = (1/2) m u^2 = (1/2) × 40 × 5^2 = 500 Joules
Now, by work energy theorem
Work done = change in kinetic energy.
Final K.E. = initial K.E. + total work = 500 + 3920 - 2000 = 2420 Joules
Now, Final K.E. = (1/2) m v^2 [final speed being v= speed at the bottom]
⇒ 2420 = (1/2)×40×v^2
⇒ 121 = v^ 2
v = 11 m/s is her final speed
Please answer the question in the picture below! Need it urgently!!!
It’s related to force and pressure! Pls help!!
Answer:
Both the parts will weigh the same
Explanation:
A disk of mass m and moment of inertia of I is spinning freely at 6.00 rad/s when a second identical disk, initially not spinning, is dropped onto it so that their axes coincide. In a short time, the two disks are corotating. What is the angular speed of the new system
Answer:
The angular speed of the new system is [tex]3\,\frac{rad}{s}[/tex].
Explanation:
Due to the absence of external forces between both disks, the Principle of Angular Momentum Conservation is observed. Since axes of rotation of each disk coincide with each other, the principle can be simplified into its scalar form. The magnitude of the Angular Momentum is equal to the product of the moment of inertial and angular speed. When both disks begin to rotate, moment of inertia is doubled and angular speed halved. That is:
[tex]I\cdot \omega_{o} = 2\cdot I \cdot \omega_{f}[/tex]
Where:
[tex]I[/tex] - Moment of inertia of a disk, measured in kilogram-square meter.
[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.
[tex]\omega_{f}[/tex] - Final angular speed, measured in radians per second.
This relationship is simplified and final angular speed can be determined in terms of initial angular speed:
[tex]\omega_{f} = \frac{1}{2}\cdot \omega_{o}[/tex]
Given that [tex]\omega_{o} = 6\,\frac{rad}{s}[/tex], the angular speed of the new system is:
[tex]\omega_{f} = \frac{1}{2}\cdot \left(6\,\frac{rad}{s} \right)[/tex]
[tex]\omega_{f} = 3\,\frac{rad}{s}[/tex]
The angular speed of the new system is [tex]3\,\frac{rad}{s}[/tex].