Answer:
According to federal dietary guidelines, Magnesium among these nutrients is under-consumed.
The diets of most people in the United States provide less than the recommended amounts of magnesium. Old age (people above 65 to 70 mostly) and younger people are most likely to have low intakes of magnesium.
Low magnesium intake for a long period of time, however, can lead to magnesium deficiency and some severe illness due to these deficiencies.
Explanation:
3. A small statue is recovered in an archaeological dig. Its weight is measured to be 96 lb, and its volume 0.08 ft3. a. What is the statue’s weight density?
Answer:
[tex]d=1200\ lb/ft^3[/tex]
Explanation:
Given that,
The weight of the statue, m = 96 lb
The volume of the statue = 0.08 ft³
We need to find the statue’s weight density. We know that the density of an object is the mass of an object divided by its volume. So,
[tex]d=\dfrac{m}{V}\\\\d=\dfrac{96\ lb}{0.08\ ft^3}\\\\d=1200\ lb/ft^3[/tex]
So, the density of the statue is equal to [tex]1200\ lb/ft^3[/tex].
A proton moves across a magnetic field and feels a force. If an electron were to
move at the same speed in the same direction across the same magnetic field, the
force
O 1) would be smaller in the opposite direction
O2) would be the same magnitude in the same direction
3) would be the same magnitude in the opposite direction
4) would be larger in the opposite direction
5) would be smaller in the same direction
potang ina mooooooo bubu hayop kaaaaapestrng yawa
Explanation:
peste kakkkkaaaaaaaa bubu pesteng yawaaaayaka kaayu kaaaaaaa
3. A 5 gm/100 ml solution of drug X is stored in a closed test tube
at 25°C. If the rate of degradation of the drug is 0.05 day-1,
calculate the time required for the initial concentration to
drop to (a) 50% (half-life) and (b) 90% (shelf-life) of its initial
value.
Answer:
See explanation
Explanation:
The degradation of the drug is a first order process;
Hence;
ln[A] = ln[A]o - kt
Where;
ln[A] = final concentration of the drug
ln[A]o= initial concentration of the drug = 5 gm/100
k= degradation constant = 0.05 day-1
t= time taken
When [A] =[ A]o - 0.5[A]o = 0.5[A]o
ln2.5 = ln5 - 0.05t
ln2.5- ln5 = - 0.05t
t= ln2.5- ln5/-0.05
t= 0.9162 - 1.6094/-0.05
t= 14 days
b) when [A] = [A]o - 0.9[A]o = 0.1[A]o
ln0.5 = ln5 -0.05t
t= ln0.5 - ln5/0.05
t= -0.693 - 1.6094/-0.05
t= 46 days
The plum pudding model of the atom states that
Answer:
According to this model, the atom is a sphere of positive charge, and negatively charged electrons are embedded in it to balance the total positive charge.
Explanation:
Hope this helps you
Answer:
The plum pudding model of the atom states that had negatively-charged electrons embedded within a positively-charged "soup."
Explanation:
Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup." Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus. Based on these results, Rutherford proposed the nuclear model of the atom.
Part
A block of mass 2 kg is acted upon by two forces: 3 N (directed to the left) and 4 N (directed to the right). What can you say about the blocks motion?
Answer:
Explanation:
the block will move to the right side with small velocity because the force from the left side greater than force from right side. Velocity will be less because of friction and gravitational attraction.
If a block of mass 2 kg is acted upon by two forces: 3 N (directed to the left) and 4 N (directed to the right), then the block would move towards the right side.
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
As given in the problem statement If a block of mass 2 kg is acted upon by two forces: 3 N (directed to the left) and 4 N (directed to the right),
The net force acting on the block = 4 Newtons - 3 newtons
= 1 Newton
Thus, we can say that the block would move to the right side.
To learn more about Newton's second law here, refer to the link given below;
brainly.com/question/13447525
#SPJ2
dipole moment are used to calculate the
Answer:
Explanation:
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mấy bạn việt nam giúp mình với. cần gấp quá
A spherical light bulb dissipates 100W and is of 5cm diameter. Assume the emissivity is 0.8 and the irradiation is negligible. What is the surface temperature of this spherical light bulb
Answer:
[tex]T=728.9K[/tex]
Explanation:
Power [tex]P=100W[/tex]
Diameter [tex]d=5[/tex]
Radius [tex]r=2.5cm=>2.5*10^{-2}m[/tex]
Emissivity [tex]e=0.8[/tex]
Generally the equation for Area of Spherical bulb is mathematically given by
[tex]A=4\pi r^2[/tex]
[tex]A=4\pi (2.5*10^{-2}m)^2[/tex]
[tex]A=7.85*10^{-3}m^2[/tex]
Generally the equation for Emissive Power bulb is mathematically given by
[tex]E=e\mu AT^4[/tex]
Where
[tex]\mu=Boltzmann constants\\\\\mu=5.67*10^{-8}[/tex]
Therefore
[tex]T^4=\frac{E}{e\mu A}[/tex]
[tex]T^4=\frac{100}{0.8*5.67*10^{-8}*7.85*10^{-3}m^2}[/tex]
[tex]T=^4\sqrt{2.80*10^{11}}[/tex]
[tex]T=728.9K[/tex]
Suppose a proton is moving with a speed of 10 m/s in a direction parallel to a uniform magnetic field of 3.0 T. What is the magnitude and direction of the magnetic force on the proton
Answer:
the magnetic force on the proton is zero.
Explanation:
Given;
speed of the proton, v = 10 m/s
magnitude of the magnetic field, B = 3 T
The magnitude of the magnetic force on the particle is calculated as;
F = qvBsinθ
where;
θ is the angle between the velocity of the particle and the magnetic field
Since the particle is moving parallel to the magnetic field, θ = 0
F = qvBsin(0)
F = 0
Therefore, the magnetic force on the proton is zero.
Answer:
This is a trick question, in that the numbers do not matter.
Study the dependence of the magnetic force on the direction of the magnetic field and the direction of motion of the particle. When is it a maximum? When is it a minimum?
Hint: In vector notation, this is often expressed as q v x B, where q is the electric charge of the particle, v is its velocity (a vector), and B is the magnetic field vector.
(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find the first six positive frequencies that will be present in the replicated spectrum.
Answer:
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
Explanation:
Given :
Frequencies of the sinusoids,
[tex]$f_{m_1}= 65 \ Hz$[/tex] , and
[tex]$f_{m_2}= 95 \ Hz$[/tex]
Sampling rate [tex]f_s = \ 245 \ Hz[/tex]
The positive frequencies at the output of the sampling system are :
[tex]$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $[/tex]
When n = 0,
[tex]$f_{o_1}= f_{m_1} = 65 \ Hz,\ \ f_{o_2}= f_{m_2} = 95 \ Hz $[/tex]
when n = 1,
[tex]$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm f_s $[/tex]
[tex]$f_{o_1}= \pm 65 \pm 245,\ \ f_{o_2}=\pm 95 \pm 245$[/tex]
[tex]$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \ f_{o_2}= 150 \ Hz,340 \ Hz$[/tex]
When n = 2,
[tex]$f_{o_1}= \pm 65 \pm 2(245),\ \ f_{o_2}=\pm 95 \pm 2(245)$[/tex]
[tex]$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \ f_{o_2}= 395 \ Hz,585 \ Hz$[/tex]
Therefore, the first six positive frequencies present in the replicated spectrum are :
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
g A circular swimming pool has a diameter of 10 m, the sides are 4 m high, and the depth of the water is 3 m. How much work (in Joules) is required to pump all of the water over the side
Answer:
[tex]W=2.69*10^{6}J[/tex]
Explanation:
Generally the equation for momentum is mathematically given by
Diameter [tex]d=10m[/tex]
Radius [tex]r =\frac{d}{2}=>5m[/tex]
Height [tex]h=4m[/tex]
Depth [tex]d=3m[/tex]
Generally the equation for Area of the circle is mathematically given by
[tex]A=\pi r^2[/tex]
[tex]A=\pi 5^2[/tex]
[tex]A=78.5m^2[/tex]
Generally the equation for Work done is mathematically given by
[tex]d_w=d_f*x[/tex]
Where
[tex]d_f=(\rho*d_v)*g[/tex]
Therefore
[tex]d_w=((\rho*d_v)*g)*x[/tex]
[tex]d_w=((1000*78.5)*9.8)*x dx[/tex]
[tex]d_w=769300x dx[/tex]
[tex]W=\int dw[/tex]
[tex]\int dw=769300 \int^4_{3}x dx[/tex]
[tex]W=(769300\2)*7[/tex]
[tex]W=2.69*10^{6}J[/tex]
A car (m=1200kg) accelerates at 3m/s/s for d=10m. How much work has the engine done? *
A) 5400J
B) 36000J
C) 352800J
show your work please
This is the answer hope it helps
What is the voltage drop across an alarm clock that is connected to a circuit with a current of 1.10A and a resistance of 90Ω?
Answer:
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Explanation:
What is the electric field at a point 0.450 m to the left of a -5.77*10^-9 C charge?
Include a + or - sign to indicate the direction the field.
The answer is -0.13
256.2 is the answer.
Here, we are asked to calulate the electric field.
Using the following formula:
E=k*lql/r^2
Yeah! Something that can easily bamboozle you.
So:
E=electric field
k=constant (8.99x10^9)
q=charge
ll= absolute value marks ( I still remember from my Algebra Course))
r=distance
Now, let's plug in the numbers.
E=8.99x10^9*l-5.77x10^-9l/(0.450)^2
And yes! That's a LOT!
But I solved the same problem on paper 2 minutes ago:)
If you perform the calculations, you will get 256.2
Well. THIS one was correct or Acellus.
So our answer is:
256.2 N/CNewtons per CoulombHope this helps:)
When a person sits erect, increasing the vertical position of their brain by 38.6 cm, the heart must continue to pump blood to the brain at the same rate. (a) What is the gain in gravitational potential energy (in J) for 110 mL of blood raised 38.6 cm
Answer:
the gain in gravitational potential energy is 0.4369 J
Explanation:
Given the data in the question;
from the definition of density, we know that;
ρ = m / V
where ρ is density, m is the mass and V is the volume.
Now, lets make mass the subject of the formula
m = ρV ------ let this be equation 1
Now, we know that potential energy PE = mgh ------- let this be equation 2
where m is mass, g is acceleration due gravity, and h is the height.
substitute equation 1 into 2
PE = ρVgh
given that; V = 110 mL = 110 × 10⁻⁶ m³, h = 38.6 cm = 38.6 × 10⁻² m, g = 9.8 m/s², ρ = 1.05 × 10³ kg/m³
we substitute
PE = (1.05 × 10³ kg/m³) × (110 × 10⁻⁶ m³) × 9.8 m/s² × 38.6 × 10⁻² m
PE = 0.4369 J
Therefore, the gain in gravitational potential energy is 0.4369 J
A 0.2-kg stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the stone makes 150 revolutions per minute, the tension force of the string on the stone is:____.
a. 0.75 N.
b. 1.96 N.
c. 0.03 N.
d. 30 N.
e. 0.2 N.
Answer:
the tension force of the string on the stone is 30 N
Option d) 30 N is the correct answer.
Explanation:
Given the data in the question;
mass m = 0.2 kg
radius r = 0.6 m
θ = 150 revolutions = 300π rad
time t = 60 seconds
we know that; Angular speed ω = θ / t
we substitute
ω = 300π / 60
ω = 5π rad
Linear speed of stone u = ω × r
we substitute
u = 5π × 0.6
u = 3π m/s
The tension force of the string on the stone is equal to centripetal force, which aid it move in circle;
so
T = mv² / r
we substitute
T = [ 0.2 × (3π)² ] / 0.6
T = 17.7652879 / 0.6
T = 29.6 ≈ 30 N
Therefore, the tension force of the string on the stone is 30 N
Option d) 30 N is the correct answer.
Describe 3 Levers of Power and how they work.
DESCRIBE
A thermodynamic system consists of an ideal gas at a volume of 3.50 L and initial pressure of 6.2 × 104 Pa. As the volume is held constant, the pressure is increased to 8.2 × 104 Pa. What work is involved in this process?
Answer:
0 J
Explanation:
Since work done W = PΔV where P = pressure and ΔV = change in volume.
Since the volume is constant, ΔV = 0
So, Work done, W = PΔV = P × 0 = 0 J
So, the work done is 0 J.
If a 50 W light bulb and a 90 W light bulb operate from 120 V, which bulb has a greater current in it?
50 W
90 W
When the bell in a clock tower rings with a sound of 474 Hz, a pigeon roosting in the belfry flies directly away from the bell.
If the pigeon hears a frequency of 453Hz, what is its speed?
Answer:
15.44 m/s
Explanation:
Below is the given values:
The bell in rings with a sound = 474 Hz
If the frequency of pigeon hears = 453 Hz
Speed = ?
Use below formula:
Frequency = [(Vs - Vo) / Vs ] x fo
Vs = Speed of sound
Vo = Speed of observer
fo = Sound frequency
Frequency = [(Vs - Vo) / Vs ] x fo
453 = [(343 - Vo) / 343 ] x 474
453 / 474 = [(343 - Vo) / 343 ]
0.955 = (343 - Vo) / 343
0.955 x 343 = 343 - Vo
327.56 = 343 - Vo
Vo = 343 - 327.56
Vo = 15.44 m/s
At an amusement park, a 7.00 kg swimmer uses a water slide to enter the main pool. The swimmer starts at rest, slides without friction, and descends through a vertical height of 0.40 m. Find the total energy present when the swimmer is at the top of the slide.
Answer:
27.44 J
Explanation:
We can find the energy at the top of the slide by using the potential energy equation:
PE = mghAt the top of the slide, the swimmer has 0 kinetic energy and maximum potential energy.
The swimmer's mass is given as 7.00 kg.
The acceleration due to gravity is 9.8 m/s².
The (vertical) height of the water slide is 0.40 m.
Substitute these values into the potential energy equation:
PE = (7.00)(9.8)(0.40) PE = 27.44Since there is 0 kinetic energy at the top of the slide, the total energy present is the swimmer's potential energy.
Therefore, the answer is 27.44 J of energy when the swimmer is at the top of the slide.
Which of the following absorbs the energy required by photosynthesis?
Answer:
There are no options, so....
Explanation:
Chlorophyll a absorbs its energy from the Violet-Blue and Reddish orange-Red wavelengths, and little from the intermediate (Green-Yellow-Orange) wavelengths.
Assume the following vehicles are all moving at the Sam speed .it would be harder to change the velocity of which vehicle . What law is it
Answer:what the choices
Explanation:
Calculate the energy in electron volts of X-rays that have a frequency of 6.00 x 1016 Hz.
207 eV
228 eV
249 eV
Answer:
228
Explanation:
As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force
acting on it is equal to (4i - 9 j) N. If the speed of the object at the initial position is 4.0 m/s,
what is its kinetic energy at its final position?
Answer:
Answer: [tex]107.8\ J[/tex]
Explanation:
Given
Initial position of object is (4.4 i+5 j)
Final position of object is (11.6 i -2 j)
Force acting (4i-9j)
Work done is given by
[tex]\Rightarrow W=F\cdot dx\\\Rightarrow W=(4i-9j)\cdot (11.6i-4.4i-2j-5j)\\\Rightarrow W=(4i-9j)\cdot (7.2i-7j)\\\Rightarrow W=28.8+63\\\Rightarrow W=91.8\ J[/tex]
Initial kinetic energy
[tex]K_i=\dfrac{1}{2}\times 2\times 4^2\\\\K_i=16\ J[/tex]
Change in kinetic energy is equal to work done by object
[tex]\Rightarrow K_f=K_i+W\\\Rightarrow K_f=16+91.8\\\Rightarrow K_f=107.8\ J[/tex]
14. Name the 3 primary causes for the expected mass migration due to Climate
Change.
Answer: sea-level rise, extreme weather events, and drought, and water scarcity.
Explanation:
What are moana's hobbies
what mass of water will release 60000 J of energy while cooling from 40.0 degrees celsius to 20.0 degrees celsius?
Answer:
714.286g
Explanation:
Q=mcT
60000=m×4.2×(40-20)
m=714.285....g
Consider the cylindrical weir of diameter 3 m and length 6m. If the fluid on the left has a specific gravity of 0.8, find the magnitude and direction of the resultant force
This question is incomplete, the complete question is;
Consider the cylindrical weir of diameter 3m and length 6m. If the fluid on the left has a specific gravity of 1.6 and on the right has a specific gravity of 0.8, Find the magnitude and direction of the resultant force.
Answer:
- the magnitude of the resultant force is 557.32 kN
- the direction of resultant force is 48.29°
Explanation:
Given the data in the question and the diagram below,
First we work on the force on the left hand side.
Left Horizontal
[tex]F_{LH[/tex] = βgAr
here, h = 3/2 = 1.5 m, β = 1.6, g = 9.81 m/s², A = 3 m × 6 m = 18 m²
we substitute
[tex]F_{LH[/tex] = βgAh = ( 1.6 × 1000 ) × 9.81 × 18 × 1.5 = 423792 N
Left Vertical
[tex]F_{LV[/tex] = ( βgπh² / 2 ) × W
we substitute
[tex]F_{LV[/tex] = [ ( ( 1.6 × 1000 ) × 9.81 × π(1.5)² ) / 2 ] × 6 = 332845.458 N
Now we go to the right hand side
Right Horizontal
[tex]F_{RH[/tex] = βgAh
here, h' = 1.5/2 = 0.75 m, β = 0.8, g = 9.81 m/s², A = 1.5 m × 6 m = 9 m²
we substitute
[tex]F_{RH[/tex] = ( 0.8 × 1000 ) × 9.81 × 9 × 0.75 ) = 52974 N
Right Vertical
[tex]F_{RV[/tex] = ( βgπh² / 4 ) × W
we substitute
[tex]F_{RV[/tex] = [ ( ( 0.8 × 1000 ) × 9.81 × π(1.5)² ) / 4 ] × 6 = 83211.36 N
Hence
Fx = [tex]F_{LH[/tex] - [tex]F_{RH[/tex] = 52974 N - 423792 N = 370818 N
Fy = [tex]F_{LV[/tex] + [tex]F_{RV[/tex] = 332845.458 N + 83211.36 N = 416056.818 N
R = √( Fx² + Fy² ) = √[ (370818 N)² + (416056.818 N)² ] = 557323.3 N
R = 557.32 kN
Therefore, the magnitude of the resultant force is 557.32 kN
Direction of resultant force;
tanθ = Fy / Fx
we substitute
tanθ = 416056.818 N / 370818 N
tanθ = 1.121997
θ = tan⁻¹( 1.121997 )
θ = 48.29°
Therefore, the direction of resultant force is 48.29°
A meterstick of negligible mass is placed on a fulcrum at the 0.4 m mark, with a 1 kg mass hung at the zero mark and a 0.5 kg mass hung at the 1.0 m mark. The meterstick is held horizontal and released. Immediately after release, the magnitude of the net torque on the meterstick about the fulcrum is most nearly:________
a. 1 Nm
b. 2 Nm.
c. 2.5 Nm.
d. 7 Nm
e. 7.5 Nm
Answer:
The net torque is 0.98 Nm.
Explanation:
The torque is given by
Torque = force x perpendicular distance
The clock wise torque is taken as negative while the counter clock wise torque is taken as positive.
Take the torques about the fulcrum.
Torque = 1 x 9.8 x 0.4 - 0.5 x 9.8 x 0.6
Torque = 3.92 - 2.94 = 0.98 Nm