Answer:
The required emf moved across the wire is zero
Explanation:
For a moving charge particle, the magnetic force can be determined by using the formula;
[tex]\varepsilon = Bvlsin \theta[/tex]
since the wire moves in parallel, the angle [tex]\theta[/tex] between magnetic field and velocity = 0°
B = 0.500 T
v = 1.50 m/s
l = 0.200 m
∴
[tex]\varepsilon = (0.500 \ T )(1.50 \ m/s) \times (0.200 \ m)\times sin (0)[/tex]
[tex]\varepsilon = 0.15\times sin (0)[/tex]
[tex]\varepsilon = 0[/tex]
Two blocks A and B with mA = 2.9 kg and mB = 0.87 kg are connected by a string of negligible mass. They rest on a frictionless horizontal surface. You pull on block A with a horizontal force of 6 N. Determine the tension in the string connecting the two blocks.
Answer:
Tension = 1.38 N
Explanation:
Given that:
Mass of block A [tex]m_A[/tex] = 2.9 kg
mass of block B [tex]m_B[/tex] = 0.87 kg
Force F = 6 N
Assume a = acceleration of the blocks.
Then:
[tex]m_A[/tex] (a) + [tex]m_B[/tex] (a) = 6
2.9a + 0.87a = 6
3.77a = 6
a = 6/3.77
a = 1.59 m/s²
Suppose T to be the tension in the string.
If we take a look at the forces acting on the first block, then:
F - T = [tex]m_A[/tex] (a)
T = F - [tex]m_A[/tex] (a)
T = 6 - 2.9(1.59)
T = 6 - 4.62
T = 1.38 N
A ball is spun around in circular motion such that its frequency is 10 Hz.
a. What is the period of its rotation?
b. How much time will be required to complete 100 rotations?
Answer:
a = 0.1 s b. 10 s
Explanation:
Given that,
The frequency in circular motion, f = 10 Hz
(a) Let T is the period of itsrotation. We know that,
T = 1/f
So,
T = 1/10
= 0.1 s
(b) Frequency is number of rotations per unit time. So,
[tex]t=\dfrac{n}{f}\\\\t=\dfrac{100}{10}\\\\t=10\ s[/tex]
Hence, this is the required solution.
Which element is the biggest contributor to Climate Change?
Answer:
carbon dioxide (CO2)
Explanation:
the burning or combustion of these fossil fuels creates gases that are released into the atmosphere. Of these gases, carbon dioxide (CO2) is the most common and is the gas most responsible for exacerbating the green- house effect that is changing global climate patterns.
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]
A model plane has a mass of 0.75 kg and is flying 12 m above the ground
with a speed of 18 m/s. What is the total mechanical energy of the plane?
The acceleration due to gravity is 9.8 m/s2.
Answer:
Option C. 210 J.
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 0.75 Kg
Height (h) = 12 m
Velocity (v) = 18 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Total Mechanical energy (ME) =?
Next, we shall determine the potential energy of the plane. This can be obtained as follow:
Mass (m) = 0.75 Kg
Height (h) = 12 m
Acceleration due to gravity (g) = 9.8 m/s²
Potential energy (PE) =?
PE = mgh
PE = 0.75 × 9.8 × 12
PE = 88.2 J
Next, we shall determine the kinetic energy of the plane. This can be obtained as follow:
Mass (m) = 0.75 Kg
Velocity (v) = 18 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.75 × 18²
KE = ½ × 0.75 × 324
KE = 121.5 J
Finally, we shall determine the total mechanical energy of the plane. This can be obtained as follow:
Potential energy (PE) = 88.2 J
Kinetic energy (KE) = 121.5 J
Total Mechanical energy (ME) =?
ME = PE + KE
ME = 88.2 + 121.5
ME = 209.7 J
ME ≈ 210 J
Therefore, the total mechanical energy of the plane is 210 J.
An open 1-m-diameter tank contains water at a depth of 0.5 m when at rest. As the tank is rotated about its vertical axis the center of the fluid surface is depressed. At what angular velocity will the bottom of the tank first be exposed
Answer:
Angular velocity (w) = 8.86 rad/s
Explanation:
Angular velocity (w) = [tex]\sqrt{} 4ghi/R^{2}[/tex]
g= 9.81 m/s
R= 0.5
hi (initial depth) = 0.5m
Hence= [tex]\sqrt4* 9.81* 0.5/0.5^{2}[/tex] = 8.86 rad/s
3. How can a generator that otherwise produces AC
current be modified to produce DC current?
potential Energy is associated with position
True or false
Answer:
yes.
Explanation:
potential Energy is associated with position
Answer:
true
Explanation:
the energy possessed by a body by virtue of its position or configuration is called potential energy.Everything at a height from the earth surface posseses potential energy.
What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 555 nmnm
Answer:
The minimum thickness of soap film is [tex]2.086\times 10^{-7} m[/tex].
Explanation:
wavelength = 555 nm
refractive index, n = 1.33
For the destructive interference,
the thickness is given by
[tex]t =\frac{m \lambda }{2 n}[/tex]
Here, m is the order, n is the refractive index and [tex]\lambda[/tex] is the wavelength.
For minimum thickness, m = 1
So the thickness is
[tex]t =\frac{1 \times 555\times 10^{-9} }{2 \times 1.33}\\\\t = 2.086\times 10^{-7} m[/tex]
A proton accelerates from rest in a uniform electric field of 595 N/C. At one later moment, its speed is 1.15 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton. (b) Over what time interval does the proton reach this speed? (c) How far does it move in this time interval? (d) What is its kinetic energy at the end of this interval?
An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and the drag coefficient is 0.065. The mass of the aircraft is 900 kg. Calculate the effective lift area for the aircraft and the required engine thrust and power to maintain level flight.
Answer:
- the effective lift area for the aircraft is 8.30 m²
- the required engine thrust is 1275 N
- required power is 79.7 kW
Explanation:
Given the data in the question;
Speed V = 225 km/hr = 62.5 m/s
The lift coefficient CL = 0.45
drag coefficient CD = 0.065
mass = 900 kg
g = 9.81 m/s²
a) the effective lift area for the aircraft
we know that for a steady level flight, weight = lift and thrust = drag
Using the equation for the lift force
F[tex]_L[/tex] = C[tex]_L[/tex][tex]\frac{1}{2}[/tex]ρV²A = W
we substitute
0.45 × [tex]\frac{1}{2}[/tex] × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )
1081.05 × A = 8829
A = 8829 / 1081.05
A = 8.30 m²
Therefore, the effective lift area for the aircraft is 8.30 m²
b) the required engine thrust and power to maintain level flight.
we use the expression for drag force
F[tex]_D[/tex] = T = C[tex]_D[/tex][tex]\frac{1}{2}[/tex]ρV²A
we substitute
= 0.065 × [tex]\frac{1}{2}[/tex] × 1.21 × ( 62.5 )² × 8.30
T = 1275 N
Since drag and thrust force are the same,
Therefore, the required engine thrust is 1275 N
Power required;
P = TV
p = 1275 × 62.5
p = 79687.5 W
p = ( 79687.5 / 1000 )kW
p = 79.7 kW
Therefore, required power is 79.7 kW
option are
cerebrum, cortex, cerebellum, brain stem, medulla, pons, midbrain , thalamus, hypothalamus, reticular formation
1.________regulates and maintains brain's awareness level.
2. _______controls the activity of the pituitary gland.
3._______relays impulses from one part of the brain to another.
4._______regulates body temperature, hunger, and other internal body conditions.
5______connects nerve impulses from body to cortex.
Answer:
1. Reticular formation.
2. Hypothalamus.
3. Thalamus.
4. Hypothalamus.
5. Thalamus.
Explanation:
The human brain consists of various sections and these includes;
I. Anterior Cingulate Cortex (ACC) which resembles a bow or collar surrounding the frontal part of the corpus callosum. This is the frontal part of the cingulate cortex which helps to make complex cognitive functions such as impulse control, decision-making, emotions and empathy.
II. Ventral prefrontal cortex in humans are interconnected with the brain and are responsible for the processing of risk, empathy, fear and social decision-making
III. The Cerebral Cortex: this part of the brain primarily comprises of grey matter, foldable sheets of neurons and forms its outermost layer. Therefore, cerebral cortex is known as the outermost layer of the brain (cerebrum) and thus, makes up half of its weight. It is about 2.5 millimeters in thickness and as such it's able to fold.
1. Reticular formation: regulates and maintains brain's awareness level.
2. Hypothalamus: controls the activity of the pituitary gland.
3. Thalamus: relays impulses from one part of the brain to another.
4. Hypothalamus: regulates body temperature, hunger, and other internal body conditions.
5. Thalamus: connects nerve impulses from body to cortex.
Calculate the force exerted on a thresher shark's eye by the hydrostatic pressure in ocean water at a depth of 380 m. (Assume the water's mass density at this depth is 1000 kg/m3k.)
Answer:
Explanation:
Hydrostatic pressure due to a water column of height h can be given by the following expression.
P = hρg
where ρ is density of water and g is acceleration due to gravity .
Substituting the values.
P = 380 x 1000 x 9.8
= 3.72 x 10⁶ Pa.
Answer:
[tex]F=\dfrac{3.72\times 10^6\ Pa}{A}\ N[/tex]
Explanation:
Given that,
The density of water, d = 1000 kg/m³
Depth, h = 380 m
We need to find the force exerted on a thresher shark's eye by the hydrostatic pressure in ocean water. The force exerted by the hydrostatic pressure is given by :
[tex]P=\rho gh[/tex]
Put all the values,
[tex]P=1000\times 9.8\times 380\\\\P=3.72\times 10^6\ Pa[/tex]
Force exerted,
F = P/A
So,
[tex]F=\dfrac{3.72\times 10^6\ Pa}{A}\ N[/tex]
Where
A is the area of crosss section
Hence, this is the required solution.
The conversion of thermal energy into mechanical energy requires
a thermometer.
D. Beat engine
C vaporizer.
d thermostat.
write the chemical equation of rust
ANSWER
4Fe+ 3O2 → 2Fe2O.
EXPLANATION
iron (Fe) combines with oxygen (O2) to form rust, or iron oxide (Fe2O3). The equation for this reaction is: 4Fe+ 3O2 → 2Fe2O.
Select the correct answer.
A force of 100 newtons is applied to a box at an angle
acceleration of the box?
36° with the horizontal. If the mass of the box is 25 kilograms, what is the horizontal
OA
1.52 meters/second2
3.24 meters/second
C.
5.48 meters/second
Ο Ο Ο Ο
6.87 meters/seconda
7.15 meters/second
Answer:
3
Explanation:
first you find the original component of the force be a cause in 36 degrees then with ka sin 36 find the average interforce by multiplying it with hundred Newton then divide the original force by the mass 25 Kg is equals tto ma
Answer:
See image
Explanation:
Plato
Thiết kế mạch điện tử sử dụng chân ngõ ra (I/O) của arduino UNO để lái relay 12VDC/500mA sao cho chân Arduino xuất ra tín hiệu 5V thì relay đóng, ngược lại chân I/O arduino xuất tín hiệu 0V thì relay hở. Mạch lái dùng BJT
Answer:
Mình cũng không biết làm bạn ơi =)))
Explanation:
Describe how the human ear hears sound.
Answer:
sound waves enter the outer ear and travel through a narrow passageway called the ear canal, which leads to the eardrum. The eardrum vibrates from the incoming sound waves and sends these vibrations to three tiny bones in the middle ear.
Explanation:
An astronaut weighing 248 lbs on Earth is on a mission to the Moon and Mars. (a) What would he weigh in newtons when he is on the Moon
Answer:
[tex]W_m=183.495N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=248Ibs[/tex]
Mass of Weight [tex]m= 248*0.453[/tex]
[tex]m=112.344kg[/tex]
Generally the acceleration due to gravity on the Moon is one-sixth that on Earth.
Therefore
The equation for Weight on Moon is given as
[tex]W_m=m*g/6[/tex]
[tex]W_m=122.344*\frac{9.8}{6}[/tex]
[tex]W_m=183.495N[/tex]
Answer:
F ’= 40.9 lb
Explanation:
The weight of a body is the attraction of the Earth on the body
F = [tex]G \frac{m M_e}{R_e^2}[/tex]
F = mg
m = F / g
m = 248/32
m = 7.75 slug
The weight of the body on the moon is the attraction of the body for the satellite
F ’= [tex]G \frac{mM}{R^2}[/tex]
from the tables the mass and radius of the moon are M = 7.34 10²² kg and R = 1.74 10⁶ m
let's reduce the mass to the SI system
m = 7.75 slug (14.59 kg / 1 slug) = 113 kg
F ’= 6.67 10⁻¹¹ 113 7.34 10²² / (1.74 10⁶) ²
F ’= 1.82 10² N
F ’= 1.82 10² N (0.2248 lb / 1 N)
F ’= 40.9 lb
