The atoms in your body are mostly empty space . And so are the atoms in any wall. Why then is your body unable to pass through walls ?
First of all, both are not a single sheet of atom. There are many layers of atoms, so the empty part gets beside each other, so there are less empty part. Secondly, there are so many atoms that the probability that they will have empty space at the same place necessary, is negligible.
This was something from logic.
The reason I was taught in my class was that only a limited number of electrons can be in a given orbit, so atoms cannot overlap each other.
1.Lõi thép máy biến áp được ghép từcác lá thép là để:
(a) Giảm tổn hao công suất do dòng điện xoáy
(b) Giảm tổn hao công suất do từ trễ
(c) Giảm tổn hao công suất do dòng điện chạy qua dây quấn
(d) Giảm tất cảcác loại tổn hao công suất.
Answer:
Option (c)
Explanation:
1.The transformer core is assembled from steel sheets to:
(a) Reduced power loss due to eddy current
(b) Reduced power loss due to hysteresis
(c) Reduced power loss due to current flowing through the winding
(d) Reduce all types of power loss.
A transformer is a device which converts the low voltage into high and vice versa.
There are two types of a transformer.
Step up: It is used to convert low voltage into high.
Step down It is used to convert high voltage into high.
It depends on the number of turns in primary and the secondary coil.
The core of the transformer is laminated and it is in the form of sheets.
By using such type of core, the power loss due to the windings is reduced.
option (c) .
A proton traveling due west in a region that contains only a magnetic field experiences a vertically upward force (away from the surface of the earth). What is the direction of the magnetic field?
South
Explanation:
The magnetic force F on a point charge moving with a velocity v in the presence of a magnetic field B is given by
[tex]\vec{\textbf{F}} = q\vec{\textbf{v}}\textbf{×}\vec{\textbf{B}}[/tex]
and according to the right-hand rule, an upward magnetic force on a proton moving westward is only possible if the magnetic field is directed southward.
Three spheres (water, iron and ice) of the exact same volume are submerged in a tub of water. After the spheres are lined up, they are released. The spheres are made of plastic with the same density as water, ice, and iron.
Required:
a. Compare the weights of the three spheres.
b. Compare the buoyant forces on the three spheres.
c. What direction does the net force push on each of the spheres?
d. What happens to each sphere after it is released?
Answer:
(a) Iron > plastic > ice
(b) Same on all
(c) Iron downwards, plastic net force zero, ice upwards.
(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.
Explanation:
Three spheres have same volume , plastic, ice and iron.
(a) The weight is given by
Weight = mass x gravity = volume x density x gravity
As the density of iron is maximum and the density of ice is least so the order of the weight is
Weight of iron > weight of plastic > weight of ice
(b) Buoyant force is given by
Buoyant force = Volume immersed x density of fluid x g
As they have same volume, density of fluid is same so the buoyant force is same on all the spheres.
(c) Net force is
F = weight - buoyant force
So, the net force on the iron sphere is downwards
On plastic sphere is zero as the density of plastic sphere is same as water. On ice sphere it is upwards.
(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.
how many bits are required to sample an incoming signal 4000 times per second using 64 different amplitude level
Answer:
6 bits
Explanation:
The quality of digitized signal can be improved by reducing quantizing error. This is done by increasing the number of amplitude levels, thereby minimizing the difference between the levels and hence producing a smoother signal.
Also, Sampling frequently (also known as oversampling) can help in improving signal quality.
To get the number of bits, we use:
2ⁿ = amplitude level
where n is the number of bits.
Given an amplitude level of 64, hence:
2ⁿ = 64
2ⁿ = 2⁶
n = 6 bits
Would this pressure difference be greater or smaller if the scuba diver were in seawater (density 1050 kg/m3 ) and went to the same depth you calculated in question D1, took and held his breath, and then returned to the surface
Answer:
Greater.
Explanation:
This pressure difference will be greater if the scuba diver were in seawater and went to the same depth because the seawater have salts which increases the density of water as compared to freshwater. Salt in water increases the density which automatically increases the pressure on the diver so that's why we can say that the pressure will be increases for the scuba diver in seawater as compared to freshwater.
A hot air balloon is a sphere of volume 2210 m3. The density of the hot air inside is 1.13 kg/m3, while the air outside has a density of 1.29 kg/m3. The balloon itself has a mass of 240 kg. What is the TOTAL NET force acting on the balloon?
[?]N
The total net force acting on the balloon will be 24498 Newtons
Given that
Volume of the balloon = 2210 cubic meter
Density of the air inside the balloon = 1.13 kg/m3
What will be the net force exerted on the balloon ?Here force on the balloon will be equal to the weight of the air displaced by balloon
[tex]F= mass of air displaced\times gravity[/tex]
[tex]F= Density \times volume \times gravity[/tex]
[tex]F=1.13 \times 2210 \times 9.81[/tex]
[tex]F=24498 N[/tex]
The total net force acting on the balloon will be 24498 Newtons
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What's the speed of a sound wave through water at 25 Celsius?
A. 1,000 m/s
B. 1,500 m/s
C. 1,250 m/s
D. 750 m/s
Answer:
B) 1500m/s
Explanation:
Ans is 1500m/s
Which of the following is a noncontact force?
O A. Friction between your hands
O B. A man pushing on a wall
O C. Air resistance on a car
D. Gravity between you and the Sun
Answer:
Gravity between you and the sun
Which parts of The Action Potential Are Represented On The ECG?
Answer:
The phases of the cardiac action potential correspond to the surface ECG (ECG) (Figure). The P wave reflects atrial depolarization (phase 0), the PR interval reflects the conduction velocity through the AV node, the QRS complex the ventricular depolarization and QT interval the duration potential ventricular action.
b) Assume the rod is 0.60 m long and has a mass of 0.50 kg, and the clay blob has a mass of 0.20 kg and moves at an initial velocity of 8.0 m/s. Calculate the final angular velocity of the rod. Be sure to put units in your calculation and show the resulting units in your answer.
Answer:
The correct answer is "6.96 rad/s".
Explanation:
The given values are:
Length,
L = 0.6 m
Mass,
m₁ = 0.5 kg
m₂ = 0.2 kg
Initial velocity,
V = 8 m/s
Now,
The final angular velocity will be:
⇒ [tex]\omega =\frac{6m_1V}{(4m_1+3m_2)L}[/tex]
By substituting the values, we get
⇒ [tex]=\frac{6\times 0.2\times 8}{(4\times 0.2+3\times 0.5)0.6}[/tex]
⇒ [tex]=\frac{9.6}{1.38}[/tex]
⇒ [tex]=6.96 \ rad/s[/tex]
TIME REMAINING
45:43
What are possible units for impulse? Check all that apply.
kg • m
kg • meters per second
N • s
N • m
StartFraction Newtons times meters per second EndFraction
Answer:
n.m maby
Explanation:
i think or its kg m/s
Answer:
answer (B) & (C)
Explanation:
kg • /N • s
Which is the definition of refraction?
1)the blocking of light waves vibrating in a particular plane
2) the bending of a light wave as it passes at an angle from one medium to another
3) a false or distorted image causing the gradual distortion of light through hot air
the redirection of light by tiny particles as it passes through a medium
Answer:
2) The bending of a light wave as it passes at an angle from one medium to another .
Hope it is helpful to you ☺️
If the resistance in a circuit remains constant, what happens to the electric power when the current increases?
The power will increase.
B.
The power will decrease.
Ο Ο Ο Ο
There will be no power.
D
The current does not affect the power.
Answer:
Resistance is inversly proportional to the current.
V=I.R.
P=V.I
A body of mass 450g changes it speed from 5ms¹ to 25ms¹. what is the work done by the body?
Answer:
135J
Explanation:
So we know ΔKinetic Energy= ΔWork
Kinetic energy=1/2mv²
So Kf-Ki=ΔK
ΔK=1/2*0.45(25²-5²)=135J
135J=ΔWork
Two pendulums have the same dimensions (length {L}) and total mass (m). Pendulum A is a very small ball swinging at the end of a uniform massless bar. In pendulum B, half the mass is in the ball and half is in the uniform bar.
1. Find the period of pendulum A for small oscillations.
2. Find the period of pendulum B for small oscillations.
Answer:
1) [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex], 2) [tex]T_{B} \approx 1.137\cdot T_{A}[/tex], where [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex].
Explanation:
1) Pendulum A is a simple pendulum, whose period ([tex]T_{A}[/tex]) is determined by the following formula:
[tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex] (1)
Where:
[tex]l[/tex] - Length of the massless bar.
[tex]g[/tex] - Gravitational acceleration.
2) Pendulum B is a physical pendulum, whose period ([tex]T_{B}[/tex]) is determined by the following formula:
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{I_{O}}{m\cdot g\cdot l} }[/tex] (2)
Where:
[tex]m[/tex] - Total mass of the pendulum.
[tex]g[/tex] - Gravitational acceleration.
[tex]l[/tex] - Length of the uniform bar.
[tex]I_{O}[/tex] - Moment of inertia of the pendulum with respect to its suspension axis.
The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:
[tex]I_{O} = \frac{1}{2} \cdot m\cdot l^{2}+\frac{1}{24}\cdot m\cdot l^{2} + \frac{3}{4}\cdot m\cdot l^{2}[/tex]
[tex]I_{O} = \frac{31}{24}\cdot m\cdot l^{2}[/tex] (3)
By applying (3) in (2) we get the following expression:
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{\frac{31}{24}\cdot m \cdot l^{2} }{m\cdot g \cdot l} }[/tex]
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{31\cdot l}{24\cdot g} }[/tex]
[tex]T_{B} = \sqrt{\frac{31}{24} } \cdot \left(2\pi \cdot \sqrt{\frac{l}{g} }\right)[/tex]
[tex]T_{B} \approx 1.137\cdot T_{A}[/tex]
1. The period of pendulum A for small oscillations is
[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]
2. The period of pendulum B for small oscillations.
[tex]T_B=1.137.T_A[/tex]
What is simple harmonic motion?Simple harmonic motion is the periodic motion or back and forth motion of any object with respect to its equilibrium or mean position. The restoring force is always acting on the object which try to bring it to the equilibrium.
1) Pendulum A is a simple pendulum, whose period () is determined by the following formula:
[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]
Where:
l - Length of the massless bar.
g - Gravitational acceleration.
2) Pendulum B is a physical pendulum, whose period () is determined by the following formula:
[tex]T_A=2\pi\sqrt{\dfrac{I_o}{mgl}}[/tex] .............................2
Where:
m - Total mass of the pendulum.
g - Gravitational acceleration.
l - Length of the uniform bar.
Io- Moment of inertia of the pendulum with respect to its suspension axis.
The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:
[tex]I_o=\dfrac{1}{2}ml^2+\dfrac{1}{24}ml^2+\dfrac{3}{4}ml^2[/tex]
[tex]I_o=\dfrac{31}{24}ml^2[/tex]..................................3
By applying (3) in (2) we get the following expression:
[tex]T_B=2\pi\sqrt{\dfrac{\frac{31}{24}ml^2}{mgl}[/tex]
[tex]T_B=2\pi\sqrt{\dfrac{31l}{24g}}[/tex]
[tex]T_B=\sqrt{\dfrac{31}{24}}. (2\pi\sqrt{\dfrac{l}{g}})[/tex]
[tex]TB=1.137.T_A[/tex]
Thus to know more about Simple harmomnic motion follow
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A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. How far will block m drop in the first seconds after the system is released?
How long will block M move during above time?
At the time, calculate the velocity of block M
Find out the deceleration of the block M, if the connected string is
removal by cutting after the first second. Then, calculate the time
taken to contact block M and pulley.
Answer:
a) y = 0.98 t², t=1s y= 0.98 m,
b) he two blocks must move the same distance
c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m
Explanation:
For this exercise we can use Newton's second law
Big Block
Y axis
N-W = 0
N = M g
X axis
T- fr = Ma
the friction force has the expression
fr = μ N
fr = μ Mg
small block
w- T = m a
we write the system of equations
T - fr = M a
mg - T = m a
we add and resolved
mg- μ Mg = (M + m) a
a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]
a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]
a = 9.8 (6/30)
a = 1.96 m / s²
a) now we can use the kinematic relations
y = v₀ t + ½ a t²
the blocks come out of rest so their initial velocity is zero
y = ½ a t²
y = ½ 1.96 t²
y = 0.98 t²
for t = 1s y = 0.98 m
t = 2s y = 1.96 m
b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.
As the curda is in tension the two blocks must move the same distance
c) the velocity of the block M
v = vo + a t
v = 0 + 1.96 t
for t = 1 s v = 1.96 m / s
t = 2 s v = 3.92 m / s
d) the deceleration if the chain is cut
when removing the chain the tension becomes zero
-fr = M a
- μ M g = M a
a = - μ g
a = - 0.2 9.8
a = -1.96 m / s²
e) the distance to stop the block is
v² = vo² - 2 a x
0 = vo² - 2a x
x = vo² / 2a
x = 1.96² / 2 1.96
x = 0.98 m
the time to travel this distance is
v = vo - a t
t = vo / a
t = 1.96 /1.96
t = 1 s
The free-fall acceleration at the surface of planet 1 is 22 m/s^2. The radius and the mass of planet 2 are twice those of planet 1. What is the free-fall acceleration on planet 2?
Answer:
g₂ = 11 m/s²
Explanation:
The value of free-fall acceleration on the surface of a planet is given by the following formula:
[tex]g = \frac{Gm}{r^2}[/tex]
where,
g = free-fall acceleration
G = Universal Gravitational Constant
m = mass of the planet
r = radius of planet
FOR PLANET 1:
[tex]g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2[/tex] --------------------- equation (1)
FOR PLANET 2:
[tex]g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\[/tex]
using equation (1):
[tex]g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}[/tex]
g₂ = 11 m/s²
A student claimed that thermometers are useless because a
thermometer always registers its own temperature. How would you
respond?
[
Two objects attract each other with a gravitational force of magnitude 1.00 3 1028 N when separated by 20.0 cm. If the total mass of the two objects is 5.00 kg, what is the mass of each
Answer:
The mass of each object is 2kg and 3 kg.
Explanation:
Given that,
Gravitational force,[tex]F=1\times 10^{-8}\ N[/tex]
The distance between masses, d = 20 cm = 0.2 m
The total mass of the two objects, M + m = 5 kg
M = 5-m
The formula for the gravitational force is :
[tex]F=G\dfrac{Mm}{d^2}\\\\1\times 10^{-8}=6.67\times 10^{-11}\times \dfrac{(5-m)m}{(0.2)^2}\\\\\frac{1\times10^{-8}}{6.67\times10^{-11}}=\frac{(5-x)x}{(0.2)^{2}}\\\\\frac{1\times10^{-8}}{6.67\times10^{-11}}\cdot(0.2)^{2}\\\\5.99=(5-x)x\\\\x=2\ kg\ and\ 3 \ kg[/tex]
So, the mass of each object is 2kg and 3 kg.
A random sample of 22 lunch orders at Noodles & Company showed a mean bill of $10.26
with a standard deviation of $5.21. Find the 99 percent confidence interval for the mean bill of
all lunch orders.
Answer:
(7.115 ; 13.405)
Explanation:
Given :
Sample size, n = 22
Mean bill, μ = 10.26
Standard deviation, s = 5.21
To obtain the 99% confidence interval for the mean bill of all orders ;
Mean ± margin of error
Margin of Error = Tcritical * s/√n
Tcritical at 99%, df = n-1, 22 - 1 = 21
Tcritical = 2.831
Margin of Error = 2.831 * (5.21/√22) = 3.145
Confidence interval = 10.26 ± 3.145
Lower boundary = 10.26 - 3.145 = 7.115
Upper boundary = 10.26 + 3.145 = 13.405
Confidence interval :
(7.115 ; 13.405)
Electrical resistance is a measure of resistance to the flow of _?____
Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.
Hope this helps!!!!
Answer:
electric current
Explanation:
The answer is electric current
At what distance x from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?
Answer:
The value of x is 2.1 cm from the center of the coil.
Explanation:
Radius, R = 2.7 cm
Number of turns, N = 800
The magnetic field at the axis is half of the magnetic field at the center.
[tex]B_{axis}=\frac{B_{center}}{2}\\\\\frac{\mu o}{4\pi}\times \frac{2 \pi I N R^2}{\left (R^2 + x^2 \right )^{\frac{3}{2}}} = 0.5\frac{\mu o}{4\pi}\times\frac{2\pi N I}{R}\\\\\frac{R^2}{(R^2 + x^2)^\frac{3}{2}} = \frac{1}{2R}\\\\4R^6 = (R^2+x^2)^3\\\\1.6 R^2 = R^2 + x^2\\\\x^2 = 0.6 \times 2.7\times 2.7 \\\\x = 2.1 cm[/tex]
Find the current in the thin straight wire if the magnetic field strength is equal to 0.00005 T at distance 5 cm.
Answer:
Answer
Correct option is
A
5×10
−6
tesla
I=5A
x=0.2m
Magnetic field at a distance 0.2 m away from the wire.
B=
2πx
μ
0
I
=
2π×0.2
4π×10
−7
×5
=10×5×10
−7
=5×10
−6
tesla
During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race
Answer:
The maximum speed of Bolt for the 100 m race is 14.66 m/s
Explanation:
Given;
initial distance covered by Bolt, d = 200 m
time of this motion, t = 19.3 s
The second distance covered by Bolt, = 100 m
Assuming Bolt maintained the same acceleration for both races.
His acceleration can be determined from the 200 m race.
d = ut + ¹/₂at²
where;
u is his initial velocity = 0
d = ¹/₂at²
[tex]at^2 = 2d\\\\a = \frac{2d}{t^2} \\\\a = \frac{2\times 200}{19.3^2} \\\\a = 1.074 \ m/s^2[/tex]
Let the final or maximum velocity for the 100 m race = v
v² = u² + 2ad₂
v² = 2 x 1.074 x 100
v² = 214.8
v = √214.8
v = 14.66 m/s
The maximum speed of Bolt for the 100 m race is 14.66 m/s
What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure
Full Question:
What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure?
A) 0°F
B) 273 K
C) 0 K
D) 100°C
E) 273°C
Answer:
The correction Option is D) 100°C
Explanation:
The temperature above is referred to as the critical point.
it is the highest temperature and pressure at which water (which has three phases - liquid, solid, and gas) can exist in vapor/liquid equilibrium. If the temperature goes higher than 100 degrees celsius, it cannot remain is liquid form regardless of what the pressure is at that point.
There is also a condition under which water can exist in its three forms: that is
- Ice (solid)
- Liquid (fluid)
- Gas (vapor)
That state is called triple point. The conditions necessary for that to occur are:
273.1600 K (0.0100 °C; 32.0180 °F) as temperature and611.657 pascals (6.11657 mbar; 0.00603659 atm) as pressureCheers
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A car is moving with a velocity of45m/s. Is brought to rest in 5s.the distance travelled by car before it comes to rest is
Answer:
The car travels the distance of 225m before coming to rest.
Explanation:
Here,
v = 45m/s
t = 5s
d = v × t
Therefore,
d = 45 × 5
= 225m
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective
Answer:
Reflective
Explanation:
The radiation pressure of the wave that totally absorbed is given by;
[tex]P_{abs}= \frac{I}{C}[/tex]
and While the radiation pressure of the wave totally reflected is given by;
[tex]P_{ref}= \frac{2I}{C}[/tex]
Now compare the two-equation you can clearly see that the pressure due to reflection is larger than absorption therefore the sail should be reflective.
Helium gas at 20 °C is confined within a rigid vessel. The gas is then heated until its pressure is doubled. What is the final temperature of the gas?
Answer:
586 K
Explanation:
Let P is the initial pressure.
Initial temperature, T₁ = 20°C = 293 K
Final pressure, P₂ = 2P
We need to find the final temperature of the gas.
The relation between the pressure and the temperature is as follows
[tex]P\propto T\\\\or\\\\\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}[/tex]
Put all the values,
[tex]\dfrac{P}{2P}=\dfrac{293}{T_2}\\\\\dfrac{1}{2}=\dfrac{293}{T_2}\\\\T_2=2\times 293\\\\T_2=586\ K[/tex]
So, the final temperature of the gas is 586 K.
A rescue plane spots a survivor 132 m directly below and releases an emergency kit with a parachute. If the package descends at a constant vertical acceleration of 6.89 m/s2 the initial plane horizontal speed was 86.9 m/s, how far away from the survivor will it hit the waves
Answer: 19.15 meters on the waves away from the survivor.
Explanation:
