Answer:
The displacement of a body has two components: rigid-body displacement and deformation. A rigid-body displacement consists of a simultaneous translation and rotation of the body without changing its shape or size.
Explanation:
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Like charges repel and opposite charges attract.
A uniform disk turns at 3.6 rev/s around a frictionless spindle. A non rotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk . They then both turn around the spindle with their centers superposed.
What is the angular frequency in rev/s of the combination?
please express answer in proper significant figures and rounding.
Answer:
ω₁ = 2.2 rev/s
Explanation:
Conservation of angular momentum
moment of inertia uniform disk is ½mR²
moment of inertia uniform rod about an end mL²/3
We can think of our rod as two rods of mass m/2 and length R
L = ½mR²ω₀
L = (½mR² + 2(m/2)R²/3)ω₁
ω₁ = ω₀(½mR² / (½mR² + mR²/3))
ω₁ = ω₀(½ / (½ + 1/3))
ω₁ = 0.6ω₀
ω₁ = 2.16
In part B of the lab, when the current flows through the orange part of the wire from right to left, the wire deflects (or moves) ____. This is in accordance with the right-hand-rule.
This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.
We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:
F = L*(IxB)
if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:
I = i*(1, 0, 0)
And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.
To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.
Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.
For example, if the magnetic field is in the positive z-axis, we will point upwards.
Now the palm of our hand tells us in which direction the force is applied.
This is the right-hand rule.
For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.
What is the volume of a metal block 3cm long by 2cm wide by 4cm high? What would be the volume of a block twice as long, wide, and high?
Answer:
Volume of a metal block = 24 cm^3
Volume of a block twice as long, wide and high = 192 cm^3
Explanation:
Volume of a block is measured in l*w*h and in the first block, the sides are 3, 2 and 4 and 3*2*4 = 24
Second block, just double each of the lengths to get 6*4*8 = 192
what are the limitation of clinical thermometer
Answer:
Their main disadvantage is that they are fairly easy to break and if they do, it results in small splinters of glass and the release of mercury which is quite toxic if absorbed into the body.
A cylindrical swimming pool has a radius 2m and depth 1.3m .it is completely filled with salt water of specific gravity 1.03.The atmospheric preassure is 1.013 x 10^5 Pa.
a.calculate the density of salt water.
Answer:
the density of the salt water is 1030 kg/m³
Explanation:
Given;
radius of the cylindrical pool, r = 2 m
depth of the pool, h = 1.3 m
specific gravity of the salt water, γ = 1.03
The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa
Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³
The density of the salt water is calculated as;
[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]
Therefore, the density of the salt water is 1030 kg/m³
A 700N marine in basic training climbs a 10m vertical rope at constant speed in 8sec. what is power put
Answer:
875 Watts
Explanation:
P = W/t = mgh/t = 700(10)/8 = 875 Watts
what are the two main types of sound like soundwave
Answer:
acoustic energy and mechanical energy
Explanation:
each type of sounds has to be tackled in their own way.
answer bhejo please please please
Answer:
Various uses of water :
1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.
2. Water used as a universal solvent.
3. water maintains the temperature of our body.
4. Water helps in digestion in our body.
5 .water is used in factories and industries.
6. Water is used to grow plants , vegetables and crops.
Astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c relative to an observer on Earth. When Jill left Earth, the spaceship was equipped with all kinds of scientific instruments, including a meter stick. Now that Jill is underway, how long does she measure the meter stick to be?
A) 0.280 m
B) 1.00 m
C) 0.960 m
D) 1.28 m
E) 1.04 m
(B) 1.00 m
Explanation:
Since the meter stick is traveling with Jill, it will have the same speed as she does so relative to Jill, the meter stick is stationary so its length remains 1.00 m as measured by her.
When astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c and measure the meter stick to be 1 meter. Hence, option B is correct.
What is length contraction?Length contraction is defined as the phenomenon of the moving object being shorter than its appropriate length, measured in the object's rest frame.
When the object travels with the speed of light, the length of the object gets more contracted than its original length, relative to the observer. It is also known as the Lorentz-Fitgerald contraction.
Length contraction, L = L₀√(1-v²/c²), where L is the original length, L₀ is the contracted length. c² is known as the velocity of light. v² is the velocity of the speed of the object.
From the given,
speed of the spaceship = 0.280c (c is the speed of the light)
Length contraction, L = L₀ √(1-v²/c²)
The stick also travels in the spaceship. Hence, the length of the meter stick does not change. It remains at its original length of one meter. Thus, the ideal solution is option B.
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Why does cold water kept in an open container become warm on a hot summer afternoon?
Cold water kept in an open container acquires heat from the warmer surrounding becomes warm like the air around it due to the transfer of thermal energy
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We do not use water instead of mercury in a barometer
Answer:
Because
1)Water is relatively less dense. Mercury is 13.6 times more dense than water. SO, If you use water, you have to have the length of barometer of length (or height) 13.534 times the length of mercury barometer, which may be more than 11 meter in length.
2)Also mercury as compared to water, has comparatively less specific heat and good conductor of heat, could come to the same temperature of the atmosphere more quickly.
Answer:
The atmospheric pressure at sea level 76 cm of Hg=1.013× 10⁵ pascal .
Explanation:
That is if water is used in barometer tube instead of mercury the length of the tube must be greater than 10.326 cm.so we cannot replace mercury by water in the barometer.
Describe how the words Science and optics would appear when viewed in a plane mirror?
Answer:
Lateral inversion will occur in a plane mirror.
Explanation:
When words are displayed in a plane or flat mirror, the result is that if the words are displayed left, they change to right and if they were normally displayed right, they change to left. This phenomenon is known as lateral inversion. So, this will apply to the words, Science and optics. Only the sides will be interchanged.
A plane mirror reflects light, therefore, the image that is produced by it remains the same size. The image produced will not appear upside down. Only the sides will be interchanged.
If a nucleus decays by successive b, a, a emissions, its mass number will Group of answer choices decrease by seven. decrease by two. decrease by four. decrease by eight. increase by four.
Answer:
The mass number will decrease by eight (8).
Explanation:
Given;
successive beta (b), alpha (a), alpha (a) emissions.
Generally, when a radioactive element emits a beta-particle (b), its mass number doesn't increase but its atomic number increases by 1 . [tex](^{0}_{-1}\beta )[/tex]
Also when a radioactive element emits an alpha-particle (a), its mass number decreases by 4, while its atomic number decrease by 2. [tex](^4_2\alpha)[/tex]
For the given question, a successive beta (b), alpha (a), and alpha (a) emissions = (0) + (-4) + (-4) = -8
Thus, when a radioactive element emit a successive beta (b), alpha (a), alpha (a) particles, the mass number will decrease by eight (8).
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Use the values from PRACTICE IT to help you work this exercise. If the current in each wire is doubled, how far apart should the wires be placed if the magnitudes of the gravitational and magnetic forces on the upper wire are to be equal
basic source of magnetism is a) charged particles alone b)Movement of charged particles c) Magnetic dipoles d)magnetic domains
Answer:
C . Magnetic dipoles is the correct
Answer:
b). movement of charged particles.
Explanation:
These charges create the nagnetic dipoles.
Can you please help me please?
Explanation:
CH3CH2OH
That is the answer I hope this helps
The spaceship Enterprise 1 is moving directly away from earth at a velocity that an earth-based observer measures to be 0.66c. A sister ship, Enterprise 2, is ahead of Enterprise 1 and is also moving directly away from earth along the same line. The velocity of Enterprise 2 relative to Enterprise 1 is 0.34c. What is the velocity of Enterprise 2, as measured by the earth-based observer
Answer:
The answer is "0.82 c".
Explanation:
Given:
Spacecraft speed 1 is [tex]u = + 0.66 \ c[/tex]
Space velocity 2 relative to spacecraft 1 is [tex]v = + 0.34\ c[/tex]
The spacecraft velocity 2 measured by the Earth observation
[tex]\to u' = \frac{u +v}{1 + ( \frac{uv}{c^2})}[/tex]
[tex]= \frac{0.66 \ c +0.34\ c}{ 1+ (\frac{0.66\ c \times 0.33\ c }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (\frac{0.2178\ c^2 }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (0.2178 )}\\\\ = \frac{1 \ c }{ 1.2178 }\\\\=0.82\ c[/tex]
a. Do the waves made by the two faucets travel faster than the waves made by just one faucet?
b. How do you know this? Describe how the two-faucet wave pattern compares with the one-faucet pattern.
c. Describe what happens to the two-faucet wave pattern as the separation of the faucets is increased.
Answer:
asdasd dsa dasdasd sadas dasd asdasd asd asd dsa asdd 223 aasd ada dasd sa dasd dsaa sd adsd asasd
Explanation:
Which of the following behaviors would best describe someone who is listening and paying attention? a) Leaning toward the speaker O b) Interrupting the speaker to share their opinion c) Avoiding eye contact d) Asking questions to make sure they understand what's being said
Answer:
D
Explanation:
Anyone could be leaning forward toward the speaker but be distracted and I believe if you're paying attention to the speaker, you would ask questions to make sure you're understanding what they are speaking
Answer:
A
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A lightning bolt has a current of 56,000 A and lasts for 80 x 10-6 seconds (80 μs). How much charge (in Coulombs) has flowed in this bolt?
Answer:
A cloud can discharge as much as 20 coulombs in a lightning bolt.
An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]
Apply the principle of conservation of energy;
ΔK.E = ΔU
[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]
Therefore, the speed of the electron at the given position is 106.2 m/s
A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that the spring is compressed by 0.090 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise
After being released, the restoring force exerted by the spring performs
1/2 (5200 N/m) (0.090 m)² = 12.06 J
of work on the block. At the same time, the block's weight performs
- (0.260 kg) g (0.090 m) ≈ -0.229 J
of work. Then the total work done on the block is about
W ≈ 11.83 J
The block accelerates to a speed v such that, by the work-energy theorem,
W = ∆K ==> 11.83 J = 1/2 (0.260 kg) v ² ==> v ≈ 9.54 m/s
Past the equilibrium point, the spring no longer exerts a force on the block, and the only force acting on it is due to its weight, hence it has a downward acceleration of magnitude g. At its highest point, the block has zero velocity, so that
0² - v ² = -2gy
where y is the maximum height. Solving for y gives
y = v ²/(2g) ≈ 4.64 m
Polarized sunglasses:
a. block most sunlight because sunlight is polarized
b. are better but work the same way as non-polarized sunglasses
c. are polarized to filter out certain wavelengths of light
d. block reflected light because reflected light is partially polarized.
Polarized sunglasses creates filter of vertical openings for light. The light rays will reach the eyes of human vertically only.
The sun rays will not reach human eye directly which will create a shield for sun light burden on human eye.
Polarized sunglasses are best used for blocking and eliminating certain wavelengths of light.
Therefore the correct answer is option C. Polarizes Sunglasses are polarized and it filter out certain wavelengths of light.
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Which was a major effect of Pope Leo III crowning Charlemagne emperor of the Romans ?
Answer:
The crowning of Charlemagne by Pope Leo III was significant in a number of ways. For Charlemagne, it was necessary because it encouraged to give him higher reliability. It gave him the rank of a dictator, giving him the only ruler in Europe west of the Byzantine emperor in Constantinople.
Calculate the kinetic energy of a mass 2kg moving with a velocity of 0.1m/s
ANSWER-:
1/2 mv²
K.E = 1/2 mv²
K.E = 0.01 J.
Hence, the kinetic energy of a body is 0.01 Joule.
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Why are scientific models important?
Answer:
Scientific models are representations of objects, systems or events and are used as tools for understanding the natural world. Models use familiar objects to represent unfamiliar things. Models can help scientists communicate their ideas, understand processes, and make predictions.
What's the minimum Out PUT WORK
required to raise 14,0m3 of water 26.0m?
Answer:
3.57 MJ
Explanation:
ASSUMING it's fresh water with density of 1000 kg/m³
W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J
Salt water would require more.
A bar of steel has the minimum properties Se = 40 kpsi, Sy = 60 kpsi, and Sut = 80 ksi. The bar is subjected to a steady torsional stress of 15 kpsi and an alternating bending stress of 25 ksi. Find the factor of safety guarding against a static failure, and either the factor of safety guarding against a fatigue failure or the expected life of the part. For the fatigue analysis use Modified Goodman criterion.
Answer:
The correct solution is:
(a) 1.66
(b) 1.05
Explanation:
Given:
Bending stress,
[tex]\sigma_b = 25 \ kpsi[/tex]
Torsional stress,
[tex]\tau= 15 \ kpsi[/tex]
Yield stress of steel bar,
[tex]\delta_y = 60 \ kpsi[/tex]
As we know,
⇒ [tex]\sigma_{max}^' \ = \sqrt{\sigma_b^2 + 3 \gamma^2}[/tex]
[tex]= \sqrt{(25)^2+3(15)^2}[/tex]
[tex]=36.055 \ kpsi[/tex]
(a)
The factor of safety against static failure will be:
⇒ [tex]\eta_y = \frac{\delta_y}{\sigma_{max}^'}[/tex]
By putting the values, we get
[tex]=\frac{60}{36.055}[/tex]
[tex]=1.66[/tex]
(b)
According to the Goodman line failure,
[tex]\sigma_a = \sigma_b = 25 \ kpsi[/tex]
[tex]S_e = 40 \ kpsi[/tex]
[tex]\sigma_m = \sqrt{3} \tau[/tex]
[tex]=\sqrt{3}\times 15[/tex]
[tex]=26 \ kpsi[/tex]
[tex]Sut = 80 \ kpsi[/tex]
⇒ [tex]\frac{\sigma_a}{S_e} +\frac{\sigma_m}{Sut} =\frac{1}{\eta_y}[/tex]
[tex]\frac{25}{40}+\frac{26}{80}=\frac{1}{\eta_y}[/tex]
[tex]\eta_y = 1.05[/tex]
coin 1 is thrown upward from the top of 100m tower with a speed of 15m/s. coin 2 is dropped from the top of the tower 2.0second later. assume g is 10m/s. how far below the top of the tower des coin 1 pass coin 2
The height below the tower at which coin 1 pass coin 2 is 89.04 m.
The given parameters:
height of the tower, h = 100 m
initial velocity of coin 1, v = 15 m/s
time spent in air by coin 1 before coin 2 was dropped = 2s
To find:
the height below the tower at which coin 1 passes coin 2Find the maximum height attained by coin 1 before falling to the ground:
[tex]v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m[/tex]
Find the time taken for coin 1 to fall to the ground:
Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m
[tex]t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s[/tex]
But the time taken for the coin 1 to reach 11.25 m above the tower:
[tex]t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s[/tex]
Total time spent by coin 1 before reaching ground with respect to coin 2:
time = (1.5 s + 4.72 s) - 2 s
time = 4.22 s
Note: the 2 s was subtracted to keep both coins at a fair starting time below the tower.
Find the total time taken for coin 2 to fall to the ground:
Height of coin 2 above the ground = 100 m
Total time taken by coin 2 before falling to the ground is calculated as:
[tex]t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 = 4.47s[/tex]
The time at which coin 1 will pass coin 2 is 4.22 s.
Find the height below the tower when the time is 4.22 s.
[tex]h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m[/tex]
Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.
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