Answer:
1.52×10⁻³ Wb
Explanation:
Using
Φ = BAcosθ.......................... Equation
Where, Φ = magnetic Field, B = 0.510 T, A = cross sectional area of the loop, θ = angle between field and the plane of the loop
Given: B = 0.510 T, θ = 22°,
A = πr², Where r = radius of the circular loop = 3.20 cm = 0.032 m
A = 3.14(0.032²)
A = 3.215×10⁻³ m²
Substitute into equation 1
Ф = 0.510(3.215×10⁻³)cos22°
Ф = 0.510(3.215×10⁻³)(0.927)
Ф = 1.52×10⁻³ Wb
Hence the magnetic flux through the loop = 1.52×10⁻³ Wb
Final answer:
To calculate the magnetic flux through a loop, use the formula Φ = B * A * cos(θ) with the given magnetic field strength, loop's area, and angle. For a 0.510 T field and a 3.20 cm radius loop at 22°, the magnetic flux is 1.5 * 10⁻³ Wb.
Explanation:
The student asked about calculating the magnetic flux through a circular loop of wire placed in a magnetic field. To find the magnetic flux (Φ), we use the formula Φ = B * A * cos(θ), where B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop's plane. Given the field's magnitude is 0.510 T, the radius of the loop is 3.20 cm (which gives an area A = π * r²), and the angle is 22.0°, the magnetic flux can be calculated.
The area A of the loop is A = π * (0.032 m)² = 3.2*10⁻³ m². Then, we apply the cosine of the angle, cos(22°) = 0.927. So the flux Φ = (0.510 T) * (3.2*10⁻³ m²) * 0.927 = 1.5 * 10⁻³ Wb (weber).
What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 46 rpmrpm (revolutions per minute) if the wheel's diameter is 32 cmcm ?
Acceleration of a speck is 0.77 m/s²
Explanation:
Given of the solution-
Diameter (We can represent as d) = 32 cm
radius, (We can represent as r )= 32/2 = 16 cm = 0.16 m
Angular acceleration,(We can represent as α ) = 46 rpm
[tex]\alpha = 46 * \frac{2\pi }{60}[/tex]
Acceleration, a = ?
We know that the formula for the acceleration is
[tex]a = r * \alpha[/tex]
[tex]a = 0.16 * 46 * \frac{2\pi }{60}\\ \\a = 0.16 * 46 * \frac{2 * 3.14}{60}[/tex]
[tex]a = 0.77m/s^2[/tex]
Therefore, acceleration of a speck is 0.77 m/s²
A bullet is fired with a horizontal velocity of 1500 ft/s through a 6-lb block A and becomes embedded in a 4.95-lb block B. Knowing that blocks A and B start moving with velocities of 5 ft/s and 9 ft/s, respectively, determine (a) the weight of the bullet, (b) its velocity as it travels from block A to block B.
Answer:
weight of the bullet is 0.0500 lb
velocity as it travels from block A to block B is 900 ft/s
Explanation:
given data
horizontal velocity = 1500 ft/s
mass block A = 6-lb
mass block B = 4.95 lb
blocks A velocity = 5 ft/s
blocks B velocity = 9 ft/s
solution
we apply here law of conservation of momentum that is
m × v(o) + m1 × (0) + m2 × (0) = m × v(2) + m1 × v1 + m2 × v2 ................1
put here value and we get m
m = [tex]\frac{m1 \times v1 +m2 \times v2}{v(o) - v2}[/tex] ..............2
m = [tex]\frac{6 \times 5 + 4.95 \times 9}{1500 - 9}[/tex]
m = 0.0500 lb
and
when here bullet is pass through the A block then moment is conserve that is
m × v(o) + m × v1 + m1 × v1 ............3
v1 = [tex]\frac{m\times v(o)- m1\times v1}{m}[/tex]
v1 = [tex]\frac{0.0500\times 1500 - 61\times 5}{0.05}[/tex]
v1 = 900 ft/s
Singly charged positive ions are kept on a circular orbit in a cyclotron. The magnetic field inside the cyclotron is 1.833 T. The mass of the ions is 2.00×10-26 kg, and speed of the ions is 2.05 percent of the speed of the light. What is the diameter of the orbit? (The speed of the light is 3.00×108 m/s.)
Answer:
The diameter is 0.8376 m
Explanation:
Magnetic force is the force that is associated with the magnetic field, the magnitude of the magnitude force can be obtained using equation 1;
F = q v B .....................................1,
where q is the magnitude of the charge of the particle =;
v is the velocity and;
B is the magnetic field = 1.833 T ;
Here, the path of the charge is circular so the force can be also considered as the centripetal force which is represented in equation 2;
[tex]F_{c}[/tex] = m [tex]v^{2}[/tex] / r ....................................2,
m is the particle's mass = 2.00 x[tex]10^{-26}[/tex] kg
v is the speed of ion = 2.05% of 3.00 x [tex]10^{8}[/tex] = 0.0205 x 3.00 x [tex]10^{8}[/tex]
= 6.15 x [tex]10^{6}[/tex] m/s ;
Singly charged ion has a charge equal to the electron charge and the magnitude = 1.60217646 ×[tex]10^{-19}[/tex] C and;
r is the radius of the circular path.
to get the diameter of the orbit we equate equation 1 to 2 and isolate r in equation 2.
q v B = m [tex]v^{2}[/tex] / r
r = m v/q B.....................................3
r = (2.00 x[tex]10^{-26}[/tex] kg) x (6.15 x [tex]10^{6}[/tex] m/s) / (1.60217646 ×[tex]10^{-19}[/tex] C) x (1.833 T)
r = 0.4188 m
The diameter is r x 2
D = 0.4188 m x 2
D = 0.8376 m
Therefore the diameter is 0.8376
When light goes from one material into another material having a HIGHER index of refraction,
A) its speed decreases but its wavelength and frequency both increase.
B) its speed, wavelength, and frequency all decrease.
C) its speed increases, its wavelength decreases, and its frequency stays the same.
D) its speed decreases but its frequency and wavelength stay the same.
E) its speed and wavelength decrease, but its frequency stays the same.
Answer:
When light goes from one material into another material having a HIGHER index of refraction, its speed and wavelength decrease, but its frequency stays the same.
The correct option is E
Explanation:
Refraction is the bending of light ray as it crosses the boundary between the two media of different densities, thus causing a change in it's direction.
When light goes from one material of lower refractive to another material of higher refractive index, the speed of light reduces, because from law of refraction
aNg= (speed of light in air)/(speed of light in glass)
Snell's Law
So generally,
n1/n2 = v2/v1
Let n1 be incident index
Let n2 be refracted index
v2 is refracted speed
v1 is incident speed
Since given that, n1<n2
Then, the right side of the equation will always be less than 1 i.e n1/n2
Then, v2=v1• ( n1/n2)
Since n1/n2 is less that one this show that the speed v2 will reduce.
Since we know that the speed decrease,
We also know that speed, wavelength and speed is related by
v=fλ
Since, the speed is directly proportional to wavelength this shows that as the speed reduces the wavelength also reduces where frequency is the constant of proportionality
v∝λ. .
then f is constant of proportional
v=fλ
So as speed reduces, the wavelength reduces.
frequency of a light wave does not depend on the medium, while wavelength and speed do.
its speed and wavelength decrease, but its frequency stays the same.
Then the answer is E
g Case 1, a mass M M hangs from a vertical spring having spring constant k , k, and is at rest at its equilibrium height. In Case 2, the same mass has been lifted a distance D D vertically upward. If the potential energy in Case 1 is defined to be zero, what is the potential energy in Case 2
In the scenario described, increasing the height of a mass M in a vertical spring-mass system will result in both gravitational and spring potential energy. Thus, the potential energy in Case 2 is the sum of gravitational potential energy (m*g*D) and the elastic potential energy of the spring (1/2*k*D^2).
Explanation:The situation described in the question deals with principles of Physics, specifically potential energy and its application to the spring-mass system. Consider both cases with the mass M hanging from a spring with a constant k in a vertical orientation. In the first case, the system is at equilibrium and thus the potential energy is defined as zero.
In the second case, the mass M is lifted upward a vertical distance D from the equilibrium position, it now possesses gravitational potential energy in addition to the elastic potential energy of the spring. This total energy is preserved as long as no external force behaves on the system. The gravitational potential energy gained by the mass is equal to m*g*D where m is the mass, g is the acceleration due to gravity and D is the vertical distance lifted.
The potential energy stored in the spring when it is stretched or compressed by a distance 'x' is given by the formula U = 1/2*k*x^2, where 'k' is the spring constant and 'x' is the displacement from the equilibrium position (in this case is D).
So, combining both energies, the total potential energy in Case 2, when the mass has been lifted a vertical distance D is m*g*D + 1/2*k*D^2.
Learn more about Potential Energy here:https://brainly.com/question/24284560
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An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1000 K and a heat rejection at 400 K. During the heat addition the volume triples. Find the two specific heat transfers (q) in the cycle and the overall cycle efficiency
Answer:
W / n = - 9133 J / mol, W / n = 3653 J / mol , e = 0.600
Explanation:
The Carnot cycle is described by
[tex]e= 1 - Q_{c} / Q_{H} = 1 - T_{c} / T_{H}[/tex]
In this case they indicate that the final volume is
V = 3V₀
In the part of the heat absorption cycle from the source is an isothermal expansion
W = n RT ln (V₀ / V)
W / n = 8.314 1000 ln (1/3)
W / n = - 9133 J / mol
During the part of the isothermal compression in contact with the cold focus, as in a machine the relation of volumes is maintained in this part is compressed three times
W / n = 8.314 400 (3)
W / n = 3653 J / mol
The efficiency of the cycle is
e = 1- 400/1000
e = 0.600
An electric motor is plugged into a standard wall socket and is running at normal speed. Suddenly, some dirt prevents the shaft of the motor from turning quite so rapidly. What happens to the back emf of the motor, and what happens to the current that the motor draws from the wall socket?
a.) The back emf decreases, and the current drawn from the socket decreases.
b.) The back emf increases, and the current drawn from the socket decreases.
c.) The back emf decreases, and the current drawn from the socket increases.
d.) The back emf increases, and the current drawn from the socket increases.
Final answer:
The back emf of the motor decreases due to reduced speed caused by dirt in the shaft, which, in turn, causes the current drawn from the wall socket to increase.
Explanation:
When dirt prevents the shaft of an electric motor from turning rapidly, the back emf produced by the motor decreases. Since the back emf is proportional to the motor's angular velocity, any decrease in speed results in a decrease in back emf. Because the back emf opposes the applied voltage from the wall socket, a decrease in back emf implies that the voltage across the motor's coil will increase, causing the motor to draw a larger current from the socket. Therefore, the current that the motor draws from the wall socket will increase. The correct answer to the question is (c) The back emf decreases, and the current drawn from the socket increases.
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right so that it slides along the surface. Let R = 1.45 ft and let the angle at which the sphere separates from the cylinder be θs = 34°. The sphere was placed in motion at the very top of the cylinder. Determine the sphare;s initial speed.
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
You shoot a 50.3-g pebble straight up with a catapult whose spring constant is 320.0 N/m. The catapult is initially stretched by 0.190 m. How high above the starting point does the pebble fly? Ignore air resistance.
Answer:
The pebble reaches a height of 11.716 m above the starting point.
Explanation:
Given:
Mass of pebble (m) = 50.3 g = 0.0503 kg [1 g = 0.001 kg]
Spring constant (k) = 320.0 N/m
Elongation of catapult (x) = 0.190 m
Height of fly (h) = ?
Air resistance is ignored. So, conservation of energy holds true.
The energy stored in the catapult on stretching it is elastic potential energy. This elastic potential energy is transferred to the pebble in the form of gravitational potential energy. Therefore,
Elastic potential energy = Gravitational potential energy
[tex]\frac{1}{2}kx^2=mgh\\\\h=\frac{kx^2}{2mg}[/tex]
Plug in the given values and solve for 'h'. This gives,
[tex]h=\frac{(320.0\ N/m)(0.190\ m)^2}{2(0.0503\ kg)(9.8\ m/s^2)}\\\\h=\frac{11.552\ Nm}{0.986\ N}\\\\h=11.716\ m[/tex]
Therefore, the pebble reaches a height of 11.716 m above the starting point.
A remote-controlled car’s wheel accelerates at 22.7 rad/s2 . If the wheel begins with an angular speed of 10.3 rad/s, what is the wheel’s angular speed after exactly twenty full turns
Explanation:
Below is an attachment containing the solution.
A beam of electrons is accelerated from rest through a potential difference of 0.200 kV and then passes through a thin slit. When viewed far from the slit, the diffracted beam shows its first diffraction minima at ± 13.6 ∘ from the original direction of the beam.
Do we need to use relativity formulas? Select the correct answer and explanation.
a. No. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=mc2/(γ−1). The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is0.511 MeV, we have γ–1=(0.511MeV)/(0.200keV) so γ−1>>1 which means that we do not have to use special relativity.
b. Yes. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=(γ−1)mc2. The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is0.511 MeV, we have γ–1=(0.200keV)/(0.511MeV) so γ<<1 which means that we have to use special relativity.
c. Yes. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=mc2/(γ−1). The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is0.511 MeV, we have γ–1=(0.511MeV)/(0.200keV) so γ>>1 which means that we have to use special relativity.
d. No. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=(γ−1)mc2. The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is 0.511 MeV, we have γ–1=(0.200keV)/(0.511MeV) so γ−1<<1 which means that we do not have to use special relativity.
Part B
How wide is the slit?
Option d is correct; relativity formulas are not needed because the Lorentz factor (γ-1) is much less than 1, indicating that the electron's velocity is non-relativistic after being accelerated through a potential difference of 0.200 kV.
Explanation:We need to determine whether we should use relativity formulas for the electrons accelerated through a potential difference. When electrons gain kinetic energy (K) by acceleration through a potential difference (V), the energy they gain can be expressed as Ve=K=(γ-1)mc2. Here, γ stands for Lorentz factor, m is the mass of electron, and c is the speed of light. Given that the potential difference ('V') is 0.200 kV, and the rest mass energy ('Eo') of an electron is 0.511 MeV, we calculate:
γ-1= (0.200 keV) / (0.511 MeV) ≈ 0.3916 × 10-3, which implies that γ is approximately 1. So, γ-1 << 1, which in turn means that the electron's velocity is much less than the speed of light, and the relativistic effect can be neglected. Thus, we do not need to use special relativity.
The correct answer is d.
The op amp in this circuit is ideal. R3 has a maximum value of 100 kΩ and σ is restricted to the range of 0.2 ≤ σ ≤ 1.0. a. Calculate the range of vO if vI = 40 mV. b. If σ is not restricted, at what value of σ will the operational amplifier saturate?
I have attached the circuit image missing in the question.
Answer:
A) The range of vo is; -6.6V≤ vo ≤-1V
B) σ = 0.1861
Explanation:
A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.
Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0
So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0
Simplifying further; -25 vg - vΔ = 0
From the question, vg = 40mV = 0.04 V
So - 25(0.04) = vΔ
So: vΔ = - 1 V
Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0
So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0
So RΔ = 100kΩ or 100,000Ω from the question.
So, substituting for RΔ, we get,
[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0
Let's put the value of - 1 for vΔ as gotten before.
So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0
Now let's make vo the subject of the equation to get;
-1 - vo = (1 - σ)[2 + (1/σ)]
-1 - vo = 2 - 2σ + (1/σ) - 1
-vo = 1 + 2 - 2σ + (1/σ) - 1
-vo = 2 - 2σ + (1/σ)
vo = - 1 (2 - 2σ + (1/σ))
When σ = 0.2; vo = - 1(2 - 0.4 + 5) =
- 1 x 6.6 = - 6.6V
Also when σ = 1;
vo = - 1(2 - 2 + 1) = - 1V
Therefore, the range of vo is;
- 6.6V ≤ vo ≤ - 1V
B) it will saturate at vo = - 7V
So, from;
vo = - 1 (2 - 2σ + (1/σ))
-7 = - 1 (2 - 2σ + (1/σ))
Divide both sides by (-1)
7 = (2 - 2σ + (1/σ))
Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)
Multiply each term by α to get;
5σ = - 2σ^(2) + 1
So 2σ^(2) + 5σ - 1 = 0
Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861
Final answer:
The question involves understanding an ideal operational amplifier's output behavior, focusing on its output voltage changes and saturation with respect to variations in σ (sigma). Calculations for the output voltage range given a specific input and understanding the conditions under which the op-amp will saturate are key.
Explanation:
The question revolves around an operational amplifier (op-amp) circuit, exploring its behavior under certain conditions, specifically examining output voltage (vO) variations and saturation point related to the parameter σ (sigma). The op-amp is assumed to be ideal, implying infinite gain, zero input current, and that its input terminals are at the same potential.
a. Range of vO if vI = 40 mV
Given that the op-amp is ideal, the output voltage will depend on the input voltage (vI), the gain settings (σ), and the feedback resistor (R3). In practical scenarios, the gain can be adjusted by changing the value of σ or R3. However, for an ideal op-amp, the input voltage is directly proportional to the output voltage, influenced by σ. With vI = 40 mV and σ ranging between 0.2 and 1.0, the output voltage will vary accordingly, directly proportional to these parameters.
b. Saturation point of the op-amp
Saturation in an op-amp occurs when the output voltage exceeds the power supply limits, meaning the op-amp can no longer amplify the input signal. The specific value of σ at which saturation occurs depends on the supply voltage, the op-amp's maximum output voltage capability, and the configuration of the feedback network. Without specific values for the power supply or feedback network, calculating the exact σ value for saturation is not possible. Yet, in theory, as σ approaches the op-amp's gain limit or if the gain results in an output voltage beyond what the op-amp can deliver based on its power supply, saturation will occur.
A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.
How fast is the skier moving when she gets to the bottom ofthe hill?
Final answer:
The skier is moving at 11.1 m/s when she gets to the bottom of the hill. This solution is derived using the principles of energy conservation and work done by friction.
Explanation:
A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high. To determine how fast the skier is moving when she gets to the bottom of the hill, we analyze the problem using principles of energy conservation and work done by friction.
We start with the equation that equates the final kinetic and potential energies to the initial energies along with the work done by friction. This equation is 0.5 mv² + 0 = 0.5 mu² + μmgl + mgh, where v is the final velocity, μ is the coefficient of friction, g is the acceleration due to gravity (9.8 m/s²), l is the length of the rough patch, and h is the height of the hill.
Plugging in the given values,
we have: 0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.50. S
implifying, we get v² = 123.07, which leads to v = 11.1 m/s.
Thus, the skier's speed at the bottom of the hill is 11.1 m/s.
Analyzing the Light Bulb: You should have noticed that the light bulb doesn't have a single well-defined "resistance," since the current vs. voltage plot is nonlinear. Nevertheless, one can define a "voltage-dependent resistance" as R(V)=V/I(V)as the ratio of voltage to current.1Basic Behavior: According to your data, does this resistance increase or decrease with voltage? A reasonable (and correct) thought is that the impact is really with temperature, as the light bulb heats up with more power going into it. How does your data imply resistance varies with temperature?Thermal Expansion: One hypothesis you might have is that the reason is that the resistor expands slightly with increased temperature (since most materials do), and hence the cross-sectional area and length of the resistor change.Supposing the resistor increases in size by the same factor in every direction, what direction does the resistance change? (I.e., does the resistance get larger or smaller?) Is this the direction that you expect based on your answer to the previous part?
Answer:
Resistance increases with increase in temperature which depends on power supplied which also depends on voltage.
Thermal expansion will make resistance larger.
Explanation:
Light bulb is a good example of a filament lamp. If we plot the graph of voltage against current we will notice that resistance is constant at constant temperature.
The filament heats up when an electric current passes through it, and produces light as a result.
The resistance of a lamp increases as the temperature of its filament increases. The current flowing through a filament lamp is not directly proportional to the voltage across it.
tensile stress begins to appear in resistor as the temperature rises. Thus, the resistance value increases as the temperature rises. Resistance value can only decrease as the temperature rises in case of thin film resistor with aluminium substrate.
In case of a filament bulb, the resistance will increase as increase in length of the wire. The thermal expansion in this regard is linear expansivity in which resistance is proportional to length of the wire.
Resistance therefore get larger.
What determines whether the equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler or warmer water?
Final answer:
The equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler water due to the specific heat capacity of water.
Explanation:
The equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler water.
When different temperatures of water are mixed, heat is transferred between them until they reach a common equilibrium temperature. The amount of heat transferred is determined by the specific heat capacity of water, which is greater than most common substances. As a result, water undergoes a smaller temperature change for a given heat transfer. Therefore, the equilibrium temperature will be closer to the initially cooler water.
Say you have two parallel current-carrying wires, each carrying a current of 1.0 A and with a distance of 1.0 m between them. What is the magnitude of the force per unit length experienced by each wire
Answer:
[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]
Explanation:
Given,
current in the wire I₁ = 1 A
I₂ = 1 A
distance between them, r = 1 m
using Force per unit length formula
[tex]\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi r}[/tex]
[tex]\mu_0 = magnetic\ permeability\ of\ free\ space = 4\pi\times 10^{-7}[/tex]
[tex]\dfrac{F}{l}=\dfrac{4\pi \times10^{-7}\times 1\times 1}{2\pi\times 1}[/tex]
[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]
Hence, the magnitude of force per unit length is equal to [tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]
Final answer:
The magnitude of the force per unit length each 1.0 A current-carrying wire experiences when separated by a distance of 1.0 m is 2 x 10⁻⁷ N/m, calculated using the formula for magnetic force between parallel conductors.
Explanation:
When discussing two parallel current-carrying wires, we're addressing the magnetic force that arises between them due to their currents. Specifically, when each wire carries a current of 1.0 A and they are placed 1.0 m apart, the force per unit length that each wire experiences can be calculated using the formula for the magnetic force between two parallel conductors. The equation is FE = (µ0 / 2π) × (I1I2 / r), where µ0 is the magnetic constant (4π x 10-7 T·m/A), I1 and I2 are the currents in the wires, and r is the distance between the wires. Given that each wire carries 1.0 A of current and the separation is 1.0 m, we can plug these values into the formula to find that FE = (4π x 10-7 T·m/A / 2π) × (1.0 A2 / 1.0 m), resulting in a force per unit length of 2 x 10-7 N/m.
The potential difference between a pair of oppositely charged parallel plates is 398 V. If the spacing between the plates is doubled without altering the charge on the plates, what is the new potential difference between the plates? Answer in units of V.
Answer:
Explanation:
capacitance of parallel plate capacitor
c = ε A / d , d is distance between plates , A is surface area , ε is constant
As d becomes two times , Capacitance c = 1/ 2 times ie c / 2
potential V = Q / C
Q is constant , potential
v = Q / c /2
= 2 . Q / C
= 2 V
So potential difference becomes 2 times.
NEW P D = 398 X 2
= 796 V.
A 20.0-kg cannonball is fired from a cannon with muzzle speed of 1 000 m/s at an angle of 37.08 with the horizontal. A second ball is fired at an angle of 90.08. Use the isolated system model to find (a) the maximum height reached by each ball and (b) the total mechanical energy of the ball–Earth system at the maximum height for each ball. Let y5 0 at the cannon.
Answer:
(a). The maximum height by first ball is [tex]1.8545\times10^{4}\ m[/tex]
The maximum height by second ball is [tex]5.1020\times10^{4}\ m[/tex]
(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is [tex]1.0\times10^{7}\ J[/tex]
Explanation:
Given that,
Mass of cannonball = 20.0 kg
Speed = 1000 m/s
Angle with horizontal= 37.08
Fired angle = 90.08
We need to calculate the speed of the ball
Using formula of speed
[tex]v_{y}=v\sin\theta_{H}[/tex]
Put the value into the formula
[tex]v_{y}=1000\times\sin37.08[/tex]
[tex]v_{y}=602.9\ m/s[/tex]
(a). We need to calculate the maximum height by first ball
Using conservation of energy
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
[tex]h= \dfrac{v^2}{2g}[/tex]
Put the value into the formula
[tex]h=\dfrac{(602.9)^2}{2\times9.8}[/tex]
[tex]h=1.8545\times10^{4}\ m[/tex]
We need to calculate the maximum height by second ball
Using conservation of energy
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
[tex]h= \dfrac{v^2}{2g}[/tex]
Put the value into the formula
[tex]h=\dfrac{(1000)^2}{2\times9.8}[/tex]
[tex]h=5.1020\times10^{4}\ m[/tex]
(b). We need to calculate the total mechanical energy of the ball–Earth system at the maximum height for each ball
Using formula of energy
[tex]E=\dfrac{1}{2}mv^2[/tex]
[tex]E=\dfrac{1}{2}\times20\times1000^2[/tex]
[tex]E=1.0\times10^{7}\ J[/tex]
Hence, (a). The maximum height by first ball is [tex]1.8545\times10^{4}\ m[/tex]
The maximum height by second ball is [tex]5.1020\times10^{4}\ m[/tex]
(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is [tex]1.0\times10^{7}\ J[/tex]
A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
a. 79.1 N
b. 344 J
c. 344 J
d. 0 J
e. 0 J
Explanation:
a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force [tex]F_p[/tex] by the worker must be equal to the friction force [tex]F_f[/tex] on the crate, which is the product of friction coefficient μ and normal force N:
Let g = 9.81 m/s2
[tex]F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N[/tex]
b. The work is done on the crate by this force is the product of its force [tex]F_p[/tex] and the distance traveled s = 4.35
[tex]W_p = F_ps = 79.1*4.35 = 344 J[/tex]
c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35
[tex]W_f = F_fs = -79.1*4.35 = -344 J[/tex]
This work is negative because the friction vector is in the opposite direction with the distance vector
d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.
e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction
[tex]W_p + W_f = 344 - 344 = 0 J[/tex]
Answer:
(A) 79N
(B) W = 344J
(C) Wf= -344J
(D) W = 0J
(E) W = 0J
Explanation:
Please see attachment below.
Before entering a mass spectrometer, ions pass through a velocity selector consisting of parallel plates separated by 1.6 mm and having a potential difference of 148 V. The magnetic field between the plates is 0.42 T. The magnetic field in the mass spectrometer is 1.2 T.
1)(a) Find the speed of the ions entering the mass spectrometer.
m/s
2)(b) Find the difference in the diameters of the orbits of singly ionized 238U and 235U. (The mass of a 235U ion is 3.903 x 10-25 kg.)
mm
Answer:
1) v = 2.20 10⁵ m / s , 2) r = 4.86 10⁻³ m, r = 4.92 10⁻³ m
Explanation:
1) A speed selector is a section where the magnetic and electrical forces have opposite directions, so
[tex]F_{m} = F_{e}[/tex]
q v B = q E
v = E / B
V = E s
E = V / s
v = V / s B
v = 148 / (0.0016 0.42)
v = 2.20 10⁵ m / s
2) when the isotopes enter the spectrometer, we can use Newton's second law
F = m a
Acceleration is centripetal
a = v² / r
q v B = m v² / r
r = m v / qB
The mass of uranium 235 is
m = 3.903 10⁻²⁵ kg
The radius of this isotope is
r = 3,903 10⁻²⁵ 2.20 10⁵ / (92 1.6 10⁻¹⁹ 1.2)
r = 4.86 10⁻³ m
The mass of the uranium isotope 238 is
m = 238 a = 238 1.66 10-27 = 395.08 10-27 kg
The radius is
r = 395.08 10⁻²⁷ 2.20 10⁵ / (92 1.6 10⁻¹⁹ 1.2)
r = 4.92 10⁻³ m
Final answer:
The speed of the ions entering the mass spectrometer is 352.38 m/s. The difference in the diameters of the orbits of the 238U and 235U ions can be calculated using the formula for the radius of a charged particle's orbit in a magnetic field.
Explanation:
To find the speed of the ions entering the mass spectrometer, we can use the equation for the force on a charged particle in a magnetic field: F = qvB, where F is the force, q is the charge, v is the velocity, and B is the magnetic field. The force is equal to the electric force, qE, so we can set these two equations equal to each other and solve for v: qvB = qE. Since q is the charge of the ion and E is the electric field strength, we can rearrange this equation to solve for v, giving us v = E/B.
Plugging in the values given in the question, we find that the speed of the ions entering the mass spectrometer is 148 V / 0.42 T = 352.38 m/s.
To find the difference in the diameters of the orbits of the 238U and 235U ions, we can use the formula for the radius of a charged particle's orbit in a magnetic field: r = mv / (qB), where m is the mass of the ion, v is its velocity, q is its charge, and B is the magnetic field. Since the ions are singly ionized, their charges are +e, where e is the charge of an electron. Plugging in the values given in the question, we can calculate the radius of the orbits of the 238U and 235U ions, and then subtract these values to find the difference in their diameters.
The element hydrogen has the highest specific heat of all elements. At a temperature of 25°C, hydrogen’s specific heat capacity is 14300J/(kg K). If the temperature of a .34kg sample of hydrogen is to be raised by 25 K, how much heat will have to be transferred to the hydrogen?
Answer:
121550 J
Explanation:
Parameters given:
Mass, m = 0.34kg
Specific heat capacity, c = 14300 J/kgK
Change in temperature, ΔT = 25K
Heat gained/lost by an object is given as:
Q = mcΔT
Since ΔT is positive in this case and also because we're told that heat was transferred to the hydrogen sample, the hydrogen sample gained heat. Therefore, Q:
Q = 0.34 * 14300 * 25
Q = 121550J or 121.55 kJ
Cylinder A has a mass of 2kg and cylinder B has a mass of 10kg. Determinethe velocity of A after it has displaced 2m from its original starting position. Neglect the mass of the cable and pulleys and assume that both cylinders start at res
The velocity of A is 5.16m/s²
Explanation:
Given-
mass of cylinder A, mₐ = 2kg
mass of cylinder B, mb = 10kg
Distance, s = 2m
Velocity of A, v = ?
Let acceleration due to gravity, g = 10m/s²
We know,
[tex]a = \frac{mb * g - ma * g}{ma + mb} \\\\a = \frac{10 * 10 - 2 * 10}{ 2 + 10} \\\\a = \frac{80}{12} \\\\a = 6.67m/s^2[/tex]
We know,
[tex]v = \sqrt{2as}[/tex]
[tex]v = \sqrt{2 X 6.67 X 2} \\\\v = \sqrt{26.68} \\\\v = 5.16m/s^2[/tex]
Therefore, the velocity of A is 5.16m/s²
A cube has one corner at the origin and the opposite cornerat the point (L,L,L). The sides of the cube are parallel to thecoordinate planes. The electric field in and around the cube isgiven by E=(a+bx)x+cy.Part A Find the total electric flux phi E through the surface of the cube. Express your answer in terms of a,b,c and L. Phi E=Part B This part will be visible after youcomplete previous item(s). Part C What is the net charge q inside the cube? Express your answer in terms of a,b,c ,L,and epsilon. q=
Answer:
Explanation:
in this case, flux and area vectors are parallel . therefore, flux is just the dot product of electric field and area vector.
(flux=EAcos0 =EA)
flux through the face x face
Physics homework question answer, step 1, image 1
the flux through -x face
Physics homework question answer, step 1, image1
Step 2
the flux through y face,
Physics homework question answer, step 2, image 2
the flux through -y face,
Physics homework question answer, step 2, image 2
the flux through the z faces are zero,
therefore, the net flux is,
Physics homework question answer, step 2, image 3
...
Final answer:
The total electric flux IS ΦE = aL2 + (a + bL)L2 + cL3 net charge is q = ε0 ( aL2 + (a + bL)L2 + cL3 )
Explanation:
The question asks to find the total electric flux ΦE through the surface of a cube with an electric field given by E = (a + bx)x + cy, and to determine the net charge q inside the cube.
Part A: Total Electric Flux through the Cube's Surface
To determine the electric flux through the cube, we need to use Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed by the surface.
The electric flux ΦE through a surface is defined by the integral of the electric field E dotted with the differential area dA over the entire surface. Since the sides of the cube are parallel to the coordinate planes, the electric field components perpendicular to the x, y, and z faces will contribute to the flux.
In this case, only the x-component of the electric field (a + bx)x contributes to the flux through the surfaces perpendicular to the x-axis, which are at x = 0 and x = L.
The y-component cy of the electric field contributes to the flux through the surfaces perpendicular to the y-axis, which are at y = 0 and y = L. There is no z-component to the electric field, so the surfaces perpendicular to the z-axis will have no flux.
The total electric flux through the cube is thus:
ΦE = ∫( (a + bx) x + cy ) · dA
For the two faces perpendicular to the x-axis:
ΦE, x=0 = ∫( a x ) · dA = aL2 (at x = 0)
ΦE, x=L = ∫( (a + bL)x ) · dA = (a + bL)L2 (at x = L)
For the two faces perpendicular to the y-axis:
ΦE, y=0 = 0 (since cy is 0 at y = 0)
ΦE, y=L = ∫( cy ) · dA = cL3 (at y = L)
Summing these up:
ΦE = aL2 + (a + bL)L2 + cL3
Part C: Net Charge Inside the Cube
The net charge q inside the cube can be found using Gauss's Law, which states "the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of free space (epsilon_0)."
q = ΦE ε0
Substituting the expression for ΦE we have:
q = ε0 ( aL2 + (a + bL)L2 + cL3 )
You are generating traveling waves on a stretched string by wiggling one end. If you suddenly begin to wiggle more rapidly without appreciably affecting the tension, you will cause the waves to move down the string a. faster than before.b. at the same speed as before.c. slower than before.
Answer:
Option as B is correct At the same speed as before
Explanation:
As we know the relation between speed of the wave and tension in string
The speed of wave in stretched string
ν = [tex]\sqrt{\frac{T}{\mu} }[/tex]
speed of wave is the directly proportional to the square root of tension as mentioned in question tension of string is unaffected when in linear mass density is constant, so we can say that the speed of wave will be the same
Option as B is correct At the same speed as before
If you suddenly begin to wiggle more rapidly without appreciably affecting the tension, you will cause the waves to move down the string at the same speed as before (Option b).
What is a wave?A wave can be defined as a type of disturbance that contains energy independently of particle motion.
The wave can move at a velocity (frequency) that is directly proportional to the tension.In this case, tension is constant, thereby velocity of the wave will remain constant.In conclusion, if you suddenly begin to wiggle more rapidly without appreciably affecting the tension, you will cause the waves to move down the string at the same speed as before (Option b).
Learn more in:
https://brainly.com/question/12215474
A point charge of -3.0X10-5 C is placed at the origin of coordinates in vacuum. Find the electric field at the point x= 5.0 m on the x axis. Determine the acceleration of a proton (q=+e, m=1.67X10-27 kg) immersed in an electric field of strength 0.50 kN/C in vacuum. How many times greater is this acceleration than that due to gravity?
a) -10,800 N/C
b) [tex]4.79\cdot 10^{10}m/s^2[/tex]
c) [tex]4.88\cdot 10^9[/tex] times g
Explanation:
a)
The magnitude of the electric field produced by a single-point charge is given by
[tex]E=\frac{kQ}{r^2}[/tex]
where
k is the Coulomb's constant
Q is the charge producing the field
r is the distance at which the field is calculated
In this problem:
[tex]Q=-3.0\cdot 10^{-5}C[/tex] is the c harge producing the field
[tex]r=5.0 m[/tex] is the distance at which we want to calculate the field
Substituting,
[tex]E=\frac{(9.0\cdot 10^9)(-3.0\cdot 10^{-5})}{(5.0)^2}=-10,800 N/C[/tex]
where the negative sign indicates that the direction of the field is towards the charge producing the field (for a negative charge, the electric field is inward, towards the charge)
b)
The force experienced by a charged particle in an electric field is given by
[tex]F=qE[/tex]
where
q is the magnitude of the charge
E is the electric field
Moreover, the force on an object can be written as:
[tex]F=ma[/tex]
where
m is the mass
a is the acceleration
Combining the two equations,
[tex]ma=qE\\a=\frac{qE}{m}[/tex]
In this problem:
[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton
[tex]E=0.50 kN/C=500 N/C[/tex] is the strength of the electric field
[tex]m=1.67\cdot 10^{-27} kg[/tex] is the mass of the proton
Substituting, we find the acceleration of the proton:
[tex]a=\frac{(1.6\cdot 10^{-19})(500)}{(1.67\cdot 10^{-27})}=4.79\cdot 10^{10}m/s^2[/tex]
c)
The acceleration due to gravity is the acceleration at which every object near the Earth's surface falls down, in absence of air resistance, and it is given by
[tex]g=9.81 m/s^2[/tex]
On the other hand, the acceleration of the proton in this problem is:
[tex]a=4.79\cdot 10^{10} m/s^2[/tex]
To find how many times greater is this acceleration than that due to gravity, we can divide the acceleration of the proton by the value of g. Doing so, we find:
[tex]\frac{a}{g}=\frac{4.79\cdot 10^{10}}{9.81}=4.88\cdot 10^9[/tex]
So, it is [tex]4.88\cdot 10^9[/tex] times greater than g.
Final answer:
The electric field at the given point is -1.08×10 N/C, which points towards the origin. The acceleration of a proton in a 0.50 kN/C electric field is 4.79×10ⁱ³ m/s². This acceleration is approximately 4.88×10ⁱ¹ times greater than the acceleration due to gravity.
Explanation:
To calculate the electric field at a point due to a point charge, we use Coulomb's law and the expression for the electric field, E = kQ/r², where k is Coulomb's constant (8.99×10⁹ N⋅m²/C²), Q is the charge, and r is the distance from the charge. Placing a charge of -3.0×10⁻⁵ C at the origin, the electric field at the point x= 5.0 m on the x-axis is calculated as follows:
E = kQ/r²
= (8.99×10⁹)(-3.0×10⁻⁵) / (5.0)²
= -1.08×10 N/C
The electric field points towards the negative charge, which is towards the origin.
To determine the acceleration of a proton in an electric field, we use the formula F = qE, where F is the force, q is the charge of the proton (1.60×10⁻⁵ C), and E is the electric field strength. The acceleration a is then found using Newton's second law, F = ma:
F = qE
= (1.60×10⁻⁵)(0.50×10)
= 8.00×10⁻⁴ N
a = F/m
= 8.00×10⁻⁴ / 1.67×10⁻⁷
= 4.79×10ⁱ³ m/s²
To find how many times greater this is compared to the acceleration due to gravity, we simply divide the proton's acceleration by the acceleration due to gravity (9.81 m/s²).
Number of times greater = a / g
= 4.79×10ⁱ³ / 9.81
=
≈ 4.88×10ⁱ¹ times
If a trapeze artist rotates 1 each second while sailing through the air, and contracts to reduce her rotational inertia to 0.40 of what it was, how many rotations per second will result?
Explanation:
A trapeze is rotating with 1 rotation per second .
Thus its angular velocity ω = 2π n
here n is the number of rotations per second
Thus ω = 2π b because n = 1 in this case
Suppose the moment of inertia of his is = I
Then angular momentum L₁ = I ω = 2 I π
In the second case , the moment of inertia becomes = 0.4 I
Let his angular velocity is ω₀
Thus angular momentum L₂ = 0.4 I ω₀
Because no external torque is applied , therefore angular momentum will remain constant .
Thus L₁ = L₂
Therefore 2 I π = 0.4 I x 2 n₀ π
here n₀ is the number of rotations per second
n₀ = [tex]\frac{1}{0.4}[/tex] = [tex]\frac{5}{2}[/tex] = 2.5
What are (a) the x component and (b) the y component of a vector in the xy plane if its direction is 259° counterclockwise from the positive direction of the x axis and its magnitude is 5.4 m?
Answer:
|Ax| =1.03 m (directed towards negative x-axis)
|Ay|= 5.30 (directed towards negative y-axis)
Explanation:
Let A is a vector = 5.4 m
θ = 259°
to Find Ax, Ay
Sol:
according the condition it lies in 3rd quadrant
we know that Horizontal Component Ax = A cos θ
Ax = 5.4 Cos 259°
Ax = - 1.03 m
|Ax| =1.03 m (directed towards negative x-axis)
Now Ay = A sin θ
Ay = 5.4 Sin 259°
Ay = -5.30
|Ay|= 5.30 (directed towards negative y-axis)
(a) -1.030m
(b) -5.301m
Explanation:Given a vector F in the xy plane, of magnitude F and in a direction θ counterclockwise from the positive direction of the x-axis;
The x-component ([tex]F_{X}[/tex]) of vector F is given by;
[tex]F_{X}[/tex] = F cos θ ---------------------(i)
And;
The y-component ([tex]F_{Y}[/tex]) of vector F is given by;
[tex]F_{Y}[/tex] = F sin θ -----------------------(ii)
Now to the question;
Let the vector be A
Therefore;
The magnitude of vector A is A = 5.4m
The direction θ of A counterclockwise from the positive direction of the x-axis = 259°
(a) The x-component ([tex]A_{X}[/tex]) of the vector A is therefore given by;
[tex]A_{X}[/tex] = A cos θ ------------------------(iii)
Substitute the values of θ and A into equation (iii) as follows;
[tex]A_{X}[/tex] = 5.4 cos 259°
[tex]A_{X}[/tex] = 5.4 x (-0.1908)
[tex]A_{X}[/tex] = -1.030
Therefore, the x-component of the vector is -1.030m
(b) The y-component ([tex]A_{Y}[/tex]) of the vector A is therefore given by;
[tex]A_{Y}[/tex] = A sin θ ------------------------(iv)
Substitute the values of θ and A into equation (iv) as follows;
[tex]A_{Y}[/tex] = 5.4 sin 259°
[tex]A_{Y}[/tex] = 5.4 x (-0.9816)
[tex]A_{Y}[/tex] = -5.301m
Therefore, the y-component of the vector is -5.301m
An airplane is flying horizontally with a speed of 103 km/hr (278 m/s) when it drops a payload. The payload hits the ground 30 s later. (Neglect air drag and the curvature of the Earth. Take g = 10 m/s².)
At what altitude H is the airplane flying?
Answer:
H = 4500 m
Explanation:
Once dropped, the payload moves along a trajectory, that can be decomposed along two directions independent each other.Just by convenience, we choose these directions to be coincident with the horizontal (-x) and vertical (y) axes.As both movements are independent each other due to both are perpendicular, in the vertical direction, the initial speed is 0.So, in order to find the vertical displacement at any point in time, we can use the following kinematic equation, where a=-g., and H = -Δy.[tex]H = \frac{1}{2}*g*t^{2} = \frac{1}{2} * 10 m/s2*(30s)^{2} = 4500 m[/tex]
The airpane was flying at a 4500 m altitude.At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. The spring with a force constant 37.0 N/cm and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 66.0 kg are pushed against the other end, compressing the spring 0.370 m. The sled is then released with zero initial velocity.
What is the sled's speed when the spring is still compressed 0.180 m?
Answer:
2.2 m/s
Explanation:
solution:
To calculate change in stored energy at desired extension
ΔU = 1/2*k*(δx)^2
= 1/2*3700*(0.37^2-0.180^2)
= 201 N.m
use work energy theorem
ΔU = ΔK = 1/2*m*v^2 = 201
= 2.2 m/s
note:
calculation maybe wrong but method is correct.
Explanation:
Below is an attachment containing the solution.
Describe what changes you would expect to see in a plot of absorbance versus temperature in an optical melting experiment on a self-complementary duplex.
Answer:
#Check explanation below for a detailed description.
Explanation:
A plot of absorbance vs temperature is used to determine [tex]T_m[/tex] of a duplex DNA.
-The duplex DNA denatures.
-The denaturation creates two single strands which enhances Ultraviolet absorption ( increased temperatures increases absorption rates).
-Higher concentrations of [tex]Sodium \ Chloride[/tex] hampers stability of the DNA.