Answer:
13 sec
Explanation:
Hello
by definition the acceleration is a vector derived magnitude that indicates the speed variation per unit of time
[tex]a=\frac{V_{f}-V_{i}}{t_{f}-t_{i}}\\a(t_{f}-t_{i})={V_{f}-V_{i} }}\\t_{f}-t_{i}=\frac{V_{f}-V_{i}}{a} \\\\t_{f}=\frac{V_{f}-V_{i}}{a}+t_{i} \\\\[/tex]
Let
Vi=4.0 m/s
a=1.0 m/s2
Vf=17.0 m/s
ti= (0)sec
t2= unknown
[tex]t_{f}=\frac{V_{f}-V_{i} }{a}+t_{i}\\t_{f}=\frac{17-4 }{1}+0\\t_{f}=\ 13\ sec\\[/tex]
Answer: 13 sec
I hope it helps
At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 4.1 m/s, and 65-kg person feels a 455-N force pressing against his back. What is the radius of a chamber?
Answer:
[tex]r=2.4m[/tex]
Explanation:
We have to use the centripetal force equation
[tex]Fc=\frac{mv^{2} }{r}[/tex]
we need the radious so we have to isolate "r" and we get
[tex]r=\frac{mv^{2} }{Fc}[/tex]
replacing m=65 kg, v= 4.1 m/s and Fc=455N we get
[tex]r=\frac{65*4.1^{2} }{455}[/tex]
[tex]r=2.4m[/tex]
The radius of the amusement park chamber is 2.4m
Two objects, with masses m1 and m2, are originally a distance r apart, and the magnitude of the gravitational force on each one is F. The objects are moved a distance 2r apart. The new gravitational force: A) decreases 2 times
B) increases 2 times
C) decreases 4 times D) remains unchanged
E) increases 4 times
Explanation:
It is given that, two objects, with masses m₁ and m₂, are originally a distance r apart, and the magnitude of the gravitational force on each one is F. If the objects are moved a distance 2 r apart, we need to find the new gravitational force.
The gravitational force is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]...........(1)
Let the force is F' when the objects are moved a distance 2 r apart. So,
[tex]F=G\dfrac{m_1m_2}{(2r)^2}[/tex]
[tex]F'=\dfrac{1}{4}G\dfrac{m_1m_2}{r^2}[/tex]
[tex]F'=\dfrac{1}{4}F[/tex] (from equation (1))
So, it is clear that the magnitude of new force is F/4 i.e. decreases by a factor is 1/4. Hence, this is the required solution.
A1.0µF capacitor has a potential difference of 6.0 V applied across its plates. If the potential difference acro its plates is increased to 8.0 V, how much additional energy does the capacitor store? If the dielectric constant is changed from 1 to 76.5 how does this change the amount of charge stored on the capacitor plates?
Answer:
14 x 10⁻⁶ J
1377 x 10⁻⁶ J
Explanation:
C = Capacitance of the capacitor = 1 x 10⁻⁶ F
ΔV = Original potential difference across the plates = 6.0 Volts
U₀ = Original energy stored in the capacitor
Original energy stored in the capacitor is given as
U₀ = (0.5) C ΔV² eq-1
ΔV' = Potential difference across the plates after increase = 8.0 Volts
U'₀ = New energy stored in the capacitor
New energy stored in the capacitor is given as
U'₀ = (0.5) C ΔV'² eq-2
U = Additional energy stored
Additional energy stored by the capacitor is given as
U = U'₀ - U₀
U = (0.5) C ΔV'² - (0.5) C ΔV²
U = (0.5) (1 x 10⁻⁶) (8)² - (0.5) (1 x 10⁻⁶) (6)²
U = 14 x 10⁻⁶ J
[tex]k_{final}[/tex] = final dielectric constant = 76.5
[tex]k_{initial}[/tex] = initial dielectric constant = 1
Energy stored in the capacitor is directly proportional to the dielectric constant, hence increase in the energy is given as
[tex]U_{inc}=(k_{final} - k_{initial})U_{o}[/tex]
Original energy stored in the capacitor is given as
U₀ = (0.5) C ΔV² = (0.5) (1 x 10⁻⁶) (6)² = 18 x 10⁻⁶ J
[tex]U_{inc}=(k_{final} - k_{initial})U_{o}[/tex]
[tex]U_{inc} = (76.5 - 1)(18\times 10^{-6})[/tex]
[tex]U_{inc} = 1377\times 10^{-6})[/tex]
The amount of charge stored on the capacitor plates increases by a factor of [tex]$76.5}$[/tex].
The energy stored in a capacitor is given by the formula:
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
where [tex]$E$[/tex] is the energy, [tex]$C$[/tex] is the capacitance, and [tex]$V$[/tex] is the potential difference.
Initially, the energy stored in the capacitor with a potential difference of 6.0 V is:
[tex]\[ E_1 = \frac{1}{2} \times 1.0 \times 10^{-6} \times (6.0)^2 \][/tex]
[tex]\[ E_1 = \frac{1}{2} \times 1.0 \times 10^{-6} \times 36 \][/tex]
[tex]\[ E_1 = 18 \times 10^{-6} \text{ J} \][/tex]
[tex]\[ E_1 = 18 \text{ µJ} \][/tex] µJ
After increasing the potential difference to 8.0 V, the energy stored in the capacitor is:
[tex]\[ E_2 = \frac{1}{2} \times 1.0 \times 10^{-6} \times (8.0)^2 \][/tex]
[tex]\[ E_2 = \frac{1}{2} \times 1.0 \times 10^{-6} \times 64 \][/tex]
[tex]\[ E_2 = 32 \times 10^{-6} \text{ J} \][/tex]
[tex]\[ E_2 = 32 \text{ µJ} \][/tex] µJ
The additional energy stored is the difference between [tex]$E_2$[/tex] and [tex]$E_1$[/tex]:
[tex]\[ \Delta E = E_2 - E_1 \][/tex]
[tex]\[ \Delta E = 32 - 18 \][/tex]
[tex]\[ \Delta E = 14 \text{ µJ} \][/tex] µJ
Now, let's consider the effect of changing the dielectric constant from 1 to 76.5. The capacitance of a capacitor with a dielectric material is given by:
[tex]\[ C = \frac{k \epsilon_0 A}{d} \][/tex]
Since the dielectric constant [tex]$k$[/tex] is the only quantity changing, the new capacitance [tex]$C'$[/tex] is:
[tex]\[ C' = kC \][/tex]
where [tex]$C$[/tex] is the original capacitance with [tex]$k = 1$[/tex].
Therefore, [tex]$C' = 76.5 \times 1.0 \times 10^{-6} \text{ F}$[/tex].
The charge stored on a capacitor is given by:
[tex]\[ Q = CV \][/tex]
With the new capacitance and the same potential difference of 8.0 V, the new charge [tex]$Q'$[/tex] is:
[tex]\[ Q' = C'V \][/tex]
[tex]\[ Q' = 76.5 \times 1.0 \times 10^{-6} \times 8.0 \][/tex]
[tex]\[ Q' = 612 \times 10^{-6} \text{ C} \][/tex]
The original charge [tex]$Q$[/tex] with [tex]$k = 1$[/tex] and [tex]$V = 8.0 \text{ V}$[/tex] was:
[tex]\[ Q = C \times V \][/tex]
[tex]\[ Q = 1.0 \times 10^{-6} \times 8.0 \][/tex]
[tex]\[ Q = 8 \times 10^{-6} \text{ C} \][/tex]
The ratio of the new charge to the original charge is:
[tex]\[ \frac{Q'}{Q} = \frac{612 \times 10^{-6}}{8 \times 10^{-6}} \][/tex]
[tex]\[ \frac{Q'}{Q} = 76.5 \][/tex]
The focal length of a planar-convex thin lens in air is 250.0 cm. The glass it is made of has an index of 1.530. Determine the radii of curvature of its surfaces. What would happen to the radii if n was reduced to 1.500?
The radii of curvature can be determined using the Lens Maker's Equation. For a planar-convex lens, we can consider one surface as flat and another as curved. If the refractive index decreases, the radius of curvature would increase.
Explanation:To find the radii of curvature for the planar-convex thin lens in air, you can use the Lens Maker's Equation, which is 1/f = (n-1)(1/R1 - 1/R2). Here, f is the focal length, n is the refractive index of the glass, R1 and R2 are the radii of curvature for the two surfaces of the lens.
For a planar-convex lens, one surface is flat (which is the planar side) and another surface is curved (which is the convex side). So, we can consider R1 = ∞ for the flat surface and R2 = R (the required radius) for the convex surface. By substituting these values into the Lens Maker's Equation, we can solve for the radius of curvature of the convex surface.
If n was reduced to 1.500, the radius of curvature would increase because, according to the Lens Maker's Equation, radius of curvature is inversely proportional to (n-1). Thus, as n decreases, the radius of curvature increases.
Learn more about Lens Maker's Equation here:https://brainly.com/question/2264507
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A hoop of mass and radius and a solid cylinder of mass and radius are rolling along level ground with the same translational speed. Which object has the greater total kinetic energy?
Answer:
the kinetic energy of hoop will be more than kinetic energy of solid cylinder
Explanation:
kinetic energy of rolling is given as
[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/tex]
here we know that for pure rolling we have
[tex]v = R \omega[/tex]
now we also know that
[tex]I = mk^2[/tex]
here k = radius of gyration
now we have
[tex]KE = \frac{1}{2}mv^2( 1 + \frac{k^2}{R^2})[/tex]
now we know that for
hoop
[tex]\frac{k^2}{R^2} = 1[/tex]
for Solid cylinder
[tex]\frac{k^2}{R^2} = \frac{1}{2}[/tex]
now the kinetic energy of hoop will be more than kinetic energy of solid cylinder
A flat disk 1.0 m in diameter is oriented so that the plane of the disk makes an angle of π/6 radians with a uniform electric field. If the field strength is 487.0 N/C, find the electric flux through the surface. A) 1 10m N·m2/C B) 420 N·m2/C C) 61 π N·m2/C D) 1700/T N·m2/C
Final answer:
The electric flux through a surface can be calculated using the formula Φ = EA cos θ, where Φ is the electric flux, E is the electric field strength, A is the area of the surface, and θ is the angle between the electric field and the normal to the surface. In this question, the electric flux through the given disk can be calculated using the given values and the formula. The answer is 61π N·m²/C, which corresponds to option C.
Explanation:
The electric flux through a surface can be calculated using the formula:
Φ = EA cos θ
where Φ is the electric flux, E is the electric field strength, A is the area of the surface, and θ is the angle between the electric field and the normal to the surface.
In this question, the electric field strength is given as 487.0 N/C, the diameter of the disk is 1.0 m (radius = 0.5 m), and the angle between the plane of the disk and the electric field is π/6 radians.
Using the formula, we can calculate the electric flux as follows:
Φ = (487.0 N/C)(π * (0.5 m)^2) * cos(π/6)
Φ = 61π Nm²/C
Therefore, the electric flux through the surface is 61π N·m²/C, which corresponds to option C.
A wire carries a steady current of 2.60 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uniform magnetic field, vector B = 1.50 k T. If the current is in the positive x direction, what is the magnetic force on the section of wire?
Answer:
The magnetic force on the section of wire is [tex]-2.925\hat{j}\ N[/tex].
Explanation:
Given that,
Current [tex]I = 2.60\hat{i}\ A[/tex]
Length = 0.750 m
Magnetic field [tex]B = 1.50\hat{k}\ T[/tex]
We need to calculate the magnetic force on the section of wire
Using formula of magnetic force
[tex]\vec{F}=l\vec{I}\times\vec{B}[/tex]
[tex]\vec{F}=0.750\times2.60\hat{i}\times1.50\hat{k}[/tex]
Since, [tex]\hat{i}\times\hat{k}=-\hat{j}[/tex]
[tex]\vec{F}=-2.925\hat{j}\ N[/tex]
Hence, The magnetic force on the section of wire is [tex]-2.925\hat{j}\ N[/tex].
Does resistance in a circuit remain constant if the temperature changes?
Answer:
No the resistance of a given circuit does not remain constant if the temperature of the circuit changes.
Explanation:
The resistance of any resistor used in a circuit depends upon the temperature of that resistor. This can be mathematically represented as follows
[tex]R(t)=R_{0}(1+\alpha \Delta t)[/tex]
Where,
R(t) is resistance of any resistor at temperature t
[tex]R_{o}[/tex] is the resistance of the resistor at time of fabrication
α is temperature coefficient of resistivity it's value is different for different materials
This change in the resistance is the cumulative effect of:
1) Variation of resistivity with temperature
2) Change in dimensions of the resistor with change in temperature
Final answer:
Resistance in a circuit changes with temperature due to increased atomic vibrations affecting electron movement.
Explanation:
Resistance in a circuit does not remain constant if the temperature changes. As temperature increases, the resistance of a conductor typically increases due to the atoms vibrating more rapidly, causing more collisions for the electrons passing through.
This change in resistance with temperature is a common phenomenon seen in various materials. It generally increases with increasing temperature due to more frequent electron collisions within the conductor.
Ohm's Law, which states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, assumes constant temperature.
In practical terms, if you were to graph resistance against temperature, for some materials, you would notice a linear increase in resistance for small temperature changes, while for large changes, the relationship can be nonlinear.
(a) What is the acceleration of gravity on the surface of the Moon? The mass of the moon is 7.35 x 1022 kg and its radius is 1.74 x 106 m. (b) What is the acceleration of gravity on the surface (or outer limit) of Mercury? The mass of Mercury is 3.30 x 1023 kg and its radius is 2.44 x 106 m
Answer:
Part a)
a = 1.62 m/s/s
Part b)
a = 3.70 m/s/s
Explanation:
Part A)
Acceleration due to gravity on the surface of moon is given as
[tex]a = \frac{GM}{R^2}[/tex]
here we know that
[tex]M = 7.35 \times 10^{22} kg[/tex]
[tex]R = 1.74 \times 10^6 m[/tex]
now we have
[tex]a_g = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{(1.74 \times 10^6)^2}[/tex]
[tex]a_g = 1.62 m/s^2[/tex]
Part B)
Acceleration due to gravity on surface of Mercury is given as
[tex]a = \frac{GM}{R^2}[/tex]
here we know that
[tex]M = 3.30 \times 10^{23} kg[/tex]
[tex]R = 2.44 \times 10^6 m[/tex]
now we have
[tex]a_g = \frac{(6.67 \times 10^{-11})(3.30 \times 10^{23})}{(2.44 \times 10^6)^2}[/tex]
[tex]a_g = 3.70 m/s^2[/tex]
You measure the pressure at 10 feet in a column of water and determine it to be 43.36 lbs/in. What would this measurement be in N/m2? 18.2 N/m2 299,034 N/m2 ) 187,334 N/m2 19.9 N/m 2
Answer:
299034 N/m²
Explanation:
1 lbs = 4.448 N
1 in = 0.0254 m
1 in² = 0.254² m²
thus,
[tex]1\frac{lbs}{in^2} = \frac{4.448N}{0.0254^2m^2}=6894.413N/m^2[/tex]
therefore,
43.36lbs/in² in N/m² will be
= 43.36 × 6894.413
= 298941.77 N/m² ≈ 299034 N/m²
so the correct option is 299034 N/m²
An old millstone, used for grinding grain in a gristmill, is a solid cylindrical wheel that can rotate about its central axle with negligible friction. The radius of the wheel is 0.330 m. A constant tangential force of 200 N applied to its edge causes the wheel to have an angular acceleration of 0.936 rad/s2. (a) What is the moment of inertia of the wheel (in kg · m2)?
Answer:
The answer is I=70,513kgm^2
Explanation:
Here we will use the rotational mechanics equation T=Ia, where T is the Torque, I is the Moment of Inertia and a is the angular acceleration.
When we speak about Torque it´s basically a Tangencial Force applied over a cylindrical or circular edge. It causes a rotation. In this case, we will have that T=Ft*r, where Ft is the Tangencial Forge and r is the radius
Now we will find the Moment of Inertia this way:
[tex]Ft*r=I*a[/tex] -> [tex](Ft*r)/(a) = I[/tex]
Replacing we get that I is:
[tex]I=(200N*0,33m)/(0,936rad/s^2)[/tex]
Then [tex]I=70,513kgm^2[/tex]
In case you need to find extra information, keep in mind the Moment of Inertia for a solid cylindrical wheel is:
[tex]I=(1/2)*(m*r^2)[/tex]
An empty beaker weighs 40.25 g. When completely filled with water, the beaker and its contents have a total mass of 317.45 g. What volume does the beaker hold? Use ????=1.00 g/mL as the density of water.
Answer:
The beaker holds 277.2 mL
Explanation:
Empty weight of beaker = 40.25 g
Weight of beaker with water = 317.45 g
Weight of water = 317.45 - 40.25 = 277.2 g
Density of water = 1 g/mL
We have
Mass = Volume x density
277.2 = Volume x 1
Volume = 277.2 mL
The beaker holds 277.2 mL
The volume of the beaker is 277.2 mL, obtained by subtracting the mass of the empty beaker from the total mass with water and using the density of water as 1.00 g/mL.
To find the volume of the beaker, we need to calculate the mass of the water it holds. First, subtract the mass of the empty beaker from the total mass with the water: 317.45 g - 40.25 g = 277.2 g. Since the density of water is 1.00 g/mL, the mass of water in grams is numerically equal to its volume in mL. Therefore, the beaker holds a volume of 277.2 mL of water.
A positively charged bead having a mass of 1.00 g falls from rest in a vacuum from a height of 5.00 m in a uniform vertical electric field with a magnitude of 1.00 ✕ 104 N/C. The bead hits the ground at a speed of 21.9 m/s.
a) Determine the direction of the electric field (up or down)
b)Determine the Charge on the bead ___ µC
Answer:
a)
down direction.
b)
3.82 µC
Explanation:
a)
Consider the motion of the positively charged bead in vertical direction
y = vertical displacement of charged bead = 5 m
a = acceleration of charged bead = ?
v₀ = initial velocity of bead = 0 m/s
v = final velocity of bead = 21.9 m/s
using the equation
v² = v₀² + 2 a y
inserting the values
21.9² = 0² + 2 a (5)
a = 47.96 m/s²
m = mass of the bead = 1 g = 0.001 kg
F = force by the electric field
Force equation for the motion of the bead in electric field is given as
mg + F = ma
(0.001) (9.8) + F = (0.001) (47.96)
F = 0.0382 N
Since the electric force due to electric field comes out to be positive, the electric force acts in down direction. we also know that a positive charge experience electric force in the same direction as electric field. hence the electric field is in down direction.
b)
q = magnitude of charge on the bead
E = electric field = 1 x 10⁴ N/C
Electric force is given as
F = q E
0.0382 = q (1 x 10⁴)
q = 3.82 x 10⁻⁶ C
q = 3.82 µC
(a) The electric field direction is down as it contributes to the increased speed of the falling bead. (b) The charge on the bead is calculated to be 3.8 µC.
Let's address the given problem step-by-step:
(a) Determine the direction of the electric field
We know that the bead is positively charged and falls from rest in a vacuum. Gravity pulls the bead downward by itself, but the bead hits the ground at a speed greater than it would under gravity alone (21.9 m/s compared to the ~9.9 m/s due to gravitational acceleration over 5.00 meters). Therefore, the electric field must be contributing additional force downward to achieve this extra speed. Thus, the electric field must be pointing down.
(b) Determine the Charge on the bead in µC
First, calculate the work done by the electric field on the bead:
Gravitational Potential Energy:
Initial PE = mgh = 1.00*10⁻³kg * 9.8 m/s² * 5.00 m = 0.049 J
Final Kinetic Energy (KE): = 1/2 * m * v² = 0.5 * 1.00*10⁻³kg * (21.9 m/s)² = 0.239 J
Total work done by the electric field:
WE = KE - PE = 0.239 J - 0.049 J = 0.19 JUsing WE = qEd, we can solve for the charge q:
q = WE / (Ed)q = 0.19 J / (1.00*10⁴ N/C * 5.00 m)q = 0.19 J / 5.00 * 10⁴ N/CConvert the charge to µC:
q = 3.8 * 10⁻⁶ Cq = 3.8 µCSo, the charge on the bead is 3.8 µC.A projectile is fired at an upward angle of 29.7° from the top of a 108-m-high cliff with a speed of 130-m/s. What will be its speed (in m/s) when it strikes the ground below?
Answer:
79.2 m/s
Explanation:
θ = angle at which projectile is launched = 29.7 deg
a = initial speed of launch = 130 m/s
Consider the motion along the vertical direction
v₀ = initial velocity along the vertical direction = a Sinθ = 130 Sin29.7 = 64.4 m/s
y = vertical displacement = - 108 m
a = acceleration = - 9.8 m/s²
v = final speed as it strikes the ground
Using the kinematics equation
v² = v₀² + 2 a y
v² = 64.4² + 2 (-9.8) (-108)
v = 79.2 m/s
A solar heating system has a 25.0% conversion efficiency; the solar radiation incident on the panels is 1 000 W/m2. What is the increase in temperature of 30.0 kg of water in a 1.00-h period by a 4.00-m2-area collector? (cw = 4 186 J/kg×°C)
Final answer:
The increase in temperature of the water is 0.796 °C.
Explanation:
To calculate the increase in temperature of the water, we first need to calculate the total energy absorbed by the collector. The power incident on the collector can be calculated by multiplying the solar radiation intensity by the collector area:
Power incident on the collector = 1000 W/m² × 4 m² = 4000 W
The energy absorbed by the collector can be calculated by multiplying the power incident on the collector by the conversion efficiency:
Energy absorbed by the collector = 4000 W × 0.25 = 1000 J/s
Now we can calculate the increase in temperature of the water using the specific heat formula:
ΔT = Energy absorbed by the collector / (mass of water × specific heat of water)
ΔT = 1000 J/s / (30 kg × 4186 J/kg×°C) = 0.796 °C
A 2170 kg space station orbits Earth at an altitude of 5.27 x 10^5 m. Find the magnitude of the force with which the space station attracts Earth. The mass and mean radius of Earth are 5.98 x 10^24 kg and 6.37 x 10^6 m, respectively.
Answer:
F = 18195.59 N or F = 18196 rounded up
Explanation:
force = GMm/d^2
G = 6.67x10^-11
M = 5.98x10^24 kg
m = 2170kg
d = 6370000 + 527000 = 6897000m
putting all values (6.67x10^-11)(5.98x10^24)(2170)/(6897000^2) = 18195.59....
The magnitude of the force with which the space station attracts Earth is about 1.82 × 10⁴ Newton
[tex]\texttt{ }[/tex]
Further explanationNewton's gravitational law states that the force of attraction between two objects can be formulated as follows:
[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]
F = Gravitational Force ( Newton )
G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )
m = Object's Mass ( kg )
R = Distance Between Objects ( m )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
mass of space station = m = 2170 kg
radius of the orbit = R = 5.27 × 10⁵ + 6.37 × 10⁶ = 6.897 × 10⁶ m/s
mass of Earth = M = 5.98 × 10²⁴ kg
Asked:
Gravitational Force = F = ?
Solution:
[tex]F = G \frac{M.m}{R^2}[/tex]
[tex]F = 6.67 \times 10^{-11} \times \frac{5.98 \times 10^{24} \times 2170}{(6.897 \times 10^6)^2}[/tex]
[tex]F \approx 1.82 \times 10^4 \texttt{ Newton}[/tex]
[tex]\texttt{ }[/tex]
Learn moreImpacts of Gravity : https://brainly.com/question/5330244Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454The Acceleration Due To Gravity : https://brainly.com/question/4189441[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Gravitational Fields
[tex]\texttt{ }[/tex]
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the axis of the loop. Somebody carefully grabs the edges of the loop and begins pulling them apart such that the area of the loop increases at a rate of 20 cm2/s. What is the magnitude of the induced EMF in the loop?
Answer:
2 x 10⁻³ volts
Explanation:
B = magnetic of magnetic field parallel to the axis of loop = 1 T
[tex]\frac{dA}{dt}[/tex] = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²
θ = Angle of the magnetic field with the area vector = 0
E = emf induced in the loop
Induced emf is given as
E = B [tex]\frac{dA}{dt}[/tex]
E = (1) (20 x 10⁻⁴ )
E = 2 x 10⁻³ volts
E = 2 mV
A recent study found that electrons that have energies between 3.45 eV and 20.9 eV can cause breaks in a DNA molecule even though they do not ionize the molecule. If a single photon were to transfer its energy to a single electron, what range of light wavelengths could cause DNA breaks?
Answer:
5.95 nm to 33.6 nm
Explanation:
Energy of a single photon can be written as:
[tex]E = \frac{hc}{\lambda}[/tex]
where, h is the Planck's constant, c is the speed of light and λ is the wavelength of light.
Consider the lowest energy of an electron that can break a DNA = 3.45 eV
1 eV = 1.6 ×10⁻¹⁹ J
⇒3.45 eV = 5.52×10⁻¹⁹ J
[tex]E = \frac{hc}{\lambda}\\ \Rightarrow \lambda = \frac{hc}{E}= \frac {6.63\times 10^{-34} m^2kg/s \times 3\times 10^8 m/s}{5.52 \times 10^{-19} J} = 3.60\times 10^{-7} m = 360 nm[/tex]
Consider the highest energy of an electron that can break a DNA = 20.9 eV
1 eV = 1.6 ×10⁻¹⁹ J
⇒20.9 eV = 33.4×10⁻¹⁹ J
[tex]E = \frac{hc}{\lambda}\\ \Rightarrow \lambda = \frac{hc}{E}= \frac {6.63\times 10^{-34} m^2kg/s \times 3\times 10^8 m/s}{33.4 \times 10^{-19} J} = 0.595\times 10^{-7} m = 59.5 nm[/tex]
The wavelength range of light that can cause DNA breaks is approximately 59.4 nm to 360 nm. This corresponds to ultraviolet light and part of the visible spectrum.
To find the range of light wavelengths that can cause DNA breaks, we need to convert the given energy range of 3.45 eV to 20.9 eV into wavelengths.
The energy of a photon (E) is related to its wavelength (λ) by the equation:
E = hc/λ
where h is Planck's constant (6.626 × 10⁻³⁴ J·s), c is the speed of light .(3.00 × 10⁸m/s), and λ is the wavelength in meters.
First, convert the energy from electron volts (eV) to Joules (J):
1 eV = 1.602 × 10⁻¹⁹ JSo, the energy range in Joules is:
3.45 eV = 3.45 × 1.602 × 10⁻¹⁹J = 5.53 × 10⁻¹⁹ J20.9 eV = 20.9 × 1.602 × 10⁻¹⁹J = 3.35 × 10⁻¹⁸ JNext, use the energy-wavelength relation to find the wavelengths:
For 3.45 eV: λ = hc/E = (6.626 × 10⁻³⁴J·s × 3.00 × 10⁸ m/s) / 5.53 × 10⁻¹⁹J ≈ 3.60 × 10⁻⁷ m ≈ 360 nmFor 20.9 eV: λ = hc/E = (6.626 × 10⁻³⁴ J·s × 3.00 × 10⁸ m/s) / 3.35 × 10⁻¹⁸ J ≈ 5.94 × 10⁻⁸ m ≈ 59.4 nmThus, the range of wavelengths that can cause DNA breaks is approximately 59.4 nm to 360 nm, corresponding to the ultraviolet (UV) and part of the visible spectrum.
The amplitude of the voltage across an inductor can be greater than the amplitude of the generator EMF in an AC RLC circuit. Consider an RLC circuit with Epeak = 90 V, R = 21.1 capital omega, L = 1.05 H, and C = 2.6 µ F. Find the amplitude of the voltage across the inductor at resonance.
Answer:
2710.66 V
Explanation:
Eo = 90 V, R = 21.1 ohm, L = 1.05 H, C = 2.6 x 10^-6 F
At resonance, the angular frequency is given by
[tex]\omega _{0}=\frac{1}{\sqrt{LC}}[/tex]
[tex]\omega _{0}=\frac{1}{\sqrt{1.05\times 2.6\times 10^{-6}}}=605.23 rad/s[/tex]
XL = ω0 x L = 605.23 x 1.05 = 635.5 ohm
Xc = 1 / ω0 C = 1 / (605.23 x 2.6 x 10^-6) = 635.5 ohm
In case of resonance, the impedance is equal to teh resistance of circuit.
Z = R = 21.1 ohm
I0 = E0 / Z = 90 / 21.1 = 4.265 A
Voltage across inductor
VL = I0 x XL
VL = 4.265 x 635.5 = 2710.66 V
To find the amplitude of the voltage across the inductor at resonance in an AC RLC circuit, we can calculate the resonant frequency using the formula fo = 1/(2*π√(LC)). Once we have the resonant frequency, we can find the amplitude of the voltage across the inductor using the formula Vl = I * XL, where XL = 2πfL.
Explanation:To find the amplitude of the voltage across the inductor at resonance, we need to calculate the resonant frequency using the formula fo = 1/(2*π√(LC)).
Substituting the given values, fo = 1/(2 * 3.1416 * √(1.05 * 2.6 * 10^-6)) Hz.
Once we have the resonant frequency, we can find the amplitude of the voltage across the inductor using the formula Vl = I * XL, where XL = 2πfL.
Substituting the values and the calculated resonant frequency, we can find the amplitude of the voltage across the inductor.
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Select True or False for the following statements about conductors in electrostatic equilibrium. All points of a conductor are at the same potential. Charges prefer to be uniformly distributed throughout the volume of a conductor. The electric field inside the conducting material is always zero. Just outside the surface of a conductor, the electric field is always zero.
Answers :
1. All points of a conductor are at the same potential. - True
2. Charges prefer to be uniformly distributed throughout the volume of a conductor. - False
3 The electric field inside the conducting material is always zero. -True
4.Just outside the surface of a conductor, the electric field is always zero. - False
a) True
b) False
c) True
d) False
What is a conductor ?
A conductor is a substance or material that allows electricity to flow through it.
a) All points of a conductor are at the same potential is True as charge distribution on the surface of the conductor is uniform
b) Charges prefer to be uniformly distributed throughout the volume of a conductor is False because all the charge comes on the surface and get distributed uniformly on the surface of the conductor and their is no charge inside the conductor
c) The electric field inside the conducting material is always zero is True
Since , all the charge is on the surface of the conductor so , there will not be any charges inside the conductor , this is why there will not be electric field .
d)Just outside the surface of a conductor, the electric field is always zero is False as due to charge on the surface there will be electric field outside the surface of conductor .
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A rocket moves upward from rest with an acceleration of 40 m/s2 for 5 seconds. It then runs out of fuel and continues to move upward. Neglecting any air resistance, how high will it rise?
Answer:
Maximum height of rocket = 2538.74 m
Explanation:
We have equation of motion s = ut + 0.5 at²
For first 5 seconds
s = 0 x 5 + 0.5 x 40 x 5² = 500 m
Now let us find out time after 5 seconds rocket move upward.
We have the equation of motion v = u + at
After 5 seconds velocity of rocket
v = 0 + 40 x 5 = 200 m/s
After 5 seconds the velocity reduces 9.8m/s per second due to gravity.
Time of flying after 5 seconds
[tex]t=\frac{200}{9.81}=20.38s[/tex]
Distance traveled in this 20.38 s
s = 200 x 20.38 - 0.5 x 9.81 x 20.38² = 2038.74 m
Maximum height of rocket = 500 +2038.74 = 2538.74 m
A driver in a car speeding to the right at 24m/s suddenly hits the brakes and goes into a skid, finally coming to rest. The coefficient of static and kinetic friction between the tires and the road are s = .8 and k = 0.7. How far does the car skid?
Answer:
42 m
Explanation:
[tex]v_{o}[/tex] = initial velocity of the car = 24 m/s
[tex]v_{f}[/tex] = final velocity of the car = 0 m/s
μ = coefficient of kinetic friction = 0.7
g = acceleration due to gravity = 9.8 m/s²
a = acceleration due to kinetic frictional force = - μg = - (0.7)(9.8) = - 6.86 m/s²
d = distance through which the car skids
Using the kinematics equation
[tex]v_{f}^{2} = v_{o}^{2} + 2 a d[/tex]
Inserting the values
[tex]0^{2} = 24^{2} + 2 (- 6.86) d[/tex]
d = 42 m
For what position of the object will a spherical concave mirror project on the screen an image smaller than the object? a. between focus and center
b. between the focus and the mirror
c. at the center of curvature
d. beyond the center of curvature
Answer:
option (d)
Explanation:
A concave mirror always forms a real and inverted image of an object except when the object placed between pole and focus of the mirror.
When the object is placed beyond the centre of curvature, it forms a image which is smaller than the object but it is real and inverted in nature.
The position for which a concave mirror projects a screen image smaller than the object is when the object is placed beyond the center of curvature.
Explanation:For what position of the object will a spherical concave mirror project on the screen an image smaller than the object? The correct answer is d. beyond the center of curvature. When an object is placed beyond the center of curvature, the concave mirror forms a real, inverted image that is reduced in size, or smaller than the object itself.
This effect can be understood through ray diagrams where rays travelling parallel to the axis, striking the center of the mirror, and moving toward the focal point, all converge to form an image between the focal point and the center of curvature of the mirror. However, if the object is placed closer to the mirror, such as between the focus and the mirror, the produced image would be larger than the object.
Does the initial velocity of an object have anything to do with its acceleration? For example, compared to dropping an object, if you throw it downward would the acceleration be different after you released it?
Answer:
Explanation:
This is an excellent question to get an answer for. It teaches you much about the nature of physics.
The answer is no.
The distance will be quite different. The time might be different in getting to the distance. But the acceleration will be the same in either case.
How do you know? Look at one of the formulas, say
d = vi * t + 1/2*a * t^2
What does vi do? vi will alter both t and d. if vi = 0 then both d and/or t will be found. But what will "a" do? Is there anything else acting in the up or down line of action? You should answer no.
If vi is not zero, t will be less and d will take less time to get where it is going.
Yes, the initial velocity of an object affects its acceleration. If you drop an object, the initial velocity is zero, resulting in a constant acceleration solely due to gravity. If you throw the object downward, the initial velocity adds to the acceleration due to gravity, leading to a higher overall acceleration.
Explanation:Yes, the initial velocity of an object can affect its acceleration. When you drop an object, its initial velocity is zero, so the acceleration is solely due to gravity and is constant at approximately 9.8 m/s2 (assuming no air resistance). However, if you throw an object downward, it already has an initial downward velocity, which adds to the acceleration due to gravity. So, the acceleration of the object after you release it will be greater than if you had dropped it.
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16 A 20-pewton force daurected west and a 5 newton force directed north act concurently on a 5 kg object. Draw the resultant vector of the two forces, calculate its magnitude, and determine the acceleration of the object
Answer:
20.62 N
4.123 m/s^2
Explanation:
A = 20 N west
B = 5 N North
m = 5 kg
Both the forces acting at right angle
Use the formula of resultant of two vectors.
Let r be the magnitude of resultant of two vectors.
[tex]R = \sqrt{A^{2} + B^{2} + 2 A B Cos\theta}[/tex]
[tex]R = \sqrt{20^{2} + 5^{2} + 2 \times 20 \times 5 \times Cos90}[/tex]
R = 20.62 N
Let a be the acceleeration.
a = Net force / mass = R / m = 20.62 / 5
a = 4.123 m/s^2
What is an Exothermic Reaction? What is an Endothermic Reaction?
Answer:
Answer to the question:
Explanation:
Exothermic Reaction
It is called an exothermic reaction to any chemical reaction that releases energy, either as light or heat, or what is the same: with a negative variation of enthalpy; that is to say: ΔH < 0. Therefore it is understood that exothermic reactions release energy.
Endothermic Reaction
It is called an endothermic reaction to any chemical reaction that absorbs energy, usually in the form of heat.
If we talk about enthalpy (H), an endothermic reaction is one that has a variation of enthalpy ΔH> 0. That is, the energy possessed by the products is greater than that of the reagents.
A square wire, 20 cm on the side, moves at constant speed parallel to the xy-plane into a region where there is a uniform magnetic field = (0.1 T) .The induced electric current in the coil is 10 mA and its resistance is 10Ω. What is the speed of the wire?
Answer:
The speed of the wire is 5 m/s.
Explanation:
Given that,
Length = 20 cm
Magnetic field = 0.1 T
Current = 10 mA
Resistance [tex]R= 10\Omega[/tex]
We need to calculate the speed of the wire
Using formula of emf
[tex]\epsilon=Bvl[/tex]
Using formula of current
[tex]I=\dfrac{\epsilon}{R}[/tex]
Put the value of [tex]\epsilon[/tex] into the formula of current
[tex]I=\dfrac{Bvl}{R}[/tex]
[tex]v=\dfrac{IR}{Bl}[/tex]
[tex]v=\dfrac{10\times10^{-3}\times10}{0.1\times20\times10^{-2}}[/tex]
[tex]v= 5\ m/s[/tex]
Hence, The speed of the wire is 5 m/s.
The absolute potential at a distance of 2.0 m from a negative point charge is-100. V. What is the absolute potential 4.0 m away from the same point charge? A)-50. V B) -25. V C)-0.40 kV D)-020 kV E)-0.50 kV
Answer:
The absolute potential 4.0 m away from the same point charge is -50 V.
(A) is correct option.
Explanation:
Given that,
Distance = 2.0 m
Potential = -100 V
Absolute potential = 4.0 m
We need to calculate the charge
Using formula of potential
[tex]V=\dfrac{kq}{r}[/tex]
Where, V = potential
q = charge
r = distance
Put the value into the formula
[tex]-100=\dfrac{9\times10^{9}\times q}{2.0}[/tex]
[tex]q=\dfrac{200}{9\times10^{9}}[/tex]
[tex]q=-22.2\times10^{-9}\ C[/tex]
We need to calculate the potential
Using formula of potential
[tex]V=\dfrac{9\times10^{9}\times(-22.2\times10^{-9})}{4.0}[/tex]
[tex]V=-49.95\ V[/tex]
[tex]V=-50\ V[/tex]
Hence, The absolute potential 4.0 m away from the same point charge is -50 V.
Charge 1 of +4 micro-coulombs is placed at the origin, charge 2 of +18 micro-coulombs is placed at x = +0.32 m, y = -0.59 m, charge 3 of -2 micro-coulombs is placed at x = -0.79 m, y = 0 m. Measured counter-clockwise in degrees, what is the angle of the total electric force vector on charge 1?
CORRECT ANSWER: 122.36
Answer:
122.36
Explanation:
The distance (d) between Charge 1 and 2 can be calculated as:
[tex]d=\sqrt{0.32^2+0.59^2}=0.67m[/tex]
The force between them is given as
[tex]F_1=\frac{1}{4\pi \epsilon_0}\frac{4*18*10^-12}{0.67^2}= 1.44N[/tex]
The angle of this force with positive x-axis is given as
[tex]\theta_1=90^{\circ}+\tan^{-1}\frac{0.32}{0.59}=118.47^{\circ}[/tex]
Now,
The force between 1 and 3 is
[tex]F_2=\frac{1}{4\pi \epsilon_0}\frac{4*2*10^{-12}}{0.79^2}= 0.115N[/tex]
As the force is attractive it is along negative x direction so the angle will be given as = [tex]\theta_2 = 180^{\circ}[/tex]
So the negative x component of the resultant force will be calculated as
= [tex]1.44\cos(180-118.47)^{\circ}+0.115=0.801[/tex]
And the positive y component = [tex]1.44\sin(180-118.47)^{\circ}=1.26[/tex]
So the angle of the resultant with positive x axis will be
[tex]90^{\circ}+\tan^{-1}\frac{0.801}{1.26}=122.36^{\circ}[/tex]
If a material has an index of refraction of 1.61, Determine the speed of light through this medium
Answer:
1.86 x 10^8 m/s
Explanation:
n = 1.61
The formula for the refractive index is given by
n = speed of light in vacuum / speed of light in material
n = c / v
v = c / n
v = (3 x 10^8) / 1.61
v = 1.86 x 10^8 m/s
The speed of light in a material with an index of refraction of 1.61 is calculated as approximately 1.86 * 10^8 m/s, using the equation v = c/n where c is the speed of light in vacuum and n is the index of refraction.
Explanation:The speed of light in a given material can be calculated using the index of refraction of the material, as defined by the equation n = c/v, where n is the index of refraction, c is the speed of light in a vacuum, and v is the speed of light in the material.
Given that the index of refraction for the material in question is 1.61, and the speed of light in vacuum, c = 3.00 * 10^8 m/s, the speed of light v in this medium would therefore be calculated by rearranging the equation to v = c/n.
By substituting the given values into the equation, v = 3.00 * 10^8 m/s / 1.61, we find that the speed of light in the material is approximately 1.86 * 10^8 m/s.
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