A 2.50-kg textbook is forced against a horizontal spring of negligible mass and force constant 250 N/m, compressing the spring a distance of 0.250 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \mu_kμ k ​  = 0.30. Use the work–energy theorem to find how far the text-book moves from its initial position before coming to rest.

Answer:

3888.23 N

Explanation:

m = mass of the person = 78.03 kg

g = standard acceleration due to gravity = 8.23 m/s²

a = acceleration of the aircraft = 5.05 g = 5.05 x 8.23 = 41.6 m/s²

F = upward force the person experience

Force equation for the motion of person is given as

F - mg = ma

Inserting the values

F - (78.03)(8.23) = (78.03)(41.6)

F = 3888.23 N

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.

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]

Newton'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]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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.

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

Answer : The atomic weight of the element is, 121.75 amu

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]

As we are given that,

Mass of isotope 1 = 120.9038 amu

Percentage abundance of isotope 1 = 57.25 %

Fractional abundance of isotope 1 = 0.5725

Mass of isotope 2 = 122.8831 amu

Percentage abundance of isotope 2 = 100 - 57.25 = 42.75 %

Fractional abundance of isotope 2 = 0.4275

Now put all the given values in above formula, we get:

[tex]\text{Average atomic mass of element}=\sum[(120.9038\times 0.5725)+(122.8831\times 0.4275)][/tex]

[tex]\text{Average atomic mass of element}=121.75amu[/tex]

Therefore, the atomic weight of the element is, 121.75 amu

The atomic weight of an element with two isotopes of 120.9038 amu at 57.25% abundance and 122.8831 amu is calculated as 121.75 amu after rounding to two decimal places.

The atomic weight of an element with two isotopes can be calculated using the weighted average formula. This takes into account the atomic masses of the isotopes and their relative abundances.

Since one isotope has an atomic weight of 120.9038 amu with 57.25% abundance and we know the atomic weight of the other isotope is 122.8831 amu, we can calculate the average atomic mass as follows:

The percent abundance of the second isotope must be 100% - 57.25% = 42.75%, since the two abundances must add up to 100%.

Atomic weight = (isotope1_mass * isotope1_abundance) + (isotope2_mass * isotope2_abundance)

Substituting our values in gives us:
Atomic weight = (120.9038 amu * 0.5725) + (122.8831 amu * 0.4275)

Atomic weight = (69.1866935 amu) + (52.53009825 amu)

Atomic weight = 121.71679175 amu

After rounding to two decimal places, the average atomic weight of the element is 121.72 amu which is closest to option 4, 121.75 amu.

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.

Answer:

[tex]N = 3[/tex]

so these are

either 3 to 2 or 2 to 1 or 3 to 1

Explanation:

As we know that total number photons emitted from nth state to ground state is given by

[tex]N = ^nC_2[/tex]

[tex]N = \frac{n(n-1)}{2}[/tex]

here it shows that if electron has transition from nth excited state to ground state then total number of photons is given by above equation

here we know that higher state is n = 3

so total number of photons is given as

[tex]N = \frac{3(3-1)}{2}[/tex]

[tex]N = 3[/tex]

so these are

either 3 to 2 or 2 to 1 or 3 to 1

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:

Using WE = qEd, we can solve for the charge q:

Convert the charge to µC:

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

Answer:

Centripetal force, F = 18486.75 N

Explanation:

It is given that,

Mass of the sensor, m = 6 kg

It is attached at the end of 100 m long wind turbine, d = 100 m

So, the radius of the wind turbine, r = 50 m

Angular velocity, [tex]\omega=1.25\ rev/s=7.85\ rad/s[/tex]

The centripetal force is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

Since, [tex]v=r\omega[/tex]

[tex]F=mr\omega^2[/tex]

[tex]F=6\ kg\times 50\ m\times (7.85\ rad/s)^2[/tex]

F = 18486.75 N

So, the centripetal force is 18486.75 N. Hence, this is the required solution.

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.

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].

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

Answer:

6.6 x 10^5 Nm^2/C

Explanation:

E = 125000 N/C

Area, A = length x width = 2.5 x 5 = 12.5 m^2

θ = 65 degree

electric flux, φ = E A Cosθ

φ = 125000 x 12.5 x Cos 65

φ = 6.6 x 10^5 Nm^2/C

Here, we are required to evaluate the electric flux through the rectangle.

The electric flux through the rectangle is;

φ = 6.6 × 10^5 Nm²/C

The electric flux through a body is given as;

electric flux, φ = E A Cosθ

where;.

The Area of the rectangle, A = 2.5 × 5

Therefore, A = 12.5m².

Therefore, the electric flux, φ = E A Cosθ

φ = 125000 × 12.5 × Cos65

φ = 660,341 Nm²/C.

φ = 6.6 × 10^5 Nm²/C

Read more:

https://brainly.com/question/24290406

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

Hey there!

a) The electric field is in direction of decreasing value of potential so the electric field here will points towards 0v from 9 v. But the charge on particle is negative so the force will be towards the 9v point. and so the charge will move towards the 9 V point.

b) Electric potential is a location-dependent quantity that expresses the amount of potential energy per unit of charge at a specified location while.the electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy. Electric potential is always a relative quantity because no one can find out absolute potential at a point, so in reality only potential difference exists. We can assume potential at certain point (say at infinity it is zero) only then we are able to define potential at every point.

c) The knowlendge of electric potential helps us in finding the work done on the particle which is used in work energy theorem to find out the chanrge in kinetic energy and other stuff.

Hope this helps!

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

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.

Answer:

Magnitude of the net force acting on the kayak = 39.61 N

Explanation:

Considering motion of kayak:-

Initial velocity, u =  0 m/s

Distance , s = 0.40 m

Final velocity, v = 0.65 m/s

We have equation of motion v² = u² + 2as

Substituting

   v² = u² + 2as

    0.65² = 0² + 2 x a x 0.4

    a = 0.53 m/s²

We have force, F  = ma

Mass, m = 75 kg

    F  = ma = 75 x 0.53 = 39.61 N

Magnitude of the net force acting on the kayak = 39.61 N

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]

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]

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.

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.

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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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):

So, the energy range in Joules is:

Next, use the energy-wavelength relation to find the wavelengths:

Thus, 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.

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.

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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Answer: a) [tex]^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n[/tex]

b) [tex]^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}[/tex]

Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

a) The given reaction is [tex]^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + x^{1}_{0}n[/tex]

Now,  as the mass on both reactant and product side must be equal:

[tex]235+1=87+146+x[/tex]

[tex]x=3[/tex]

Thus three neutrons are produced and nuclear equation will be: [tex]^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n[/tex]

b) For the another fission reaction:

[tex]^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{A}_{56}\textrm{Ba}+^{94}_{Z}\textrm{X}+2^1_0\textrm{n}[/tex]

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 2

A = 140

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = 56 + Z + 0

Z = 36

As Krypton has atomic number of 36,Thus the nuclear equation will be :

[tex]^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}[/tex]

Final answer:

The fission of uranium-235 can yield multiple products including bromine-87, lanthanum-146, and additional neutrons. Part a explains the specific fission process forming these products, while part b outlines a scenario producing a nucleus with 56 protons and another with 94 nucleons, alongside more neutrons.

Explanation:

The question relates to the process of nuclear fission, specifically the fission of uranium-235 (U-235) when it captures a neutron. Nuclear fission involves a large nucleus splitting into smaller nuclei along with the release of energy and additional neutrons. These neutrons can initiate further fission reactions, creating a chain reaction.

For part a (forming bromine-87 and lanthanum-146):

U-235 + n → Br-87 + La-146 + 3n

This equation shows that upon absorbing a neutron, U-235 splits into bromine-87, lanthanum-146, and releases 3 more neutrons.

For part b (forming a nucleus with 56 protons and another with a total of 94 nucleons):

A nucleus with 56 protons corresponds to iron (Fe), due to the atomic number. A second nucleus having a total of 94 nucleons with an unspecified proton-to-neutron ratio could refer to an element like plutonium-94, considering the total number of protons and neutrons. However, the provided information primarily focuses on the fission products rather than specifics on this second nucleus. Additionally, 2 more neutrons are produced. A precise equation for this scenario cannot be formulated without more specific information on the second fragment.

Final answer:

The work done by the force of push is 12 J. The work done by the force of friction is 7.08 J. The work done on the book by the force of gravity is zero.

Explanation:

To find the work done by the force of push, we can use the formula:

Work = Force x Distance

In this case, the force of push is 10 N and the distance is 1.2 m. So the work done by the force of push is:

Work = 10 N x 1.2 m = 12 J

To find the work done by the force of friction, we can use the same formula. In this case, the force of friction is 5.9 N and the distance is also 1.2 m. So the work done by the force of friction is:

Work = 5.9 N x 1.2 m = 7.08 J

The work done on the book by the force of gravity is zero since the book is being moved horizontally and gravity acts vertically.

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.

Answer and Explanation:

The two necessary design conditions before an ion chamber can be considered a Bragg-Grey chamber are:

Such a chamber can be used for the purpose of calculation and measurement of the absorbed ionizing radiation dose inside the chamber. Basically, ionization chambers are used as detectors for X-ray absorption spectroscopy.