(c) It is suggested that one side of the copper sheet cools to a lower temperature than the
other side.
Explain why this does not happen.
[2]

Answer:

a) 889 N / m

b) 133 N

The force experienced at P if a 1.5uC was placed is -1.67 N

To find the electric field at point P, first calculate the electric field due to the 5.0 μC charge and the electric field due to the -2.0 μC charge, and then add them together using vector addition.

Electric field due to the 5.0 μC charge:

E1 = kq₁/r₁²

where q₁ = 5.0 μC, r₁ = distance between charge and point P

r₁ = √(0.6²) = 0.6 m

E₁ = (9 x 10⁹ Nm²/C²) x (5.0 x 10⁻⁶ C) / (0.6)²

E₁ = 1.25 x 10⁶ N/C (directed downwards along the negative y-axis)

Electric field due to the -2.0 μC charge:

E₂ = kq₂/r₂²

where q₂ = -2.0 μC, r₂ = distance between charge and point P

r₂ = √(0.74² + 0.6²) = 0.945 m

E2 = (9 x 10⁹ Nm²/C²) x (-2.0 x 10⁻⁶C) / (0.945)²

E2 = -2.36 x 10⁶ N/C (directed upwards along the positive y-axis)

Total electric field at point P:

E = E₁ + E₂

E = 1.25 x 10⁶ N/C - 2.36 x 10⁶ N/C

E = -1.11 x 10⁶ N/C (directed upwards along the positive y-axis)

To find the force on a 1.5 μC charge placed at point P, use the formula:

F = qE

where q = 1.5 μC

F = (1.5 x 10⁻⁶ C) x (-1.11 x 10⁶ N/C)

F = -1.67 N (directed upwards along the positive y-axis)

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Answer:

radiation

Explanation:

Answer:

The brain is a part of the body that is particularly sensitive to high temperature. Hence some ungulates, like the Thomson's gazelle, use a counter-current heat exchanging structure known as the carotid rete to keep the brain cooler than the body.The cooled arterial blood then continues toward the brain.

Answer: -200

Explanation: is it math

Two forces f1=(8i+3j)N and f2=(4i+6j) are acting on 5kg object then  the magnitude of the resultant force is 15 N and the direction of the resultant force is approximately 36.87 degrees from the positive x-axis.

The acceleration of the object in the x-component (\(a_x\)) is 2.4  \(m/s^{2}\), and the acceleration in the y-component (\(a_y\)) is 1.8  \(m/s^{2}\).

The magnitude of the acceleration of the object is 3  \(m/s^{2}\).

To find the magnitude and direction of the resultant force, we need to add the two given forces together.

Given:

f1 = (8i + 3j) N

f2 = (4i + 6j) N

To find the resultant force (\(F_res\)), we simply add the corresponding components:

\(F_res\) = f1 + f2

= (8i + 3j) + (4i + 6j)

= (8 + 4)i + (3 + 6)j

= 12i + 9j

The magnitude of the resultant force (\(|F_res|\)) can be found using the Pythagorean theorem:

\(|F_res|\)= \(\sqrt{(12^2) + (9^2)}\)

= \(\sqrt{144 + 81}\)

= \(\sqrt{225}\)

= 15 N

So, the magnitude of the resultant force is 15 N.

To find the direction of the resultant force, we can use trigonometry. The direction can be represented by the angle θ between the positive x-axis and the resultant force vector. We can calculate θ using the inverse tangent function:

θ = arctan(9/12)

= arctan(3/4)

≈ 36.87 degrees

Therefore, the direction of the resultant force is approximately 36.87 degrees from the positive x-axis.

Now let's calculate the acceleration of the object in the x and y components. We know that force (F) is related to acceleration (a) through Newton's second law:

F = ma

For the x-component:

\(F_x\)= 12 N

m = 5 kg

Using  \(F_x\)= \(ma_x\), we can solve for \(a_x\):

12 N = 5 kg * \(a_x\)

\(a_x\)= 12 N / 5 kg

\(a_x\) = 2.4 \(m/s^{2}\)

For the y-component:

\(F_y\) = 9 N

m = 5 kg

Using \(F_y\) = \(ma_y\), we can solve for \(a_y\):

9 N = 5 kg * \(a_y\)

\(a_y\) = 9 N / 5 kg

\(a_y\)= 1.8  \(m/s^{2}\)

So, the acceleration of the object in the x-component (\(a_x\)) is 2.4  \(m/s^{2}\), and the acceleration in the y-component (\(a_y\)) is 1.8  \(m/s^{2}\).

To find the magnitude of the acceleration (|a|), we can use the Pythagorean theorem:

|a| = \(\sqrt{(a_x^2) + (a_y^2)}\)

= \(\sqrt{(2.4^2) + (1.8^2}\)

= \(\sqrt{5.76 + 3.24}\)

= \(\sqrt{9}\)

= 3  \(m/s^{2}\)

Therefore, the magnitude of the acceleration of the object is 3  \(m/s^{2}\)

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The instantaneous current at the same moment in the RL circuit can be expressed as I = Imaxsin(150), where Imax represents the maximum current.

1. Given that the instantaneous voltage at a specific moment in the RL circuit is V = Vmaxsin(150).

2. We can express the current at the same moment using Ohm's Law, which states that V = IR, where V is voltage, I is current, and R is resistance.

3. In an RL circuit, the resistance is represented by the symbol R, and it is typically associated with the resistance of the wire or any resistors in the circuit.

4. However, the given equation does not explicitly mention resistance.

5. Since we are considering an RL circuit, it suggests the presence of inductance (L) along with resistance (R).

6. In an RL circuit, the voltage across the inductor (VL) can be expressed as VL = L(di/dt), where L is the inductance and di/dt represents the rate of change of current.

7. At any given instant, the total voltage across the circuit (V) can be expressed as the sum of the voltage across the resistor (VR) and the voltage across the inductor (VL).

8. Therefore, V = VR + VL.

9. Since the given equation represents the instantaneous voltage (V), we can deduce that V = VR.

10. By comparing V = VR with Ohm's Law (V = IR), we can conclude that I = Imaxsin(150), where Imax represents the maximum current.

The specific values of Vmax, Imax, and the phase angle have not been provided in the question, so we are working with the general expression.

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Answer:

 E_interior = 0

Explanation:

As the sphere is metallic, the electrical charges are distributed on its surface, as far away as possible from each other.

If we apply Gauss's law, as the charge is on the surface, when drawing a spherical Gaussian surface, we see that there is no charge inside, therefore there is no electric field inside the metallic sphere.

          E_interior = 0

Answer:

40Kg

Explanation:

F=m×a

24N÷0.6m/s^2=40

The speed of the puck is 3.67 m/s.

To find the speed of the puck, we can use the concept of centripetal force. The tension in the string provides the necessary centripetal force to keep the puck moving in a circle. At the same time, the tension in the string also supports the weight of the suspended mass.

Using Newton's second law, we can write two equations of motion: one for the puck and one for the suspended mass. For the puck, the net force acting on it is the tension in the string, which is equal to the centripetal force required to keep it moving in a circle. Thus, we can write:

= m1 * v^2 / R

where T is the tension in the string, v is the speed of the puck, and R is the radius of the circle.

For the suspended mass, the net force acting on it is its weight minus the tension in the string, which must be zero since the mass is in equilibrium. Thus, we can write:

T = m2 * g

where g is the acceleration due to gravity.

Combining these two equations, we can solve for the speed of the puck:

v = sqrt(T * R / m1) = sqrt(m2 * g * R / m1)

Substituting the given values, we get:

v = sqrt(1.0 kg * 9.81 m/s^2 * 0.9 m / 0.21 kg) = 3.67 m/s

Therefore, the speed of the puck is 3.67 m/s.

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Answer:

42.5km/s^2

Explanation:

Fnet=m⋅a

We know that

\(a=\frac{v-u}{t}\)

Here,  

v=4m/s,  u=0m/s,  t=8s

\(a=\frac{4m/s-0m/s}{8s} =0/5m/s^2\)

Also, we have  m=85kg

Fnet=85  kg⋅0.5  m/s2=42.5N

Acceleration = (change in speed) / (time for the change)

Acceleration = (4 m/s) / (8 seconds)

Acceleration = 0.5 m/s²

Force = (mass) x (acceleration)

Force = (85 kg) x (0.5 m/s²)

Force = 42.5 Newtons

Answer:

\(\sf\longmapsto42.5 N\)

Explanation:

We would need to use Newton's second law ofmotion, which states that-

\(\sf \longmapsto \: F _{net} = m•a \)

We know that,

\(\sf \longmapsto \: a = \frac{v - u}{t} \)

Here,v = 4 m/s, u = 0 m/s, t = 8 s

\(\sf \longmapsto \: a = \frac{4 \: m/s - 0 \: m/s}{8s} = 0.5 \: m / {s}^{2} \)

Also, we have

m = 85 kg

\(\sf \longmapsto \: F _{net} = 85 \: kg \: • \: 0.5 m / {s}^{2} \: \: = 42.5 \: N\)

The crate is being pulled by a total force of 160 N. the box is being subjected to a 350 N frictional force. The crate is being movement to a 96.99 N force of friction.

A rope pulls a 60 kg 60 kg container across the floor with a constant force applied of 250 N. Under the impact of this coercion and the compressive stress the floor exerts on the crate, it is seen that the speed of the crate increases from 1 m/s to 3 m/s to 3 m/s in a period of 2 seconds.\($a = \frac{\Delta v}{\Delta t} =\frac{9}{ \text m/s - 1} \text m/s^2, \text s = 4, \text{and} \ \text m/s 2$\)

The following formula can be used to determine the net force on the crate:

\(\rm F_{net }= ma = (40kg)(4m/s) = 160 N\)

As a result, the crate is being pulled by a total force of 160 N.

The degree of kinetic friction is inversely correlated with the level of normal force and the level of irregularity in between sliding surfaces. The magnitude of static friction is inversely related to the amount of normal force and the measure of roughness between of sliding surfaces.\((40 \text kg) = $N = mg 9.81 \text m/s^2= 392.4 \text N$\)

As the coefficient of friction factor between the crate and the floor is not provided, we are unable to directly calculate it. Yet, we are aware that the 350 N force exerted by the rope must be balanced by a force of friction that is both equal to and directed in the opposite direction. Hence, it is possible to compute the frictional force's magnitude as follows:

\(355 \text N, $F friction\)

As a result, the box is being subjected to a 350 N frictional force.

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Lateral rotation is an anatomical term of motion used to describe rotation along the long axis of a joint. In lateral rotation, this movement is away from the midline of the body and occurs in the transverse plane.

On the other hand, medial rotation is a rotational movement towards the midline. It is sometimes referred to as internal rotation.

According to this question, twisting one's lower limb at the hip while standing, so that the toes point off to the side is an example of lateral rotation while doing the above but in such a way that it points towards the other foot is an example of medial rotation.

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Answer:

The value is \(p = - 2.63 \ Diopters\)

Explanation:

From the question we are told that  

      The value of the far point is  \(a = 40 \ cm = 0.4 \ m\)

      The distance of the lens to the eye is  \(b = 2 \ cm = 0.02\)

Generally

        \(1 Diopter = > 1 m^{-1}\)

Generally the power spectacle lens needed is mathematically represented as

           \(p = \frac{1}{d_o } + \frac{1}{d_i}\)

Here \(d_o\) is the object distance which for a near sighted person is \(d_o = \infty\)

And  \(d_i\) is the image distance which is evaluated as

        \(d_i = b - a\)

=>     \(d_i = 0.02 - 0.4\)

=>     \(d_i = -0.38 \ m\)

So

         \(p = \frac{1}{\infty } + \frac{1}{-0.38}\)

=>      \(p = 0 - 2.63\)

=>      \(p = - 2.63 \ Diopters\)

Answer:

Explanation:

I hope it help you;)

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

 Goodluck

Explanation:

Answer:

\(5.76\times 10^{-18}\ \text{N}\) perpendicular to the velocity and magnetic field

Explanation:

B = Magnetic field = \(4\times 10^{-5}\ \text{T}\)

\(\theta\) = Angle the magnetic field makes with the horizontal = \(30^{\circ}\)

v = Velocity of electron = \(9\times 10^5\ \text{m/s}\)

q = Charge of electron = \(1.6\times 10^{-19}\ \text{C}\)

Magnetic force is given by

\(F=qvB\sin\theta\\\Rightarrow F=1.6\times 10^{-19}\times 9\times 10^5\times 4\times 10^{-5}\sin30^{\circ}\\\Rightarrow F=2.88\times 10^{-18}\ \text{N}\)

The magnitude of the magnetic force is \(2.88\times 10^{-18}\ \text{N}\) and the direction is perpendicular to the velocity and magnetic field.

Given:

The thermal conductivity of the wall, k=0.77 W/(m·°C)

The dimensions of the wall, 4.3 m by 5 m

Thus the area of the wall, A=4.3×5=21.5 m²

The thickness of the wall, d=19 cm=0.19 m

The time period, t=7.7 h=7.7×3600=27720 s

The temperature inside the building, T₁=29 °C

The temperature outside the building, T₂=4 °C

To find:

The amount of the heat that flows through the wall in 7.7 h

Explanation:

The amount of the heat lost through the wall is given by,

On substituting the known values,

Final answer:

The heat that flows through the wall in the given time period is 60.38 MJ

Given

distance of the object  u = -40 cm.

radius of curvature of the refracting surface  R = -20 cm.

The height of the object  h = 1 cm.

To Find

The size of the image

Solution

We will now write the following data using the sign convention:

The object's distance is u = -40 cm.

The refracting surface has a radius of curvature of R = -20 cm.

The object's height is h = 1 cm.

We know that the lens maker formula is: n2vn1u=n2n1R.

v denotes the image formation distance.

The refractive index of the second medium is n2=1.33, and the refractive index of the first medium is n1=1.

In the preceding equation, we will now substitute the known values.

1.33v−1−40cm=1.33−1−20cmv=32.05cm

We know that the magnification ratio can be expressed as vu=Hh.

H denotes the height of the created image.

We'll now swap the given and obtained values.

⇒32.05cm divide by 40cm=H divide by 1cm

⇒H=0.6cm

As a result, the best choice is (B).

Further information: The rules of reflection apply to reflection from a concave mirror. The normal to the point of incidence is drawn along the radius of the mirror, that is, by connecting the point of incidence to the centre of curvature.

The development of an image in a concave mirror is mostly determined by the distance between the object and the mirror. The concave mirror produces both real and virtual images.  A virtual and magnified picture is generated when the object is placed very close to the mirror.

The complete question is -

Find the size of the image formed in the situation shown in figure.

photo

A) 0.5 cm

B) 0.6 cm

C) 1.2 cm

D) 1 cm

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Explanation:

(a) To find the work done by the frictional force on the block, we can use the work-energy principle which states that the work done on an object is equal to the change in its kinetic energy. Therefore,

Work done by frictional force = Change in kinetic energy

The initial kinetic energy of the block is given by:

K1 = (1/2)mv1^2 = (1/2)(5 kg)(7 m/s)^2 = 122.5 J

The final kinetic energy of the block is given by:

K2 = (1/2)mv2^2 = (1/2)(5 kg)(4 m/s)^2 = 40 J

Therefore, the change in kinetic energy is:

ΔK = K2 - K1 = 40 J - 122.5 J = -82.5 J

Since the work done by the frictional force is negative, we have:

Work done by frictional force = -|-82.5 J| = 82.5 J

Therefore, the magnitude of the work done by the frictional force on the block is 82.5 J.

(b) To find the magnitude of the average frictional force on the block, we can use the equation:

Average frictional force = Work done by frictional force / Distance traveled

The distance traveled by the block during the 6 seconds is given by:

d = vit + (1/2)at^2

where vi is the initial velocity, t is the time, a is the acceleration, and d is the distance.

Since the surface is smooth before the rough patch, the block moves with constant velocity and the acceleration is zero. Therefore,

d = vit = (7 m/s)(6 s) = 42 m

After the rough patch, the block slows down from 7 m/s to 4 m/s in 6 seconds. Therefore, the acceleration of the block is given by:

a = (v2 - v1) / t = (4 m/s - 7 m/s) / 6 s = -0.5 m/s^2

Using this acceleration, we can find the distance traveled by the block after the rough patch:

d' = vit + (1/2)at^2 = (4 m/s)(6 s) + (1/2)(-0.5 m/s^2)(6 s)^2 = 12 m

Therefore, the total distance traveled by the block is:

d_total = d + d' = 42 m + 12 m = 54 m

Now, we can find the average frictional force on the block:

Average frictional force = Work done by frictional force / Distance traveled

= 82.5 J / 54 m

≈ 1.53 N

Therefore, the magnitude of the average frictional force on the block is approximately 1.53 N.

a. Total energy is 0.098 J

b. Potential and Kinetic Energies is 0.032 sin^2(8.08t) J

c. Velocity at x is -0.808 sin(8.08t) m/s

d. Potential and Kinetic Energies at x is 0.016 sin^2(8.08t) J

We can use the following formulas for the energy, velocity, and potential and kinetic energies of a simple harmonic oscillator:

where ω = √(k/m) is the angular frequency.

Given that k = 19.6 N/m, A = 0.100 m, x = -(0.100 m) cos 8.08t, and v = (0.808 m/s) sin 8.08t, we can find the values of E, U, and K as follows:

(a) Total Energy:

E = 1/2 k A^2 = 1/2 * 19.6 * 0.1^2 = 0.098 J

(b) Potential and Kinetic Energies:

U = 1/2 k x^2 = 1/2 * 19.6 * (-0.1 cos(8.08t))^2 = 0.098 cos^2(8.08t) J

K = 1/2 m v^2 = 1/2 * (0.1) * (0.808 sin(8.08t))^2 = 0.032 sin^2(8.08t) J

(c) Velocity at x = 0.050 m:

When x = 0.050 m, cos(8.08t) = -0.5, so we have:

v = -ωA sin(ωt) = -ω(0.1) sin(8.08t) = -0.808 sin(8.08t) m/s

(d) Potential and Kinetic Energies at x = A/2:

When x = A/2 = 0.050 m, cos(8.08t) = -0.5, so we have:

U = 1/2 k x^2 = 1/2 * 19.6 * (0.050)^2 = 0.0245 J

K = 1/2 m v^2 = 1/2 * (0.1) * (0.808 sin(8.08t))^2 = 0.016 sin^2(8.08t) J

Note that the sum of potential and kinetic energies at any point in time is equal to the total energy, which is constant.

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Answer:

F = 160.0 N

Explanation:

Given: Soccer payer with a mass = 80 kg, force = 20 N

To find: force

Formula: \(F=ma\)

Solution: It is summarized by the equation: Force (N) = mass (kg) × acceleration (m/s²). Thus, an object of constant mass accelerates in proportion to the force applied.

F = m × a

F = 20 kg - 10 = 2  

F = 80 × 2 = 160

F = 160.0 N

Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.

This same distance from the Earth's core to a location where the gravitational acceleration is g/16 is roughly 20,000 miles (32,000 kilometers).

The pace at which speed changes is known as acceleration. So because direction of an object's velocity is shifting even while it follows a circular course, it continues to accelerate.

According to the formula a = v/t, kinetic energy (a) is the product of the change in momentum (v) and the shift in time (t).

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We are given that two objects of same mass have the same velocity but in opposite directions. The kinetic energy of each object is given by:

Where:

Since the kinetic energy depends on the mass and the magnitude of the velocity, this means that the object will have the same kinetic energy regardless of the direction of the velocity. Therefore, option C is correct.

Answer:

a) When moving body applies brake then velocity and acceleration would be in opposite direction

b) When body starts to increase velocity then velocity and acceleration would be in same direction

c) When body is circulating then velocity and acceleration would be perpendicular to each other

Explanation:

a) When body applies brake then its velocity starts decreasing, in this case its acceleration would try to stop the moving body. So direction of velocity would be same as direction of motion of body but direction of acceleration would be in opposite direction

b) When body starts to increase velocity, its acceleration would make the body to move faster. So direction of velocity would be the direction of motion of body and acceleration would also be in same direction

c) When body moves in circular path then its acceleration would be towards centre of circle and velocity would try to snap the body out of circle to straight line which in tangent to circle.

Answer:962 hz

Explanation: got it right on acellus

The frequency of the siren when the cyclist is stationary will be 963 Hz.

The frequency of a wave is the number of cycles per second.

FRom the doppler effect,

f =f'[ (V+Vs/V-Vo) ]

Here, doppler shifted frequency f' = 894 Hz, Source is stationary, Vs =0. Velocity of observer Vo = 24.5m/s and velocity of sound wave V = 343m/s

Substitute the value into the expression , we get

f = 894 x [(343/343 - 24.5)}

f = 963 Hz

Thus, the frequency of the siren when the cyclist is stationary is 963 Hz.

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Answer:

In a material, the magnetic behavior depends on the alignment of magnetic moments of the atoms. Magnetic moments are generated by the motion of the electrons in the atoms. When the magnetic moments of atoms in a material are aligned in a specific pattern, it creates a magnetic field which results in the material being magnetic.

In many materials, the magnetic behavior arises due to the alignment of magnetic domains, which are regions of atoms with magnetic moments aligned in the same direction. When many domains with aligned magnetic moments are present in a material, the material becomes magnetic.

The magnetic behavior of a material depends on the number of electrons and the arrangement of those electrons in the atoms. In particular, for an atom to have a magnetic moment, it must have unpaired electrons, meaning electrons that are not paired with another electron with the opposite spin. When these unpaired electrons in the atoms are aligned, they generate a magnetic moment. If all electrons are paired, there will not be a net magnetic moment, so the material will not be magnetic.

So, in summary, the magnetic behavior of a material is determined by the alignment of magnetic moments of atoms. When the magnetic moments of many atoms in a material align in the same direction, it creates a magnetic field, leading to a material being magnetic. This alignment is usually present in magnetic domains consisting of atoms with unpaired electrons.

The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².

To calculate the average acceleration of the motorcycle, we can use the formula:

Average acceleration = (final velocity - initial velocity) / time

First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.

Converting the final velocity:

Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s

Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:

Average acceleration = (21.67 m/s - 0 m/s) / time

To find the time taken to reach this velocity, we need to use the formula for average speed:

Average speed = total distance/time

Rearranging the formula:

time = total distance / average speed

Plugging in the values:

time = 50 m / 21.67 m/s ≈ 2.31 seconds

Now we can calculate the average acceleration:

Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²

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