When 1.50 ✕ 10^5 J of heat transfer occurs into a meat pie initially at 20.0°C, its entropy increases by 465 J/K. What is its final temperature (in degrees)?

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:

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:

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:

0.081 m /s

Explanation:

According to the conservation of momentum, the momentum of a system is conserved when no external force is applied on a body.

momentum of the system before throwing = momentum of the system after throwing

Let v be the velocity of quarterback after throwing the football.

80 x 0 + 0.43 x 0 = 80 x v + 0.43 x 15

0 = 80 v + 6.45

v = - 0.081 m /s

The negative sign shows that he is moving in backward direction.

The quarterback will be moving backward at approximately 0.080625 m/s just after releasing the football due to the conservation of momentum.

The student's question involves practicing Problem-Solving Strategy 11.1 for conservation of momentum problems. Specifically, the student is asked to calculate how fast an 80-kg quarterback will be moving backward just after releasing a football that has a mass of 0.43 kg and is thrown horizontally at a speed of 15 m/s. To solve this, we use the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it.

Before the throw, the total momentum of the system (quarterback and football) is zero because both are stationary with respect to horizontal motion. After the throw, the combined momentum must still be zero. We can set up the equation mQB × vQB + mball × vball = 0, where mQB is the mass of the quarterback, vQB is the quarterback's velocity after the throw, mball is the mass of the football, and vball is the velocity of the football. This simplifies to vQB = -(mball × vball) / mQB.

Plugging in the numbers gives us vQB = -(0.43 kg × 15 m/s) / 80 kg, which yields a velocity of vQB ≈ -0.080625 m/s. The negative sign indicates that the quarterback is moving in the opposite direction of the throw, which is backward.

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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]

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:

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.

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.

Answer:

Object 1 is hotter.

Explanation:

Object 1 T=[tex]0^{\circ}C[/tex]  

Object 2 T=0 K

Object 3 T=[tex]0^{\circ}F[/tex]

Relation between Celcius ,Kelvin and Fahrenheit

[tex]\dfrac{C-0}{100}=\dfrac{K-273}{100}=\dfrac{F-32}{180}[/tex]

K=C+273,  [tex]K=\dfrac{5}{9}(F-32)+273[/tex].

So now we will convert all in one unit.

Object 1 T=273 K

Object 2 T=0 K

Object 3 T=255.22 K

From above we can say that Object 2 is coolest and object 1 is hottest.

So  Object 1 is hotter.

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.

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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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]

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:

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:

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

Answer:

The equilibrium vibrational frequency that causes the shift is [tex]0.56\times10^{14}\ Hz[/tex]

Explanation:

Given that,

Wavelength of Raman line [tex]\lambda'=4768.5\ A[/tex]

Wavelength [tex]\lambda=4358.3\ A[/tex]

We need to calculate the frequency

Using formula of frequency

[tex]f =\dfrac{c}{\lambda}[/tex]

For 4748.5 A

The frequency is

[tex]f'=\dfrac{3\times10^{8}}{4748.5\times10^{-10}}[/tex]

[tex]f' =6.32\times10^{14}\ Hz[/tex]

For 4358.3 A

The frequency is

[tex]f=\dfrac{3\times10^{8}}{4358.3\times10^{-10}}[/tex]

[tex]f=6.88\times10^{14}\ Hz[/tex]

We need to calculate the shift

[tex]\Delta f=f-f'[/tex]

[tex]\Delta f=(6.88-6.32)\times10^{14}\ Hz[/tex]

[tex]\Delta f=0.56\times10^{14}\ Hz[/tex]

Hence, The equilibrium vibrational frequency that causes the shift is [tex]0.56\times10^{14}\ Hz[/tex]

Answer:

Explanation:

A proton is positively charged in nature.

Let an electric field strength is E.

The force on a charged particle placed in an electric field is given by

F = q x E

Where q is the charge

Here, the proton is positively charged, so the direction of force acting on the proton is same as teh direction of electric field strength.

Thus, the motion of proton is parallel to the electric field and the motion is accelerated.

Answer:

Change in ball's momentum is 1.5 kg-m/s.

Explanation:

It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

[tex]\Delta p=m(v-u)[/tex]

[tex]\Delta p=0.15\ kg(-3.5\ m/s-6.5\ m/s)[/tex]

[tex]\Delta p=-1.5\ kg-m/s[/tex]

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

The change in the magnitude of the ball's momentum is 1.125 kg·m/s.

The change in the magnitude of an object's momentum can be calculated by subtracting its initial momentum from its final momentum. In this case, the initial momentum of the ball is calculated by multiplying its mass (0.15 kg) by its initial speed (6.5 m/s), and the final momentum is calculated by multiplying its mass by its final speed (3.5 m/s).

So, the change in the magnitude of the ball's momentum is calculated as:

|pf| - |pi| = (0.15 kg)(3.5 m/s) - (0.15 kg)(6.5 m/s) = -0.15 kg·m/s - 0.975 kg·m/s = -1.125 kg·m/s.

Since momentum is a vector quantity, the change in its magnitude is given by its absolute value, which is 1.125 kg·m/s. Therefore, the correct answer is A. 1.125 kg·m/s.

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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.

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.

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Answer:

v = 1 m/s

Explanation:

from the principle of conservation of momentum, we have following relation

initial momentum = final momentum

[tex]m_{1}v_{1}+m_{2}v_{2} = (m_{1}+m_{2})v^{2}[/tex]

where

m1 = 1.14 kg

v1 = 2.0 m/s

m2 = 1.14 kg

v2  = 0 m/s

putting all value in the above equation

[tex]1.14 *2.0+ 0 =(1.14+1.14)v^{2}[/tex]

[tex]v =\frac{1.14*2.0}{1.14+1.14}[/tex]

v = 1 m/s

By applying the conservation of momentum principle, the speed of the two blocks immediately after the collision is found by equating the momentum before and after the collision.

In this physics problem, we need to determine the speed of the two blocks attached by a spring immediately after the collision. The principle of conservation of momentum can be applied to this. Before the collision, only block m1 is moving, so the total momentum is m1*v1. After the collision, the two blocks move together at speed v', resulting in total momentum of (m1 + m2) * v'. According to the principle of conservation of momentum, the momentum before collision equals to momentum after collision. Hence, m1*v1 = (m1 + m2)*v'.

Solving this equation will yield the value of v'. Substituting the values of m1 = 2.27 kg, m2 = 1.14 kg, and v1 = 2.00 m/s will result in v' = (2.27 kg * 2.00 m/s) / (2.27 kg + 1.14 kg), which gives the speed of the blocks immediately after collision.

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Answer:

a) 1.34 Volts

b) 3.08 W

c) 2.68 W

Explanation:

Given:

Emf of the cell, E = 1.54 V

current, i = 2.0 A

internal resistance, r = 0.100Ω

(a) Terminal voltage (V) = E - v

where,

v is the potential difference across the resistance 'r'

now,

according to the Ohm's Law, we have

v = i × r  

substituting the values in the above equation we get

v = 2.0 × 0.100 = 0.2 Volts

thus,

Terminal voltage (V) = (1.54 - 0.2) = 1.34 V

(b) Now, the Total power (P) is given as

P = E × i = (1.54 × 2.0) = 3.08 W  

(c) Power into its load = [terminal voltage, v] * i  

= (1.34 × 2.0) = 2.68 W

Final answer:

Terminal voltage of a carbon-zinc cell is 1.34 V, producing 2.68 W of electrical power, with 2.68 W going to the load.

Explanation:

Terminal voltage: The formula to calculate terminal voltage is V = EMF - I * r, where V is the terminal voltage, EMF is the electromotive force, I is the current, and r is the internal resistance of the cell. So, V = 1.54 V - 2.00 A * 0.100 Ω = 1.34 V.

Electrical power: The electrical power produced by the cell is given by P = V * I, where P is power, V is voltage, and I is current. Substituting the values, P = 1.34 V * 2.00 A = 2.68 W.

Power to load: The power delivered to the load is equal to the voltage supplied to the load times the current flowing through it. Hence, the power to the load is P = V_load * I = 1.34 V * 2.00 A = 2.68 W.

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

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

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

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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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:

Answer to the question:

Explanation:

A black hole is a finite region of space within which there is a mass concentration high and dense enough to generate a gravitational field such that no material particle, not even light, can escape it.

Answer:

The number of atoms are [tex]1.86\times10^{8}[/tex].

Explanation:

Given that,

Diameter [tex]D = 1.40\times10^{2}\ pm[/tex]

[tex]D=1.40\times10^{2}\times10^{-12}\ m[/tex]

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

[tex]Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}[/tex]

[tex]Number\of\atoms =1.86\times10^{8}[/tex]

Hence, The number of atoms are [tex]1.86\times10^{8}[/tex].

The distance traveled by a particle with velocity function

over the first t seconds can be obtained by integrating that function from 0 to t using integration by parts.

The distance traveled by a particle can essentially be obtained by integrating the velocity function over a given timeframe. In this case, the velocity function is given by

To calculate the distance traveled over the first t seconds, we need to integrate this function from 0 to t. Therefore, the integral ∫v(t)dt from 0 to t will provide the required solution. Use the method of integration by parts, choosing u=t^2 and

then complete the integration process following the standard procedure.

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Answer:

The normal force acting on the block is 980 N.

Explanation:

Mass of box = 100 kg

Coefficient of static friction = 0.40

Coefficient of kinetic friction = 0.20

We need to calculate the normal force

We know that,

When the object is placed on the flat surface then the normal force is equal to the weight of the object.

So. The normal force = mass x acceleration due to gravity

[tex]N=mg[/tex]

Where, N = Normal force

m = mass of object

g = acceleration due to gravity

[tex]N=100\times9.8[/tex]

[tex]N=980\ N[/tex]

Hence, The normal force acting on the block is 980 N.

Final answer:

The normal force acts perpendicular to the floor supporting the box and is equal to 980 N.

Explanation:

The question asks about the normal force acting on a 100-kg box at rest on the floor. To find the normal force on the box, we can use the equation that relates the normal force to the weight of the object, which is W = mg, where m is the mass and g is the acceleration due to gravity.

Given that the mass of the box is 100 kg and the acceleration due to gravity is 9.80 m/s², the normal force can be calculated as:

Normal Force (N) = Mass (m) × Acceleration due to gravity (g)

Normal Force = 100 kg × 9.80 m/s² = 980 N

The normal force acts perpendicular to the floor supporting the box and is equal to 980 N. It is often confused with the applied force P, but the normal force is related to the box's weight, not the horizontal force applied.

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

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

Answer:

[tex]Q = 1.45 \times 10^{-7} C[/tex]

Explanation:

Here flux passing through each surface is given

so total flux through whole cube is given by

[tex]\phi = 2060 + 1590 + 1690 + 3430 + 1870 + 5760[/tex]

[tex]\phi = 16400 Nm^2 C[/tex]

now we also know that total flux through a closed surface depends on the total charge enclosed in the surface

So we will have

[tex]\frac{Q}{\epsilon_0} = 16400[/tex]

[tex]Q = (8.85 \times 10^{-12})(16400)[/tex]

[tex]Q = 1.45 \times 10^{-7} C[/tex]

The total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].

The electric flux [tex](\( \Phi \))[/tex] through a closed surface is given by Gauss's Law:

[tex]\[ \Phi = \frac{Q}{\varepsilon_0} \][/tex]

where:

- [tex]\( \Phi \)[/tex] is the electric flux,

- [tex]\( Q \)[/tex] is the total charge enclosed by the closed surface,

- [tex]\( \varepsilon_0 \)[/tex] is the permittivity of free space [tex](\( \varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \))[/tex].

For a closed box, the total electric flux [tex](\( \Phi_{\text{total}} \))[/tex] is the sum of the electric flux through each of its six surfaces. Let's denote the electric flux through each surface as [tex]\( \Phi_i \) where \( i = 1, 2, \ldots, 6 \)[/tex].

[tex]\[ \Phi_{\text{total}} = \sum_{i=1}^{6} \Phi_i \][/tex]

Now, we can set up an equation using the given values:

[tex]\[ \Phi_{\text{total}} = 12060 \, \text{Nm}^2/\text{C} + 1590 \, \text{Nm}^2/\text{C} + 1690 \, \text{Nm}^2/\text{C} + 3430 \, \text{Nm}^2/\text{C} + 1870 \, \text{Nm}^2/\text{C} + 5760 \, \text{Nm}^2/\text{C} \][/tex]

[tex]\[ \Phi_{\text{total}} = 28800 \, \text{Nm}^2/\text{C} \][/tex]

Now, use Gauss's Law to find the total charge [tex](\( Q \))[/tex] enclosed by the closed surface:

[tex]\[ Q = \Phi_{\text{total}} \cdot \varepsilon_0 \][/tex]

[tex]\[ Q = 28800 \, \text{Nm}^2/\text{C} \cdot 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \][/tex]

[tex]\[ Q \approx 2.544 \times 10^{-4} \, \text{C} \][/tex]

So, the total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].