he impulse exerted on the bowling pin equals the negative of the of the ___ projectile during collision.
Acceleration
Its velocity would change during a very short time interval
Its direction would change
Change in momentum

Answer:

  15√2 N

Explanation:

The acceleration is given by ...

  a = F/m = 5t/5 = t . . . . meters/second^2

The velocity is the integral of acceleration:

  v = ∫a·dt = (1/2)t^2

This will be 9 m/s when ...

  9 = (1/2)t^2

  t = √18 . . . . seconds

And the force at that time is ...

  F = 5(√18) = 15√2 . . . . newtons

Answer:

q=9.43×10^-5C

Explanation:

F=kq^2/r^2

500= 9×10^9 × q^2/ (0.4)^2

500N×0.16m=9×10^9Nm^2C^2 × q^2

80/(9×10^9)= q^2

√(8.8889×10^9) = q

q= 9.43×10^-5C

since they are identical both charges, q=9.43×10^-5C

The magnitude of the two charges is ​9.43×10^-5 C.

To calculate the magnitude of an electric force, it is necessary to use the following expression:

                                          \(F = k \frac{q_{1} \times q_{2} }{d^{2}}\)

Assuming that the constant is:

                                          \(k = 9\times 10^{9}\)

We can apply the values ​​in the formula above, obtaining:

                                     \(500 = 9 \times 10^{9} \times \frac{q^{2}}{0.4^{2}}\)

                                         \(q = \sqrt{0.88\times10^{-8}}\)

                                          \(q = 9.43 \times 10^{-5} C\)

So, the magnitude of the two charges is   ​9.43×10^-5 C.

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Answer:

34.24 %

Explanation:

Since I₀ is the intensity of the un-polarized light, the intensity I₁ of the light polarized by the 1st sheet is (by the one-half rule) I₁ = I₀/2.

The intensity of polarized light I from a polarized source I' is I = I'cos²Ф where Ф is the angle between the direction of I' and I. Since the second sheet has its transmission axis at 25° with respect °o the 1st sheet, the intensity of light I₂ from the second sheet is I₂ = I₁cos²25°.

Also, the 3rd sheet has its transmission axis 47° with respect to the 1st sheet. So, the angle between the transmission axis of the 2nd sheet and 3rd sheet is 47° - 25° = 22°. So, the intensity I₃ from the 3rd sheet is I₃ = I₂cos²22°

Finally, the 4th sheet has its transmission axis 10° with respect to the 3rd sheet. So, the intensity I₄ from the 4th sheet is I₄ = I₃cos²10°.

So,  I₄ = I₃cos²10°

I₄ = I₂cos²22°cos²10°

I₄ = I₁cos²25°cos²22°cos²10°

I₄ = (I₀/2)cos²25°cos²22°cos²10°

I₄/I₀ = cos²25°cos²22°cos²10°/2

I₄/I₀ = (cos25°cos22°cos10°)²/2

I₄/I₀ = (0.9063 × 0.9272 × 0.9848)²/2

I₄/I₀ = 0.8275²/2

I₄/I₀ = 0.6848/2

I₄/I₀ = 0.3424

So, as a percentage,

I₄/I₀ × 100% = 0.3424 × 100% = 34.24 %

Answer:

The expression for destructive interference in thin films allows to find the result for the smallest thickness of the films is:

         t = 2.03 10⁻⁸ mm

Explanation:

Given parameters

Incident wavelength lamo = 535 nm

Refractive index of the film n = 1.32

To find

The minimum thickness for destructive interference

The interference phenomenon occurs when the path of two rays scattered by an obstacle have different optical paths. In the case of thin films we must take into account:

The reflected wave has a phase change of 180º when it goes from a medium with a lower refractive index to a medium with a higher index.

Inside the film medium the wavelength is modulated by the refractive index.

In the attachment we see an outline of these events and the expression for destructive interference remains.

               2 n t = m λ₀

Where n is the refractive index, t the thickness of the film, λ₀ the wavelength in the vacuum and m an integer indicating the order of interference.  

             t =  

The first destructive interference occurs for m = 1, let's calculate.

       t =    

       t = 202.65 nm

Let's reduce this amount to millimeters.

        t = 202.65 nm  

        t = 2,027 10⁻⁸ mm

In conclusion, using the expression for destructive interference in thin films we can find the result for the smallest thickness of the films is:

         t = 2.03 10⁻⁸ mm

Answer:

Energy: The ability to do work

Kinetic energy: Kinetic energy is the energy an object has due to its motion

Potential energy: potential energy is the energy that is stored in an object

Work: Work is the transfer of energy by a force acting on an object as it is displaced

Friction: Friction acts as a resisting force which is generated , when two solid surfaces slide against one another

The final temperature of the water and copper pellets is 43.8°C.

We can use the equation:

\(mc\Delta T = mwCw\Delta T\)

Substituting the values given in the problem:

m c = 28 g

m w = 90 mL = 90 g

c = 0.385 J/g°C

Cw = 4.18 J/g°C

ΔT = T f - 25°C

where T f is the final temperature of the water and copper pellets.

Simplifying the equation:

28 g x 0.385 J/g°C x (T f - 300°C) = 90 g x 4.18 J/g°C x (T f - 25°C)

Solving for T f:

T f = (90 g x 4.18 J/g°C x 25°C + 28 g x 0.385 J/g°C x 300°C) / (90 g x 4.18 J/g°C + 28 g x 0.385 J/g°C)

T f = 43.8°C

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--The complete Question is, What are the final temperature of the water and the copper pellets after thermal equilibrium is reached, assuming no heat is lost to the surroundings?  --

1.) The relative humidity is 100 percent. 2.)To saturate the air at 20°C, 0.018g of water should evaporate.

(i) To calculate the relative humidity at room temperature, compare the amount of water vapour in the air to the maximum amount of water vapour that the air can hold at that temperature.

The saturated vapour pressure at that temperature determines the maximum amount. The relative humidity is stated as a percentage and can be calculated using the formula:

(Actual vapour pressure / Saturated vapour pressure) times 100% = Relative humidity

The saturation vapour pressure at room temperature (20°C) is 17.54mmHg. The relative humidity is 100% if the actual vapour pressure in the air equals the saturation vapour pressure.This signifies that the air has held the most water vapour it can at that temperature.

(ii) Using the ideal gas law, we can calculate the amount of water that should evaporate to saturate the air at 20°C. PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin, according to the ideal gas equation.

This equation can be rearranged to calculate the amount of moles of water:

n = (PV) / (RT)

We know that the required pressure is 17.54mmHg, which is equal to 2.338 kPa. We also know that the temperature is 20° C, which is 293.15 K.The volume is not specified, but we can assume it is constant and hence ignore it. The gas constant is 8.31 joules per mol-K.

n = ((2.338 kPa) / (8.31 J/mol-K * 293.15K))

To calculate the mass of water, multiply the number of moles by the molar mass of water, which is about 18.015 g/mol.

0.01796 g = 0.000997 mol * 18.015 g/mol mass of water

To saturate the air at 20°C, around 0.018g of water should evaporate.

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With the use of formula, the extension which is increase in length is 4.08 mm

Young's modulus can simply be defined as the ratio of tensile stress to tensile strain. Where stress is force per cross-sectional area.

Given that a 180-kg load is hung on a wire of length of 3.70 m, cross-sectional area 2.00 10-5 m2, and Young's modulus 8.00 1010 N/m2.

The given parameters are;

Young's modulus = stress / strain

stress = F/A

strain = e/L

Young's modulus = F/A ÷ e/L

Young's modulus = FL/Ae

Substitute all the parameters into the formula

8 × 10^10 = (1764 × 3.7) / 2 × 10^-5e

Cross multiply

1600000e = 6526.8

e = 6526.8 / 1600000

e = 4.08 × 10^-3 m

e = 4.08 × 10^-3 × 1000 mm

e = 4.08 mm

Therefore, its increase in length is 4.08 mm

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The approximate heat transfer through 1.65 cm of air with a temperature difference of 23.0 °C is approximately 24.4 J/s.

To approximate the heat transfer through the air layer in the double-paned window, we can assume that the glass layers have a negligible impact on the heat flow. The heat transfer can be calculated using Fourier's Law of Heat Conduction, which states that the heat flow (Q) is proportional to the temperature difference (ΔT) and inversely proportional to the thickness (L) and thermal conductivity (k) of the material.

First, we need to calculate the effective thermal conductivity of the air layer due to its thickness and the thermal conductivity ratio between air and glass. Let's denote the thermal conductivity of air as k_air and the thermal conductivity of glass as k_glass. Since glass has a thermal conductivity roughly 36 times greater than air, we have k_glass = 36 * k_air.

Next, we calculate the effective thermal conductivity of the air layer as:

k_eff = (k_air * L_air) / (L_air + k_glass)

Substituting the given values, we have:

k_eff = (k_air * 0.0165 m) / (0.0165 m + 0.005 m) = 0.01309 * k_air

Now, we can calculate the heat flow per second through the air layer using the formula:

Q = (k_eff * A * ΔT) / L_air

Substituting the given values, we get:

Q = (0.01309 * k_air * 0.760 m^2 * 23.0 K) / 0.0165 m = 24.4 J/s

Therefore, the approximate heat transfer through 1.65 cm of air with a temperature difference of 23.0 °C is approximately 24.4 J/s.

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Answer:

Tolerance: \(\pm 10\%\)

Explanation:

Resistor Color Codes

Resistor Color Coding uses colored bands to quickly identify the resistive value or resistors and its percentage of tolerance.

Since the question does not provide a specific color table, we'll use the table attached below.

The colors of the resistor shown in the question are:

First band: orange

Second band: blue

Third band: brown

Fourth band: silver

The colors relate to the following numbers respectively:

3, 6, 10Ω, \(\pm 1\%\)

The first two colors form the number 36

The third color is the multiplier: 36*10Ω = 360Ω

And the fourth color is the tolerance or the possible variation of the resistance \(\pm 1\%\)

Resistance: 360Ω

Tolerance: \(\pm 10\%\)

The answer is 1) The pressure at point D is 80 kPa. 2) The temperature at point D is 800 K. 3) The net work done on the gas over four cycles is zero. 4) The internal energy of the gas at point A is 100 J. 5) The total change in internal energy during four complete cycles is zero.

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(a) The length of the spring just as the box is ready to move is 5.1cm

(b) The length of the spring when it is pulled forward at a constant velocity of 1.75 m/s is 47.0 cm

(c) The length of the spring when it is pulled forward at a constant velocity of 2.75 m/s is 73.8 cm

The length of the spring just as the box is ready to move is calculated by applying the principle of conservation of energy.

work done by static friction = elastic potential energy of the spring

μmgx = ¹/₂kx²2

μmg = kx

x = 2μmg/k

where;

k is spring constant  = 2.5 N/cm x 100 cm/m = 250 N/mμ is coefficient of static friction

g is the acceleration due to gravity

x is an extension of the spring

x = 2μmg/kx = (2 x 0.65 x 18 x 9.8) / 250x

= 0.051 m

x = 5.1 cm.

The length of the spring when it is pulled forward at a constant velocity of 1.75 m/s.elastic potential energy of spring  = kinetic energy of the box

¹/₂kx² = ¹/₂mv² kx² = mv² x²

= mv²/kx

= √(mv²/k)

where;

m is the mass of the box

v is the speed of the box

x is the length of the spring

x = √(18 x 1.75²/250)x = 0.470 mx = 47.0 cm

The length of the spring when it is pulled forward at a constant velocity of 2.75 m/s.x = √(mv²/k)x = √(18 x 2.75²/250)

x = 0.738 m

x = 73.8 cm

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Answer:

0.5417

Explanation:

The frictional force is equal to the coefficient of kinetic friction times the normal force, so

Then, solving for the coefficient of kinetic friction, we get:

Now, we can replace the frictional force Ff by 65 N and the normal force Fn by 120 N

Therefore, the coefficient of kinetic friction is 0.5417

Answer:

Explanation:

(i) μs = F/N = mgsinθ/mgcosθ = tanθ

  tanθ = 0.3

  θ = 16.7°

(ii) a = F/m

    a = (mgsinθ - (μk)mgcosθ) / m

    a = g(sinθ - (μk)cosθ)

    a = 9.8(sin16.7 - (0.2)cos16.7)

    a = 0.94 m/s²

(iii) s = ½at²

     t = √(2s/a)

     t = √(2(1)/0.94)

     t = 1.5 s

Answer: yes

Explanation:

Answer:

9.70 kilograms

Explanation:

I got a 100% on my test

The mass of the object is 9.7 kg. Option (A) is correct.

Every day, we experience the normal force. For instance, when a book is placed on a table, the usual response force prevents the book from falling through the table, even though gravity is pulling the book downward. That implies that there is a force acting to push it upward. The word "normal force" refers to the force. Normal in this context refers to being perpendicular to the ground.

Given parameters:

the normal force acting on the object is: N =  95.0 newtons.

coefficient of static friction: μ = 0.110.

We have to find mass of the object: m = ?

We know that:

Normal force = weight of the object

⇒ N = mg

⇒ m = N/g = 95.0/9.8 kg = 9.7 kg

Hence, the mass of the object be 9.7 kg.

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Answer:

Explanation:

To solve this problem we need to know the direction in which the ball was moving to start with.

The answer will be different depending n the original angle of the ball's movement.

It might be reasonable to assume that the ball is meant to approach along the x-axis,

but if so, the initial speed of 6.42m/s would be irrelevant to the answer.

So I will solve the problem for the general case of two objects colliding at arbitrary angles, and

tell you how to specialize it for any assumption about the initial conditions.

Let

m1 = 7.5 kg be the mass of the ball,

m2 = 1.6 kg be the mass of the pin,

v1 = 6.42 m/s be the velocity of the ball before the strike,

v2 = 0 m/s be the velocity of the pin before the strike,

α1 be the angle of v1,

α2 be the angle of v2,

w1 be the velocity of the ball after the strike,

w2 = 14.8 m/s be the velocity of the pin after the strike,

β1 be the angle of w1,

β2 = -47° be the angle of w2.

By conservation of momentum:

m1v1 + m2v2 = m1w1 + m2w2

Since the velocities are vectors, the addition is vector addition, and the equality is vector equality.

"Vector equality" means that the x-coordinates are equal and the y-coordinates are equal.

The problem cares only about y-coordinates, specifically the y-coordinate of w1, which is w1sin(β1).

(In general, the y-coordinate of any vector is obtained by multiplying the vector's norm by the sine of its angle.)

Conservation of momentum in the y-coordinate is then

m1v1sin(α1) + m2v2sin(α2) = m1w1sin(β1) + m2w2sin(β2)

Expressing the sought quantity

w1sin(β1) = (m1v1sin(α1) + m2v2sin(α2) - m2w2sin(β2))/m1

Substituting known quantities:

w1sin(β1) = (7.5×6.42×sin(α1) + 1.6×0×sin(α2) - 1.6×14.8×sin(-47°))/7.5

= (48.15×sin(α1) + 17.3)/7.5

In the above expression we do not know α1.

If we assume that the ball is approaching along the x-axis then α1 = 0, and

w1sin(β1) = 17.3/7.5 = 2.3

Under that assumption the y-component of the ball's final velocity is 2.3 m/s;

being positive, it is opposite the direction of the pin.

From the calculations, the angular momentum is  10 Kg m^2/s while the new force is 50N.

The angular velocity is defined as the product of the mass, velocity and the radius of the material.

Thus;

L = mvr = 2 Kg * 1m * 5 m/s = 10 Kg m^2/s

b) Given that;

F= Gm^2/r^2

If the distance is halved then the force is reduced by 1/4 thus the new force is 50N.

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Answer: I believe C is incorrect

Explanation:

I'm sorry if it wasn;t tho!!

When a charged object is at rest, the electric field lines emanating from it are radially symmetric and point outward in all directions, forming a pattern that depends on the magnitude and sign of the charge. If the charged object is set in motion, it creates a changing electric field that propagates outward from the object at the speed of light.

The direction of magnetic field lines is depicted using arrows or lines that indicate the direction of the magnetic field at each point in space. The convention for drawing magnetic field lines is that they always form closed loops, and the direction of the field is tangential to the bar at each point.

Yes, electric fields can affect magnetic fields and vice versa. Electric and magnetic fields are intimately related and form two sides of the same coin, as described by Maxwell's equation of electromagnetism.

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The x-component of the net force exerted by the first (q₁ = -11.0 nC) and second charge (q₂ = 36.0 nC) on a third one (q₃ = 50.5 nC), located between q₁ (x₁ = -1.740 m) and q₂ (x₂ = 0 m) at x₃ = -1.190 m is -5.00x10⁻⁶ N.    

The x-component of the net force exerted by the first and second charge on the third charge is given by:

\( \vec{F}_{net_{x}} = \vec{F}_{13} + \vec{F}_{23} \)   (1)  

Where:

We can calculate the force with Coulomb's law

\( \vec{F}_{13} = \frac{Kq_{1}q_{3}}{d_{13}^{2}} \) (2)  

Where:

The Coulomb's constant is:

\( K = \frac{1}{4\pi \epsilon_{0}} = \frac{1}{4\pi 8.854\cdot 10^{-12} C^{2}/N*m^{2}} = 8.99 \cdot 10^{9} N*m^{2}/C^{2} \)    

Since the third charge is placed between q₁ and q₂ at x₃ = -1.190 m, the magnitude of the distance d₁₃, knowing that x₁ = -1.740 m, is:

\( d_{13} = |d_{1}| - |d_{3}| = (1.740 - 1.190) m = 0.55 m \)

Hence, the force exerted by the first charge on the third is (eq 2):

\( \vec{F}_{13} = \frac{Kq_{1}q_{3}}{d_{13}^{2}} = \frac{8.99 \cdot 10^{9} N*m^{2}/C^{2}*(-11.0 \cdot 10^{-9} C)(50.5 \cdot 10^{-9} C)}{(0.55 m)^{2}} = -1.65 \cdot 10^{-5} N \)  

The minus sign is because the resultant force is in the negative x-direction (to the left).

This force is equal to:

\( \vec{F}_{23} = \frac{Kq_{2}q_{3}}{d_{23}^{2}} \)   (3)

Where:

The magnitude of the distance d₂₃, knowing that x₂ = 0, is:

\( d_{23} = |d_{3}| - |d_{2}| = (1.190 - 0) m = 1.190 m \)  

So, the force exerted by the second charge on the third is (eq 3):

\( \vec{F}_{23} = \frac{Kq_{2}q_{3}}{d_{23}^{2}} = \frac{8.99 \cdot 10^{9} N*m^{2}/C^{2}*36.0 \cdot 10^{-9} C*50.5 \cdot 10^{-9} C}{(1.190 m)^{2}} = 1.15 \cdot 10^{-5} N \)    

The positive sign means that the resultant force is in the positive x-direction (to the right).  

Finally, the x-component of the net force is (eq 1):

\( \vec{F}_{net_{x}} = \vec{F}_{13} + \vec{F}_{23} = (-1.65 \cdot 10^{-5} + 1.15 \cdot 10^{-5}) N = -5.00 \cdot 10^{-6} N \)

The minus sign means that the net force exerted by these two charges on the third charge is in the negative x-direction.  

Therefore, the x-component of the net force exerted by the first and second charge on a third charge is -5.00x10⁻⁶ N.

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The temperature gradient in the flow of direction is 294525 W.

A temperature gradient is the gradual variance in temperature with distance. The slope of the gradient is consistent within a material. A gradient is established anytime two materials at different temperatures are in physical contact with each other.

Q= T/( L/ KA)

Q= ( 1500 − 450) / 0.15 / 9.35v * 4.35)

   = 294525 W

Units of measure of temperature gradients are degrees per unit distance, such as °F per inch or °C per meter.

Many temperature gradients exist naturally, while others are created. The largest temperature gradient on Earth is the Earth itself. Q= T/Ka.

Therefore, The temperature gradient in the flow of direction is 294525 W.

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Answer:

I'm not sure but I think its aluminum

When filled with extra air its mass increased to 0.428kg also the top of polythene container mass connected to the perplex box of volume 1000cm³ and the number of times of air inside was 7.2 times. The density of the air filled in the polythene container is approximately 1.25 kg/m³.

The density of air filled in the polythene container can be determined by considering the change in mass and volume of the container before and after filling it with air. Given that the mass of the empty container is 0.419 kg and the mass of the container when filled with extra air is 0.428 kg, and the volume of the perplex box is 1000 cm³.

Calculate the mass of the air inside the container by subtracting the mass of the empty container from the mass of the container when filled with air:

Mass of air = Mass of filled container - Mass of empty container

= 0.428 kg - 0.419 kg

= 0.009 kg

Calculate the volume of the air inside the container using the given number of times the air inside is 7.2:

Volume of air = Volume of perplex box * Number of times air inside

= 1000 cm³ * 7.2

= 7200 cm³

Convert the volume of air to cubic meters (m³) by dividing by 1000000:

Volume of air = 7200 cm³ / 1000000

= 0.0072 m³

Calculate the density of air using the formula:

Density = Mass / Volume

Density = 0.009 kg / 0.0072 m³

≈ 1.25 kg/m³

Therefore, the density of the air filled in the polythene container is approximately 1.25 kg/m³.

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Olga's inertia moment approximately 42.7kgm². If Olga was ever to tuck her legs in, she would have a  10.6kgm² moment of inertia and overall body length would be cut in half.

A feature of substance known as inertia is the ability to maintain either a situation of rest or homogeneous motion in a straight line without the assistance of an external entity. A body's inertia is a quality that makes it resist any effort to move it by any force.

The unifying force of the earth is inertia. Literally. Without it, everything would not possess the magnetic forces required to create its current configuration. The heat and dynamic energy created by moving particles work to resist inertia.

(a) moment of inertia,

I = 1/3 mL²

m = 50.0 kg

L = 1.6 m

I = 1/3 (50.0kg) (1.6m)²

= 42.7kgm²

(b) Olga's body is just half the length

L = 1.6m / 2

= 0.8m

I = 1/3(50.0kg) (0.8m)²

= 10.6kgm²

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Answer:

Voltage, V = 12 V.

Explanation :

It is given that,

The power for the high beam is, P = 60.0 watts

Current flowing, I = 5 A

Car headlights have both low beams and high beams. The high beams may be necessary when it is very dark outside.

So, the voltage required for the high beam is 12 V.