Find the pressure exerted on the floor by a 100N box whose bottom area is 40cm x
50cm.

11 A is the amount of current that crosses the wire.

According to Ohm's law,

If all other physical parameters and temperatures remain constant, the voltage across a conductor is directly proportional to the current flowing through it.

V ∝ I

V= IR where,

V = Voltage across a conductor

I = Current flowing through the conductor

R= Resistance and also known as constant of proportionality

Given,

Resistance of wire(R)  = 10 ohms

Voltage applied(V) = 110V

Current(I) = ?

According to the formula

110V  = I × 10Ω

I = 110/10 = 11 A

Hence, current intensity  of 11 A crosses a wire of 10 ohms if the voltage applied to the extremes of the wire is 110 V

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A mechanical wave transfers energy through matter.

A mechanical wave is a wave that is an oscillation of matter and is responsible for the transfer of energy through a medium.

Answer: Like all waves, mechanical waves transport energy. This energy propagates in the same direction as the wave. A wave requires an initial energy input; once this initial energy is added, the wave travels through the medium until all its energy is transferred. In contrast, electromagnetic waves require no medium, but can still travel through one.

A wave is a transmission of a disturbance from one point to another. Wave involves transmission of energy from one point to another through a materials medium.

Mechanical waves vibrate in matter and transfer energy. Explore the production and propagation of mechanical waves and discover the factors that affect mechanical waves including elasticity, tension, temperature, and density

Therefore,

A mechanical wave transfers energy through matter.

A complete question is is given below:

What best describes a possible transfer of energy?

a. A mechanical wave transfers energy through matter.

b. A longitudinal wave transfers energy perpendicular to wave motion.

c. A transverse wave transfers energy parallel to wave motion.

d. A sound wave does not need a medium to transfer energy through.

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The tension on the left wire 642N and the tension on right wire is 208N

Given that the force of scaffold is 200- N and the length of the scaffold is 18m

We need to find the tension in each wire

Tension Force is the force that is transmitted through a cable, rope, wire or string when it is pulled tight by forces acting from opposite ends

Here

∑ [tex]T_b=0[/tex]

∴ [tex]W_B(15m)+W_S(9m)-T_L(18m) = 0[/tex]

∴ [tex]T_L = \frac{W_B(15m)+W_S (9m)}{18m}\\ T_L = \frac{650(15)+200(9)}{18}\\ T_L = 642 N[/tex]

Hence the tension on the left side is 642N

∑[tex]T_a[/tex] = 0

[tex]T_R(18m)-W_B(3m)-W_S(9m)=0[/tex]

∴[tex]T_R =\frac{W_B(3m)+W_S(9m)}{18m}[/tex]

∴[tex]T_R = \frac{650(3)+200(+9)}{18}\\[/tex]

∴[tex]T_R = 208N[/tex]

Hence the tension on the right side is 208N

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Answer:

The acceleration of the object is 3 m/s2.

Explanation:

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Answer: An archer pulls her bowstring back 0.400m by exerting a force that increases uniformly from zero to 230N. The equivalent spring constant of the bow will be 575N/m

Explanation: To find the answer, we have to know about the Simple Harmonic Motion.

                   [tex]F[/tex]∝[tex]-x[/tex]

                   [tex]F=-kx[/tex] where, k is called the spring constant.

                       [tex]k=-F/x= (230)/ (0.400) =575 N/m.[/tex]

Thus, we can conclude that, the equivalent spring constant of the bow will be 575N/m.

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Light can travel in the air or a vacuum and the speed is constant whether the light source is stationary or moving which is denoted as option B.

This is referred to the portion of the electromagnetic spectrum which is perceived by the human eyes.

It has a constant speed and can travel in an air or vacuum. This theefrore makes it the most appropriate choice ion this scenario.

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B. The green ball has more mass, so it had more kinetic energy, the kinetic energy transferred to sound energy.

This is the energy possessed by a moving object, and it depends on mass and speed of the object.

K.E = 0.5mv²

where;

Thus, we can conclude that, the green ball has more mass, so it had more kinetic energy, the kinetic energy transferred to sound energy.

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The resistance of the resistor is .

Given:-  A series combination of a 48.0-mF capacitor and a resistor are connected to a  battery. After one second, the voltage across the capacitor is .

To Find:- We have to find the value of the resistance of the resistor.

By using the concept of the resistance it will be solved.

According to the problem,

The value of the resistance of the resistor is .

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Answer:

Its P path describes the change of state i.e from solid to liquid.

What is solid ?

"A solid is that state of matter which has a fixed shape, mass, and volume. It suffers very small changes in volume by changing the temperature. It can not be compressed, e.g. — Sand, Wood, Copper, Ice, etc."

What is liquid ?

"A liquid is a sample of matter that conforms to the shape of a container in which it is held, and which acquires a defined surface in the presence of gravity. The term liquid is also used in reference to the state, or condition, of matter having this property."

The minimum height of the dive needed to achieve the given speed is v = 69 m/s is 242.9 m.

Given information:

The mass of peregrine falcon is,  m = 480

The final speed reached by the peregrine falcon in a vertical dive is, v = 69 m/s

It is given that the falcon is diving vertically downward. It can be compared with the same situation as the free-falling object under the effect of gravity only. So, the initial velocity of the falcon will be u = 0 m/s  as the motion starts with rest.

The value of the gravitational acceleration of gravity is, g = 9.80 m/s²

Now, using the third equation of motion, the minimum height  required for the final speed will be,

v² - u² = 2gh

69² - 0² = 2 × 9.8 × h

h = 242.9m.

Therefore, the minimum height of the dive needed to achieve the given speed is 242.9 m.

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The vector force the wall is applying to the boy is -15 N; option D

A reaction force is a force which is equal in magnitude but oppositely directed to a force acting in a particular direction.

The reaction force on the boy is equal in magnitude to the weight of the boy but has an opposite direction.

In conclusion, the vector force applied by the wall on the boy is -15 N.

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The height of the cliff above the ground is 51.76 m.

h = vt + ¹/₂gt²

where;

h = 0 + ¹/₂gt²

h = ¹/₂gt²

h = ¹/₂(9.8)(3.25)²

h = 51.76 m

Thus, the height of the cliff above the ground is 51.76 m.

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Answer:

See below

Explanation:

Kinetic Energy = 1/2 * m * v^2

                        = 1/2 * 200 * (15)^2 = ________ joules

The kinetic energy of the 200 kg object moving with a speed of 15 m/s is 11250 Joules (J).

Given:

Mass of the object (m) = 200 kg

Speed of the object (v) = 15 m/s

For the kinetic energy (KE) of an object, we use the formula:

KE = (1/2) × m × v²

where:

KE is the kinetic energy,

m is the mass of the object,

v is the speed of the object.

The kinetic energy of the object:

KE = (1/2) × 200 × (15 )²

KE = (1/2) × 200 × 225

KE = 11250

KE = 11250 Joules (J)

Therefore, the kinetic energy of the 200 kg object moving with a speed of 15 m/s is 11250 Joules (J).

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The total force on the ground due to the bricks is 47,040 N.

The total force on the ground due to weight of the bricks is calculated as follows;

Fₙ = mg

where;

Fₙ = (1500 x 3.2) x 9.8

Fₙ = 47,040 N

Thus, the total force on the ground due to the bricks is 47,040 N.

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Answer:

The decrease in energy for a lightning flash, with a potential difference between a cloud and the ground, is [tex]1.5 \times 10^{9}\text{ V}[/tex] and the quantity of charge transferred is 23.7 C is [tex]$3.555 \times 10^{10} \text{ J}$[/tex].

Explanation:

In a lightning flash, the potential difference between a cloud and the ground is [tex]1.5 \times 10^{9}\text{ V}[/tex] and the quantity of charge transferred is 23.7 C.

It is required to determine the decrease in energy of the transferred charge.

It is known that the decrease in energy is calculated by multiplying the potential difference by the quantity of charge. Mathematically it is expressed as:

[tex]\Delta U=q \Delta V[/tex]

In this problem, the value of q is 23.7 C and the value of [tex]\Delta V[/tex] is [tex]1.5 \times 10^{9}\text{ V}[/tex].

Therefore the decrease in energy is:

[tex]$$\begin{aligned}\Delta U&=q \Delta V\\&=23.7\times 1.5 \times 10^{9} \text{ J}\\&=3.555\times 10^{10} \text{ J}\end{aligned}$$[/tex]

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Answer:

Explanation:

Step 1

1 of 2

\textbf{Given}Given

n_{w}=1.33n

w

​

=1.33

$\theta_w=31\text{ ^\circ

∘

}$

$\theta_b=27\text{ ^\circ

∘

}$

\textbf{Approach}Approach

In this problem we are going to use Snell's law.

\textbf{Solution}Solution

The definition of Snell's law of refraction is

\begin{align} {n_1\cdot\sin \theta_1}={n_2\cdot \sin \theta_2} \end{align}

n

1

​

⋅sinθ

1

​

=n

2

​

⋅sinθ

2

​

​

​

where indexes 11 and 22 represent two different mediums. Since the motion is in the water we write

\begin{align} &{n_{w}\cdot\sin\theta_{w}}={n_{b}\cdot\sin\theta_{b}} \\ &{n_b}={n_{w}\cdot \frac{\sin\theta_{w}}{ \sin \theta_{b}} } \\ &{n_b}={1.33\cdot \frac{\sin31^\circ}{ \sin27^\circ} } \\ &\boxed{{n_b}=1.5} \end{align}

​

n

w

​

⋅sinθ

w

​

=n

b

​

⋅sinθ

b

​

n

b

​

=n

w

​

⋅

sinθ

b

​

sinθ

w

​

​

n

b

​

=1.33⋅

sin27

∘

sin31

∘

​

n

b

​

=1.5

​

​

The direction of the light can be changed by reflection from a reflecting surface and by the refraction of the light through a medium of different refractive index.

What is reflection?

The phenomenon of the light in which the light ray bounces back from the surface on which they fall is called the reflection.

Now when the light ray falls on a surface that is well polished like the mirror, then the light ray changes its direction. So by using the combination of mirrors, the direction of the light ray can be changed.

What is refraction?

Refraction is a phenomenon in which light rays change their path slightly when it is passing through a transparent medium with a different refractive index.

The concave and convex lens uses this property of light. So by using the convex and concave lens, the direction of the light ray can be changed.

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Both balls will take the same amount of time to hit the ground if they are dropped and fired from the same height.

What is Projectile motion:

For a ball fired horizontally from same height and time, there is only a vertical acceleration on it towards the ground.

This acceleration is equal to the acceleration due to gravity (g = 9.81 m/s^2).

A ball dropped from the same height will also experience the same vertical acceleration downwards as the horizontal acceleration.

Therefore,

Both balls will take the same amount of time to hit the ground if they are dropped and fired from the same height.

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The electric field due to charge q1 and q2 at the middle point of the line joining the charges q1 and q2 is 8.64*10^(4) N/C.

Note: You have not given the point where the resultant magnetic field has to be calculated. Most probably it is asked to find the electric field at the middle point of the line joining the point charges.

Electric field: The electric force exerted on a unit charge is called the electric field. Electric field due to a charge is calculated using the formula,

E=kq/r^2

where k is a constant whose value is 9*10^(9) N m^2/ C^(2), q is the charge and r is the distance from the charge to the point where the electric field has to be calculated. In the given case, the electric field has to be calculated at the middle point of the line joining charges q1 and q2.

Calculation of electric field due to charge q1:

Given that q1=+12 nC or q1=+12*10^(-9) C and the distance of the charge q1 from the center d1=10.0/2 cm or d1=5*10^(-2) m, the magnitude of the electric field E1 due to charge q1 is,

E1=kq1 /d1^(2)

E1=9*10^(9)*12*10^(-9) / (5*10^(-2))^2

E1=4.32*10^(4) N/C

The direction of the electric field E1 at the middle point is towards the negative charge.

Calculation of electric field due to charge q2:

Given that q2=-12 nC or q1=-12*10^(-9) C and the distance of the charge q2 from the center d2=10.0/2 cm or d2=5*10^(-2) m, the magnitude of the electric field E2 due to charge q2 is,

E2=kq2 /d2^(2)

E2=9*10^(9)*12*10^(-9) / (5*10^(-2))^2

E2=4.32*10^(4) N/C

The direction of the electric field E2 at the middle point is towards the negative charge.

Total electric field:

The total electric field is given by the addition of the electric field. The direction of the electric field is the same for both charges, hence total electric field E is,

E=E1+E2

E=4.32*10^(4) +4.32*10^(4)

E=8.64*10^(4) N/C

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The x and y components of the force will be 63.6 N and 63.6 N  respectively.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

Force, F=90 N

Angle,θ = 45°

The force of 90N at an angle of 45 degrees;

The force is resolved in the x and y direction as;

[tex]\rm F_x = 90 cos 45 ^ 0 \\\\ F_x = 63.6 \ N[/tex]

[tex]\rm F_y = 90 sin 45^0 \\\\ F_y =63.6 \ N[/tex]

Hence x and y components of the force will be 63.6 N and 63.6 N  respectively.

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Magnitude of electric force = 15 x 10⁻¹⁰ N

Magnitude of weight force = 64.092 x 10⁻⁶ N

Given

Diameter of rain drop = 2.5mm

Radius = 1.25 mm = 1.25 × 10⁻³ m/sec

Drop has a charge = +15pC

15pC = 15×10⁻¹²C

Electric field (E) = 100 N/C

Electric force on the charge

= charge x electric field = qE

= 15 x 10⁻¹² x 100

= 15 x 10⁻¹⁰ N.

This force will act in upward direction as the force on the positive charge will always acts in the direction of the electric field.

Volume of rain droplet(V) = 4/3 π R³

Volume V = 4/3 x 3.14 x ( 1.25 x 10⁻³)³

                 = 6.54 × 10⁻⁹m³

As Density of water = 1000 kg / m³

Density =Mass/Volume

So, Mass of rain droplet = 1000 x 6.54 x 10⁻⁹ kg

= 6.54 x 10⁻⁶ kg .

Now, Weight(W) = mg

= 6.54 x 10⁻⁶ x 9.8

= 64.092 x 10⁻⁶ N.

As the electric force is 15 × 10⁻¹⁰ N

So, weight force is more than the electric force.

Hence, the magnitude of the electric force on the droplet is less as compare to the weight force.

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Assumed that resistors are connected in series .

Current=I

Max Planck utilized the concept of energy quantization in order to develop a model for blackbody radiation that accurately fit experimental results.

Electromagnetic waves emitted by using a blackbody are referred to as blackbody radiation. A blackbody is bodily found out through a small hole within the wall of a hollow space radiator.

The depth of blackbody radiation depends on the wavelength of the emitted radiation and the temperature T of the blackbody.

When the temperature of a black frame increases, it is discovered that the wavelength similar to most electricity changes from 0.26 μm to 0. Thirteen μm Then the ratio of the emissive power of the body at the respective temperature is.

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The centripetal force on the ball is 140.9 N; option A

The centripetal force on the ball is given by the formula below:

where;

The linear velocity, v = wr

v = 5 * 2π/4.4 * 1.4 = 9.93 m/s

F = 10² * 2/1.4 = 140.9 N

In conclusion, the centripetal force is derived from the mass, velocity and radius of the path.

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Answer:

The suction pressure decreases and the superheat increases when the liquid line is restricted and the supply of refrigerant to the metering device is reduced.

Explanation:

1. The five components of refrigeration are:

       The compressor limits the vapor released by the refrigerant. This            

       causes a rise in pressure (in refrigerant), which then pushes the  

       vapor into the coils on the outside of the refrigerator.

2. Now when the cooler air meets the warm gas present in the coils, it

   gets converted into liquid form.

3. Thus, when the liquid form is at high pressure, the refrigerant then  

   cools down as it flows through the coils placed in the fridge ( in both

   freezing and normal sections).

4. The refrigerant also absorbs the warm air present in the fridge, which  

   causes it to evaporate and flow back through the compressor and the

   cycle repeats in the same form.

Thus, when the liquid line is restricted and the supply of refrigerant to the metering device is reduced it causes a decrease in suction pressure and an increase in superheat.

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         The apparatus must be Positively Charged.

Charge interaction:

        1. opposite charge objects attract each other

        2. like charge object repel each other

        3. neutral object with any charges will attract each other.

  So, In the question it said that after rubbing on apparatus pith ball

   become negative charge which means it has gained the electron.

          Thus, the apparatus must have lose electron and would be

          positively charged.

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The kinetic coefficient of friction of the crate is 0.1506

Let's assume that both workers are applying positive pressure to the crate. A body cannot change its state of motion whether it is at rest or moves uniformly, according to Newton's First Law (at constant velocity). The magnitude of the friction force must therefore equal the total of the two external forces. The crate's equilibrium equations are as follows:

∑Fₓ = P + T - μ . N = 0

∑[tex]F_{y}[/tex] = N - W = 0

Where:

[tex]P[/tex] - Pushing force, measured in newtons.

[tex]T[/tex] - Tension, measured in newtons.

μ- Coefficient of kinetic friction, dimensionless.

[tex]N[/tex]- Normal force, measured in newtons.

W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P + T - μ . W = 0

We finally clear the coefficient of kinetic friction and apply the definition of weight :

μ [tex]=\frac{P + T}{m.g}[/tex]

We know that P = 390 N, T = 260 N, m = 440 kg and [tex]g= 9.807 \frac{m}{s^2}[/tex] ;

μ = 0.1506

Therefore, the kinetic coefficient of friction of the crate is 0.1506

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Flexible cord sets and power-supply cords shall not be concealed behind building , structural ceilings, suspended or dropped ceilings, floors or walls, or run through doorways, windows, or similar openings.

Hence Option (b) All of these is correct

Flexible cord and flexible cable usage are now outlined in Section 400.10, while prohibited uses are listed in Section 400.12.

Note 5 under 400.12 now includes a new exclusion for flexible wires and cords (uses not permitted).

Flexible cables, flexible cord sets, and power supply cords are not permitted in areas that are "concealed by walls, floors, or ceilings or positioned above suspended or dropped ceilings," according to Note 5.

The following uses for flexible cables, flexible cord sets, and power supply cords are prohibited as well:

(1) As an alternative to a structure's fixed wiring.

(2) Run-through openings in structural ceilings, floors, suspended ceilings, or dropped ceilings.

(3) When run through windows, doorways, or comparable openings.

(4) If attached to building surfaces.

Hence, Option (b) All of these is correct .

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The ballpoint from a pen at a distance of about  4000  kilometers.

Alpha Centauri is the closest star device. Distance from Earth 4,365 al (forty 000 billion kilometers. The solar (yellow famous person) distance from Earth is 150 000 000 km, the exoplanet distance from Alpha Centauri b (orange famous person) 6 000 000 km.

Shifts in starlight from Proxima Centauri, found over extra than 2 years, screen its 1/3 planet. Astronomers have observed a 3rd planet orbiting Proxima Centauri, the star closest to the solar.

Proxima Centauri, the nearest megastar to our own, remains forty,208,000,000,000 km away. (Or about 268,770 AU.)

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Explanation:

Hello !

look at the attachment above ☝️ and if you have any questions your welcome.

a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

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