Please help and if you have the answer if you can please explain how you got it :)!

Answer:

The force is  [tex]F = 21.48 \ N[/tex]

Explanation:

From the question we are told that

   The diameter of the wire is  [tex]d = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]

     The  pressure is  [tex]P = 120 \ a.t.m = 120 * 101.3 *10^{3} = 12156000 Pa[/tex]

Generally the radius of the of the wire is  

     [tex]r = \frac{d}{2}[/tex]

=>    [tex]r = \frac{ 1.5 *10^{-3}}{2}[/tex]  

=>    [tex]r = 7.5 *10^{-4} \ m[/tex]

The Area is evaluated as

     [tex]A = \pi r^2[/tex]

=>    [tex]A = 3.142 * 7.5 *10^{-4}[/tex]

=>    [tex]A = 1.7673*10^{-6} \ m^2[/tex]

Generally pressure is mathematically represented as

     [tex]P = \frac{F}{A }[/tex]

=>   [tex]F = P* A[/tex]

=>    [tex]F = 12156000 * 1.767*10^{-6}[/tex]

=>    [tex]F = 21.48 \ N[/tex]

Complete question :

A 12 m x 15 m house is built on a 12-cm-thick concrete slab.

What is the heat-loss rate through the slab if the ground temperature is 5°C while the interior of the house is 25°C

Answer:

3kW

Explanation:

Given the following :

Dimension of house :

Length = 12m

Width = 15m

Thickness of concrete slab (t) = 12cm

t in metres :

100cm = 1m

12cm = (12/100)m

= 0.12m

Ground temperature (Tg) = 5°C

Interior temperature = (Th) = 25°C

Thermal conductivity of concrete (K) is approximately 1 Wm/k

Using the relation:

Q = KA * [ (Th - Tg) / d]

A = Length * width = (12 *15) = 180

Q = (1 * 180) * [(25°C - 5°C) / 0.12]

Q = 180 * (20/0.12)

Q = 180 * 16.6666

Q = 3,000W = 3kW

The heat-loss rate is 3kW

Given that,

Dimension of house :

Length = 12m

Width = 15m

Thickness of concrete slab (t) = 12cm

We know that

100cm = 1m

so,

12cm = (12/100)m

= 0.12m

And,

Ground temperature (Tg) = 5°C

Interior temperature = (Th) = 25°C

Q = KA * [ (Th - Tg) / d]

A = Length * width = (12 *15) = 180

Q = (1 * 180) * [(25°C - 5°C) / 0.12]

Q = 180 * (20/0.12)

Q = 180 * 16.6666

Q = 3,000W

= 3kW

learn more about the temperature here: https://brainly.com/question/16940730

Answer:

Explanation:

To focus object at .7m , the image distance can be measured as follows

object distance u = .7m

focal length f = .05 m

image distance v = ?

from lens formula

[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f}[/tex]

[tex]\frac{1}{v} +\frac{1}{.7} = \frac{1}{.05}[/tex]

[tex]\frac{1}{v} =\frac{1}{.05} -\frac{1}{.7}[/tex]

v = .054 m

= 54 mm

when the object is at infinity , image is formed at focus ie at distance of

50 mm .

So lens position from sensor  where image is formed , varies from 54 mm to 50 mm .

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

Answer:

True

Explanation:

J. Henry Alston was known as a famous African American psychologist. He was known through his thorough study of the sensations of heat and cold.

He thereby became the first African American to publish his research findings on the perception of heat and cold in a major US psychology journal and was an important figure in the field.

Answer:

C. the motion of a spacecraft under gravitational influence

Answer:

The mass of the rule is 56.41 g  

Explanation:

Given;

mass of the object suspended at zero mark, m₁ = 200 g

pivot of the uniform meter rule = 22 cm

Total length of meter rule = 100 cm

0                          22cm                                  100cm

-------------------------Δ------------------------------------

↓                                                                       ↓

200g                                                                 m₂  

Apply principle of moment

(200 g)(22 cm - 0)     = m₂(100 cm - 22 cm)

(200 g)(22 cm) = m₂(78 cm)

m₂ =  (200 g)(22 cm)  / (78 cm)

m₂ = 56.41 g  

Therefore,  the mass of the rule is 56.41 g                                            

Answer:

6235.5J

Explanation:

Using ( nစ)p= ncp x change in temp

But cp= ( 1+ f/2)R

So cp= ( 1+ 3/2R

Cp= 5R/2

So = n x 5R/2x 150k

= 2 x 5/2x 8.314 x150

= 6235.5J

Answer:

2500 J

Explanation:

Q=(3/2)nRΔT

Q=(3/2)*2 mol*(8.314 J/mol*k)*100 k

Q=2494 J

Answer:

f = 127.48 Hz ,  T = 7.844 1⁻³ s ,  Ф = 3.38 ,     λ = 12.49 m

Explanation:

The general equation for the motion of a wave in a string is

          y = A sin (kx -wt + fi)

the expression they give is

         y = ym sin (0.503x + 801 t + 3.38 )

the veloicda that accompanies time is

      w = 801   rad / s

angular velocity is related to frequency

      w = 2π f

      f = w / 2π

      f = 801 / 2π

      f = 127.48 Hz

The period is the inverse of the frequency

      T = 1 / f

       T = 1 / 127.48

      T = 7.844 10⁻³ s

the csntnate of phase fi is the independent term

      Ф = 3.38

the wave vector accompanies the position k = 0.503 cm

       ka = 2pi /λ

       λ = 2 π / k

       λ = 2 π / 0.503

       λ = 12.49 m

Answer:

52.7 m

Explanation:

Given that

speed of the vehicle, v = 95 km/h

time of inattentiveness, t = 2 s

distance travelled, s = ?

Since we have the speed in km/h and the time in s, it would be best if we converted one of them to make sure we have all units in the same rank.

95 km/h = 95 * 1000/3600 m/s

95 km/h = 95000/3600 m/s

95 km/h = 26.38 m/s

Now, we use our derived speed in m/s

Speed of a moving vehicle is given by,

v = s/t, where

v = speed in m/s

s = distance travelled, in m

t = time spent, in s

if we make d the subject of formula by rearranging the equation, we have

s = v * t

distance travelled, s = 26.38 * 2

distance travelled, s = 52.7 m

therefore, during this inattentive period, 52.7 m was travelled.

Answer:

[tex]F_x=208.25\ N[/tex]

Explanation:

Given that,

Mass of a crate is 22 kg

It moved up along the 15 degrees incline without tipping.

We need to find the corresponding magnitude of force P. The force P is acting in horizontal direction.

It means that the horizontal component of force is given by :

[tex]F_x=F\cos\theta\\\\F_x=mg\cos\theta\\\\F_x=22\times 9.8\times \cos(15)\\\\F_x=208.25\ N[/tex]

So, the horizontal component of force is 208.25 N.

Explanation:

∑F = ma

60 N = (40 kg) a

a = 1.5 m/s²

Answer:

0

Explanation:

The angular acceleration of the tires is -2.2 rad/s².

If the car continues to decelerate at this rate, the time required to stop is 27.66 s.

The total distance traveled by the car before stopping is 210.96 revolutions.

The given parameters;

The angular acceleration of the tire is calculated as follows;

[tex]\omega _f^2 = \omega _i ^2 + 2\alpha \theta\\\\(\frac{16.67}{0.42} )^2 = (\frac{25.56}{0.42} )^2 + 2( 77 \ rev \times \frac{2 \pi \ rad}{1 \ rev} ) \alpha \\\\1575.33 = 3703.59 \ + \ 967.736 \alpha \\\\-2128.26 = 967.736 \alpha\\\\\alpha = \frac{-2128.26}{967.736} \\\\\alpha = - 2.2 \ rad/s^2[/tex]

When the car stops, the final angular speed = 0. The time for the motion is calculated as;

[tex]\omega _f = \omega _i + \alpha t\\\\0 = \omega _i + \alpha t\\\\0 = 60.86 + (-2.2)t\\\\0 = 60.86 - 2.2t\\\\2.2t = 60.86\\\\t = \frac{60.86}{2.2} \\\\t = 27.66 \ s[/tex]

The total distance traveled by the car before stopping;

[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2\\\\\theta = (60.86 \times 27.66) \ + \ (0.5 \times -2.2\times 27.66^2)\\\\\theta = 841.8 \ rad\\\\\theta = 841.8 \ rad \times\frac{1 \ rev}{2\pi \ rad} = 133.96 \ rev[/tex]

total distance = 133.96 + 77 = 210.96 revolutions.

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Answer:

1.3

Explanation:

it will taken to as in from of standard form

Answer:

1.3 * 10^-5

Explanation:

We are learning about scientific notation.

When a number becomes a decimal followed before with zeroes, we know that the value of that number is decreasing. So instead of usually doing a positive exponent, we will do a negative exponent indicating we are going back.

So let's not only count the amount of zeroes followed before 13, but the decimal.

0.000013

The original number "1.3" went back 5 spaces, therefore making our exponent 5.

1.3 * 10^-5

Answer:

The  correct option is  H

Explanation:

From the question we are told that

    The index of refraction of  coating is  [tex]n_1[/tex]

       The  index of refraction of material  is  [tex]n_2[/tex]

Generally the condition for constructive for a thin film interference is mathematically represented

            [tex]2 * t = [ m + \frac{1}{2}] \frac{\lambda}{n_1 }[/tex]

Here  t represents the thickness

For minimum thickness  m =  0

So

           [tex]2 * t =0 + \frac{1}{2}\frac{\lambda}{n_1 }[/tex]

=>        [tex]t =\frac{\lambda}{4n_1 }[/tex]

Answer:

its B

Explanation:

Answer:

It's B

Explanation: hope it helps ^w^

Answer:

The force per unit between the two parallel wires with same current flowing in the same direction is 5 x 10⁻⁶ N/m repulsive force.

Explanation:

Given;

current though the two parallel wires, I₁ and I₂ = 5A

distance between the two wires, R = 1 m

The force per unit of the wires is calculated as;

[tex]\frac{F}{L} = \frac{\mu_o I_1I_2}{2\pi R}[/tex]

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

Substitute in the given values into the equation and determine the force per unit length (F/L).

[tex]\frac{F}{L} = \frac{\mu_o I_1I_2}{2\pi R} \\\\ \frac{F}{L} = \frac{4\pi *10^{-7}*5*5}{2\pi *1}\\\\ \frac{F}{L} = 5*10^{-6} \ N/m \ (repulsive)[/tex]

Therefore , the force per unit between the two parallel wires with same current flowing in the same direction is 5 x 10⁻⁶ N/m repulsive force.

Answer:

A static observer hears a sound whose frequency is appreciably different from actual frequency because here the change in frequency of the sound due to doppler effect.

Explanation:

Given that,

Frequency = f

We know that,

The sound source of frequency f moves with constant velocity through a medium that is at rest.

A static observer hears a sound whose frequency is significantly different from actual frequency due to doppler effect.

We know that,

The doppler effect is defined as

[tex]f=f_{0}(\dfrac{v+v_{0}}{v-v_{s}})[/tex]

Where, f₀ = actual frequency

f = observe frequency

v = speed of sound

[tex]v_{o}[/tex] = speed of observer

[tex]v_{s}[/tex] = speed of source

Hence, A static observer hears a sound whose frequency is significantly different from actual frequency because here the change in frequency of the sound due to doppler effect.

Answer:

The distance of separation is [tex]d = 9.04 *10^{-6 } \ m[/tex]

Explanation:

From the question we are told that

    The wavelength is [tex]\lambda = 633\ nm = 633 *10^{-9} \ m[/tex]

     The  distance of the screen is [tex]D = 5.0 \ m[/tex]

     The  distance between the fringes is  [tex]y = 35 \ cm = 0.35 \ m[/tex]

Generally the distance between the fringes is mathematically represented as

       [tex]y = \frac{ \lambda * D }{d }[/tex]

Here d is the distance of separation between the slit

=>    [tex]d = \frac{ \lambda * D }{y }[/tex]

=>    [tex]d = \frac{ 633 *10^{-9} * 5 }{ 0.35 }[/tex]

=>   [tex]d = 9.04 *10^{-6 } \ m[/tex]

Answer:

After the bullet emerges the block moves at 0.99 m/s

Explanation:

Given;

mass of bullet, m₁ = 22 g = 0.022 kg

initial speed of the bullet, u₁ = 240 m/s

final speed of the bullet, v₁ = 150 m/s

mass of block, m₂ = 2.0 kg

initial speed of the block, u₂ = 0

Let the final speed of the block = v₂

Apply principles of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.022 x 240 + 2 x 0 = 0.022 x 150 + 2v₂

5.28 = 3.3 + 2v₂

5.28 - 3.3 = 2v₂

1.98 = 2v₂

v₂ = 1.98 / 2

v₂ = 0.99 m/s

Therefore, after the bullet emerges the block moves at 0.99 m/s

The body moves at a speed of 2.61m/s after the bullet emerges.

According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.

The formula for calculating the collision of a body is expressed as:

p = mv

m is the mass of the body

v is the velocity of the body

Based on the law above;

[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]

v is the final velocity of the body after the collision

Substitute the given parameters into the formula as shown:

[tex]0.022(240) + 2(0) = (0.022+2)v\\ 5.28 = 2.022v\\v=\frac{5.28}{2.022}\\v= 2.611m/s[/tex]

This shows that the body moves at a speed of 2.61m/s after the bullet emerges.

Learn more here: https://brainly.com/question/9537044

Answer :

(a). The speed of the block is 0.395 m/s.

(b). No

Explanation :

Given that,

Diameter = 20.0 cm

Power = 26.0 MW

Mass = 110 kg

diameter = 20.0 cm

Distance = 100 m

We need to calculate the pressure due to laser

Using formula of pressure

[tex]P_{r}=\dfrac{I}{c}[/tex]

[tex]P_{r}=\dfrac{P}{Ac}

Put the value into the formula

[tex]P_{r}=\dfrac{26.0\times10^{6}}{\pi\times(10\times10^{-2})^2\times3\times10^{8}}[/tex]

[tex]P_{r}=2.75\ N/m^2[/tex]

We need to calculate the force

Using formula of force

[tex]F=P\times A[/tex]

[tex]F=P\times \pi r^2[/tex]

Put the value into the formula

[tex]F=2.75\times\pi (0.01)^2[/tex]

[tex]F=0.086\ N[/tex]

We need to calculate the acceleration

Using formula of force

[tex]F=ma[/tex]

Put the value into the formula

[tex]0.086=110\times a[/tex]

[tex]a=\dfrac{0.086}{110}[/tex]

[tex]a=0.000781\ m/s^2[/tex]

[tex]a=7.81\times10^{-4}\ m/s^2[/tex]

(a). We need to calculate speed of the block

Using equation of motion

[tex]v^2=u^2+2ad[/tex]

Put the value into the formula

[tex]v=\sqrt{2\times7.81\times10^{-4}\times100}[/tex]

[tex]v=0.395\ m/s[/tex]

(b). No because the velocity is very less.

Hence, (a). The speed of the block is 0.395 m/s.

(b). No

Answer:

The frequency changes by a factor of  0.27.

Explanation:

The frequency of an object with mass m attached to a spring is given as

[tex]f[/tex] = [tex]\frac{1}{2\pi } \sqrt{\frac{k}{m} }[/tex]

where [tex]f[/tex] is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

[tex]f_{n}[/tex] = [tex]\frac{1}{2\pi } \sqrt{\frac{k}{14m} }[/tex]

simplifying, we have

[tex]f_{n}[/tex] = [tex]\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }[/tex]

[tex]f_{n}[/tex] = [tex]\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }[/tex]

if we divide this final frequency by the original frequency, we'll have

==> [tex]\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }[/tex]  ÷  [tex]\frac{1}{2\pi } \sqrt{\frac{k}{m} }[/tex]

==> [tex]\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }[/tex]  x  [tex]2\pi \sqrt{\frac{m}{k} }[/tex]

==> 1/3.742 = 0.27

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

(b). P₂ = 236500 Pa

Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L =  90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s            (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s               (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁²  (inside)      =       P₂ + pgh₂ + 1/2PV₂²   (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

Answer:

B. The maximum angle decreases

Explanation:

If θ be the maximum angle of a slope that allows a crate placed on it to remain at rest , following condition exists .

tanθ = μ , θ is called angle of repose . μ is coefficient of static friction .

So the tan of angle of repose θ is proportional to coefficient of static friction.

If coefficient of static friction is less than .7 , naturally angle of repose will also become less ,ie,  it at lower angle of inclination , the object will start slipping .

Answer:

The average force of wind on the building is 4.728 x 10¹² N

Explanation:

Given;

speed of the air wind, v = 124 km/h

dimension of the building, 42 m wide by 73 m high

density of the air, ρ = 1.3 kg/cm³ =

speed of the air in m/s = 124/3.6 = 34.44 m/s

Area of the building, A = 42 m x 73 m = 3066 m²

density of the air in (k.g/m³);

[tex]\rho = \frac{1.3 \ kg}{cm^3} *(\frac{100\ cm}{1 \ m} )^3\\\\\rho = \frac{1.3 \ kg}{cm^3} *\frac{10^6\ cm^3}{1 \ m^3} = \frac{1.3*10^6 \ kg}{m^3}[/tex]

The average force of wind on the building;

F = mass flow rate x velocity

F = (ρvA) x V

F = ρAv²

F = 1.3 x 10⁶ x 3066  x (34.44)²

F = 4.728 x 10¹² N

Therefore, the average force of wind on the building is 4.728 x 10¹² N

Answer:

L = L0 ( 1 - v^2/c^2))1/2     where L0 is the proper length

L = 10 L-y (1 - .9^2)^1/2 = 4.36 L-y   length of pole measured by ship

t = 4.36 L-y / .9 c = 4.84 y  since the ship travels at .9 c

Answer:

[tex] \boxed{\sf Shortest \ possible \ time = 7 \ seconds} [/tex]

Given:

Distance travelled (s) = 8.4 cm

velocity (v) = 1.2 cm/s

To Find:

Shortest possible time (t) in which a bacterium travel a distance 8.4 cm across a Petri Dish

Explanation:

[tex] \boxed{ \bold{\sf Time \ (t) = \frac{Distance \ travelled \ (s)}{Velocity \ (v)}}}[/tex]

Substituting values of Distance travelled (s) & Velocity (v) in the equation:

[tex] \sf \implies t = \frac{8.4}{1.2} [/tex]

[tex] \sf \implies t = \frac{7 \times \cancel{1.2}}{ \cancel{1.2}} [/tex]

[tex] \sf \implies t = 7 \: s[/tex]

Answer:

GPS receivers provide location in latitude, longitude, and altitude. They also provide the accurate time. GPS includes 24 satellites that circle Earth in precise orbits.