During takeoff, the sound intensity level of a jet engine is 170 dB at a distance of 40 m. What is the sound intensity level at a distance of 1.0 km?

Answer:

0.2 mm

Explanation:

Assuming the Caliper has 1 mm marks on the main scale. Again, we assume that the 20 division coincides with the 16 main scale division, then we can find the least count by saying

For the given Vernier calipers,

Let one main scale division be = 1 mm.

Since there are 20 Vernier scale divisions that happen to be equal to 16 Main Scale Divisions, we can calculate that

1 Vernier Scale Division = 16/20 Main Scale Division

1 Vernier Scale Division = 0.8 mm.

The least count is finally calculated by saying that

Least Count = 1 Main Scale Division - 1 Vernier Scale Division

Least Count = 1 - 0.8

Least Count = 0.2 mm

Therefore, from the assumptions made above, the least count is 0.2 mm

The net force on the balloon is zero, so it has no acceleration. So by Newton's first law, if the balloon has some velocity in a given direction, it will continue moving in that direction at a constant speed. If it's motionless, it stays motionless.

Answer:

2.48*10^-5 N/m

Explanation:

The magnitude force per unit length on parallel wires separated by a distance D and currents I1 & I2.

F/l = µ.I1.I2/2πd

Where

µ = 4π *10^-7

I1 = 1.80 A

I2 = 3.45 A

d = 5 cm = 0.05 m

F/l = (4π*10^-7 * 1.8 * 3.45) / (2 * π * 0.05)

F/l = (1.2568*10^-6 * 6.21) / 0.3142

F/I = 7.8*10^-6 / 0.3142

F/l = 2.48*10^-5 N/m

Therefore, the magnitude of the force per unit length is 2.48*10^-5 N/m

Answer:

for what?

Explanation:

d=S x T

or

d=vt+1/2at2

srry if wrong but

hope this helps

take care

Answer:

[tex]a = 16\ m/s^2[/tex]

Explanation:

When the velocity changes uniformly, the object has a constant acceleration. The acceleration, the velocities, and the distance are related by the equation:

[tex]v_f^2=v_o^2+2ax[/tex]

Where:

vf = final velocity

vo = initial velocity

a = acceleration

x = distance

Solving for a:

[tex]\displaystyle a=\frac{v_f^2-v_o^2}{2x}[/tex]

The car travels a distance of x=200 m and the velocities are:

vo = 216 Km/h

vf = 360 Km/h

Both velocities must be converted to meters by seconds.

vo = 216 Km/h *1000/3600 = 60 m/s

vf = 360 Km/h *1000/3600 = 100 m/s

The acceleration is:

[tex]\displaystyle a=\frac{100^2-60^2}{2*200}[/tex]

[tex]\displaystyle a=\frac{10000-3600}{400}[/tex]

[tex]\displaystyle a=\frac{6400}{400}[/tex]

[tex]\mathbf{a = 16\ m/s^2}[/tex]

Answer:

During an eccentric contraction, when an active muscle, fiber or myofibril is actively stretched, its force increases substantially during the stretch.

Answer:

9.78 m/s²

Explanation:

To solve this, we use the gravitational formula

g = GM/r², where

g = acceleration due to gravity

G = gravitational constant

M = mass of the planet

r = radius of the planet

From the question, we got that the mass of the planet is

M = 9ME, where ME = 5.95*10^24

M = 9 * 5.95*10^24

M = 5.355*10^25 kg

Also, the Radius of the planet, R = 3RE, where RE = 6.37*10^6

R = 3 * 6.37*10^6

R = 1.911*10^7 m

On applying the values of both R and M to the equation, we get

g = GM/r²

g = (6.67*10^-11 * 5.355*10^25) / (1.911*10^7)²

g = 3.57*10^15/3.65*10^14

g = 9.78 m/s²

Therefore, the acceleration due to gravity on the planet is 9.78 m/s²

Please vote brainliest if it helped you <3

Answer:

Vigorous activity

Explanation:

Vigorous activity

Answer:

1.6441×10^7m

Explanation:

The distance above the surface of the earth can be calculated by finding the difference between the distance between the masses of the object and the radius of the earth which is expressed below

r= RE + h

Where

h=distance above the surface of the earth

r= distance between the masses

RE= radius of the earth= 6.3781×106 m

acceleration due to gravity can be expressed as

(g)= GmE/r^2.

Where r= distance between the masses of object

mE= mass of the earth= 5.972 × 10^24 kg

G =gravitational constant= 6.67×10^-11 m3 s-2 kg-1),

g= acceleration due to the earth's gravity= 0.765

Making r subject of the formula

r=√GmE/g

r=√[(6.67×10^-11) ×5.972 × 10^24 )]/0.765

=√5.207×10^15

=2.282×10^7m

Substitute the values in below expresion

h= r - RE

=(2.282×10^7)-(6.3781×106 m)

=1.6441×10^7m

Hence, distance above the surface of the earth is 1.6441×10^7m

Answer:

D. 2.93*10^-10N

Explanation:

A.pex

Answer:

5.3 × 10^(-8) m

Explanation:

We are given;

Shear force; F = 400 N

Length of cube; L_o = 30 cm = 0.3 m

Shear modulus; S = 2.5 × 10^(10) N/m²

Now,the resulting relative displacement can be gotten from the formula;

F = A × S × Δx/L_o

Where Δx is resulting relative displacement

A is area.

Area of cube = (L_o)² = 0.3² = 0.09

Thus, making Δx the subject, we have;

Δx = (F × L_o)/(A × S)

Plugging in the relevant values;

Δx = (400 × 0.3)/(0.09 × 2.5 × 10^(10))

Δx = 5.3 × 10^(-8) m

Answer:

[tex]a=8.34\ m/s^2[/tex]

Explanation:

Given that,

The mass of a man, m = 60 kg

When the elevator rises, the scale rads 818 N

We need to find the acceleration of the elevator. Let it is a. The equation of motion can be written as :

F - mg = ma

Put g = 9.8 m/s²

[tex]a=\dfrac{F-mg}{m}\\\\a=\dfrac{818-60(9.8)}{60}\\\\=8.34\ m/s^2[/tex]

So, the acceleration of the elevator is [tex]8.34\ m/s^2[/tex]

Answer:

Standing Wave

Explanation:

Destructive interference. Occurs when two or more waves combine to produce a wave with a smaller displacements

Answer:

the students where dided

Answer:

Explanation:

The formula for calculating the coefficient of kinetic friction between the crate and the ramp is expressed as;

η = F/R

F is the frictional force

R is the normal reaction

since F = mgsinθ (Force acting parallel to the ramp)

R = mgcosθ

Substitute into the formula

η = mgsinθ/mgcosθ

η = sinθ/cosθ

η = tanθ

Given

θ = 35°

η = tan35°

η = 0.7

Hence the coefficient of kinetic friction between the crate and the ramp is 0.7

Answer:

[tex]\theta=5.71^{o}[/tex]

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

[tex]\sum F_{y}=0[/tex]

[tex]-W_{y}+N=0[/tex]

[tex]N=W_{y}[/tex]

so:

[tex]N=mgcos(\theta)[/tex]

now we can go ahead and do a sum of forces in the x-direction:

[tex]\sum F_{x}=0[/tex]

the sum of forces in x is 0 because it's moving at a constant speed.

[tex]-f+W_{x}=0[/tex]

[tex]-\mu_{k}N+mg sen(\theta)=0[/tex]

[tex]-\mu_{k}mg cos(\theta)+mg sen(\theta)=0[/tex]

so now we solve for theta. We can start by factoring mg so we get:

[tex]mg(-\mu_{k} cos(\theta)+sen(\theta))=0[/tex]

we can divide both sides into mg so we get:

[tex]-\mu_{k} cos(\theta)+sen(\theta)=0[/tex]

this tells us that the problem is independent of the mass of the object.

[tex]\mu_{k} cos(\theta)=sen(\theta)[/tex]

we now divide both sides of the equation into [tex]cos(\theta)[/tex] so we get:

[tex]\mu_{k}=\frac{sen(\theta)}{cos(\theta)}[/tex]

[tex]\mu_{k}=tan(\theta)[/tex]

so we now take the inverse function of tan to get:

[tex]\theta=tan^{-1}(\mu_{k})[/tex]

so now we can find our angle:

[tex]\theta=tan^{-1}(0.10)[/tex]

so

[tex]\theta=5.71^{o}[/tex]

Answer:

V=voltage

I=current

R=resistance

Explanation:

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

There are 3 forces acting on the stoplight:

• its weight W, with magnitude W = 100 N, pointing directly downward

• two tension forces T₁ and T₂ with equal magnitude T₁ = T₂ = T = 1000 N, both making an angle of θ with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ F = T₁ sin(θ) + T₂ sin(180° - θ) - W = 0

We have sin(180° - θ) = sin(θ) for all θ, so the above reduces to

2T sin(θ) = W

2 (1000 N) sin(θ) = 100 N

sin(θ) = 0.05

θ ≈ 2.87°

If y is the vertical distance between the stoplight and the ground, then

tan(θ) = (15 m - y) / (100 m)

Solve for y :

tan(2.87°) = (15 m - y) / (100 m)

y = 15 m - (100 m) tan(2.87°)

y ≈ 9.99 m

Heya!!

For calculate velocity, lets applicate formula

                                                 [tex]\boxed{d = v * t}[/tex]

                                               Δ   Being   Δ

                                         d = Distance = 3790 m

                                             t = Time = 249 s

                                             v = Velocity = ?

⇒ Let's replace according the formula and clear "v":

[tex]\boxed{v= 3790\ m / 249\ s}[/tex]

⇒ Resolving

[tex]\boxed{ v = 15,22 \ m/s }[/tex]

Result:

The velocity is 15,22 meters per second.

Good Luck!!

Answer:

There is NO net force acting on the hanging weight. The net force is ZERO.

Explanation:

If the object (the weight) is NOT moving, and in static equilibrium hanging from the support, that means that the net force on it is ZERO. There are two forces acting, one is that due to gravity and the other one the tension of the string, Both of these forces must be equal in magnitude but pointing in different directions, so they cancel each other.

Hello!!

For calculate the GPE let's applicate the formula:

[tex]\boxed{GPE = m g h}[/tex]

[tex]\textbf{Being:}[/tex]

[tex]\sqrt{}[/tex] GPE = Gravitational potential energy = ?

[tex]\sqrt{}[/tex] m = Mass = 100 kg

[tex]\sqrt{}[/tex] g = Gravity = 10 m/s²

[tex]\sqrt{}[/tex] h = Height = 40 m

⇒ [tex]\text{Then let's \textbf{replace it according} we information:}[/tex]

[tex]GPE = 100 \ kg * 10 \ m / s ^{2} * 40 \ m[/tex]

⇒ [tex]\text{Let's resolve it: }[/tex]

[tex]GPE = 40000 \ J[/tex]

[tex]\textbf{Result:}\\\text{The gravitational potential energy is \textbf{40 000 Joules}}[/tex]

Answer:

What ever, thanks for the points

Explanation:

Answer:

D i think, sorry if i got it wrong

Explanation:

Answer:

Answer B (Increased by 2%)

Explanation:

Recall that the period (T) of a pendulum doesn't depend on the pendulum's mass, but depends on the pendulum's length (L) and on the local acceleration of gravity (g) via the formula:

[tex]T=2\pi\,\sqrt{\frac{L}{g} }[/tex]

There fore, if the length of the pendulum is increased by 4% (0.04 in decimal form), then the new length becomes: L + 0.04 L = 1.04 L

and therefore the period will change by:

[tex]T'=2\pi\,\sqrt{\frac{1.04\,L}{g} } = \sqrt{1.04} \,T\approx 1.02\,T[/tex]

Which means that the period was increased to about 2 % :

T + 0.02 T = 1.02 T

Answer:

3.39

Explanation:

Given that

Volume, v = 3.92*10^3 cm³ we convert the volume from cm³ to m³ and we have 0.00392

Pressure, P = 1.25*10^5 Pa

Average Kinetic Energy, K.E = 3.6*10^-22 J

We use the gas law formula,

PV = nRT

Making n subject of the formula, we have

n = PV/RT

Solving for n, we have

n = (1.25*10^5 Pa * 0.00392 m³) / 8.314 * T

n = 4.9*10^8 / 8.314 * T

n = 490 / 8.314T

n = 58.94/T

Note that average kinetic energy is given as

K.E(avg) = 3/2K.T,

3/2 K.T = 3.6*10^-22 J

where K = 1.38*10^-23

T = (3.6*10^-22 J * 2) / (3 * 1.38*10^-23)

T = 17.39

Substitute for T, we have

n = 58.94 / 17.39

n = 3.39

The number of moles of helium that are in the spherical balloon is equal to 3.39 moles.  

Given the following data:

Scientific data:

To determine the number of moles of helium that are in the balloon, we we would use the ideal gas law equation;

[tex]n=\frac{PV}{RT}[/tex]   ...equation 1.

Where;

First of all, we would determine the temperature of the spherical balloon by using this formula:

[tex]T=\frac{2}{3} \frac{K_E}{k} \\\\T = \frac{2}{3}\times \frac{3.60 \times 10^{-22}}{1.38 \times 10^{-23}} \\\\T = \frac{2}{3}\times 26.09\\\\T=17.39\;K[/tex]

Temperature, T = 17.39 K

Substituting the parameters into eqn. 1, we have;

[tex]n=\frac{1.25 \times 10^5\times 3.92 \times 10^{-3}}{8.314 \times 17.39}\\\\n=\frac{490}{144.58}[/tex]

n = 3.39 moles.

Read more: https://brainly.com/question/24031197

The car undergoes an average acceleration a of

a = (29.4 m/s - 17.3 m/s) / (58 s) ≈ 0.209 m/s²

so the force has a magnitude F of

F = (1200 kg) a ≈ 250 N

Answer A material that decreases the rate of a reaction. The amount of energy that reactants must absorb before a chemical reaction will start also called free energy of activation. What role does activation energy pay in chemical reactions? Activation energy is the energy absorbed before it can start a chemical reaction.

Explanation:

Answer:

a. False

Explanation:

For an object to be moving at a constant velocity, a net force of 0 N would be required.

Newton's 1st Law of Motion states that an object will remain at rest unless acted upon by an unbalanced force, and an object will remain in motion unless acted upon by an unbalanced force.

Therefore, the unbalanced force of 6 N would not allow the rock to maintain its constant speed.

The answer to this question is A) False.

That's false.

NO FORCE is required to keep an object moving at a constant speed.

Answer:

-Electric field will be the greatest at the point where it is closest to it's charges.

- No, there cannot be a point where the field will be zero

Explanation:

Usually, in field lines and vectors, the principle is that: Where the field lines are in a close manner together, the field will be strongest. However, where the field lines are in a manner far apart, the field will be the weakest.

Applying that to our question now it means that the electric field will be the greatest at the point where it is closest to it's charges.

In unlike charges, there can never be a field of zero in between the 2 opposite charges because the field vectors from both unlike charges will point towards the negative charge.

The magnitude of the electric field will be the greatest at the point where it is closest to its charges.

No, there cannot be a point where the field will be zero.

The region where an electrostatic force is experienced by a charged entity is known as the electric field at the point.

As per the principle of field lines and vectors, where the field lines are in a close manner together, the field will be strongest. However, where the field lines are in a manner far apart, the field will be the weakest.

As per the above concept, the electric field will be the greatest at the point where it is closest to it's charges. In unlike charges, there can never be a field of zero in between the two opposite charges because the field vectors from both unlike charges will point towards the negative charge.

Thus, we can conclude that there cannot be a point where the field will be zero.

Learn more about electric field here:

https://brainly.com/question/4440057

Answer:

133.33 kg/s

Explanation:

Calculation for how much coal is used per hour

Using this formula

Coal used per hour=(Net MW of electricity*1,000 / Efficiency%* Released kJ/kg

Let plug in the formula

Coal used per hour=1,500 MW*1,000/0.45*25,000 kJ/kg

Coal used per hour=1,500,000/11,250kJ/kg

Coal used per hour=133.33 kg/s

Therefore how much coal is used per hour is 133.33 kg/s

Answer:

The value is [tex]\eta = 54.4 \%[/tex]

Explanation:

From the question we are told that

    The work input is  [tex]W = ( 245 + A ) \ J[/tex]

     The heat delivered is [tex]Q = (142 + B) \ J[/tex]

      The value of A is  A =  14

        The value of B  is  B  = 72

Generally the efficiency of the heat engine is mathematically represented as

          [tex]\eta = \frac{W}{Q_t}[/tex]

Here  [tex]Q_t[/tex] is the total out energy produce by the heat engine and this is mathematically represented as

           [tex]Q_t= Q + W[/tex]

=>         [tex]Q_t= 245 + A + 142 + B[/tex]

=>         [tex]Q_t= 390 + A+B[/tex]

So

               [tex]\eta = \frac{245 + A }{390 + A+ B}[/tex]

=>          [tex]\eta = 0.544[/tex]

=>          [tex]\eta = 0.544 *100[/tex]

=>          [tex]\eta = 54.4 \%[/tex]