What is the velocity of a rocket that goes 700 km north in 25 seconds?

The intensity of a certain siren is 0.060 W/m^2 from a distance of 25 meters. What is the intensity of this siren at a distance

Nady [450]

<h2>The intensity of siren is 0.015 W m⁻²</h2>

Explanation:

The intensity of sound is inversely proportional to the square of distance between source and the observer .

Thus Intensity at a distance 25 m is I₁ ∝ $\frac{1}{r_1^2}$ I

Similarly the intensity at a distance 50 m is I₂ ∝ $\frac{1}{r_2^2}$ II

Dividing II by I , we have

$\frac{I_2}{I_1}$ = $\frac{r_1^2}{r_2^2}$ = $\frac{25^2}{50^2}$ = $\frac{1}{4}$

Thus I₂ = $\frac{1}{4}$ x 0.06 = 0.015 W m⁻²

7 0

3 years ago

A track begins at 0 meters and has a total distance of 100 meters. Juliet starts at the 10-meter mark while practicing for a rac

shusha [124]

Answer : The correct option is (D).

Explanation :

Given that,

A track begins at 0 meters and has a total distance of 100 meters. Juliet starts at the 10-meter mark while practicing for a race.

We have to find her position after she runs 45 meters.

From the attached figure,

Let A is the position of Juliet. O is the initial point such that OA = 10 m, AB = 45 m and OP = 100 m.

So, using simple mathematics, it is clear that the position of Juliet after running 45 meters will be 55 m. It is OB in the figure.

So, the correct option is (D) " 55 meters ".

3 0

2 years ago

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The energy delivered to the resistive coil is dissipated as heat at a rate equal to the power input of the circuit. However, not

Nuetrik [128]

Answer:

P(bat) = V²r/(R+r)²

Explanation:

Let the resistance of the coil be R

Internal resistance of the battery be r

Emf of the battery = V

Power dissipated in the internal resistance of the battery is normally given as P = I²r

where I is the current flowing in the circuit.

From Ohm's law,

V = I R(eq)

R(eq) = (R + r)

I = V/(R+r)

P = I²r

P = [V/(R+r)]²r

P = V²r/(R+r)²

Hope this Helps!!!

6 0

2 years ago

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

The masses of the two moons are determined to be 2M2M for Moon AA and MM for Moon BB . It is observed that the distance between

seraphim [82]

Answer:

F_A = 8 F_B

Explanation:

The force exerted by the planet on each moon is given by the law of universal gravitation

F = $G \frac{m M}{r^{2} }$

where M is the mass of the planet, m the mass of the moon and r the distance between its centers

let's apply this equation to our case

Moon A

the distance between the planet and the moon A is r and the mass of the moon is 2m

F_A = G \frac{2m M}{r^{2} }

Moon B

F_B = G \frac{m M}{(2r)^{2} }

F_B = G \frac{m M}{4 r^{2} }

the relationship between these forces is

F_B / F_A = $\frac{1}{2 \ 4 }$ = 1/8

F_A = 8 F_B

7 0

2 years ago

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