If the intensity of a loud car horn is 0.005 W/m^2 when you are 2 meters away from the source. Calculate the sound intensity level. A. 1.6 WB. 0.06 W
C. 97 dB
D. 223 dB
E. 179 dB

Answers

Answer 1

Answer:

Answer:

Explanation:

We have given the intensity of the loud car horn $I=0.005w/m^2$

We know that $I_O=10^(-12)w/m^2$

Now the sound intensity level is given by $\beta =10log(I)/(I_0)=10log(0.005)/(10^(-12))=96.98dB$ , which is nearly equal to 97

So the sound intensity level will be 97 dB

So option (c) will be the correct option

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An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?

Answers

Answer:

11405Volt

Explanation:

To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.

The equation that allows the calculation of this voltage is given by,

$\epsilon = BAN \omega$

Where

B = Magnetic field

A= Area

N = Number of loops

$\omega$ = Angular velocity

Our values previously given are:

$N = 140$

$A = 71.2*10^(-2)m*22.6*10^(-2)m=0.1609m^2$

$B = 4.32 T$

$\omega = 1120 rev / min$

We need convert the angular velocity to international system, then

$\omega = 1120 rev/min$

$\omega = 1120rev/min*(2\pi)/(1rev)*(1min)/(60sec)$

$\omega = 117.2rad/s$

Applying the equation for emf, we replace the values and we will obtain the value.

$\epsilon = BAN \omega$

$\epsilon = (4.32)(0.1609)(140)*117.2$

$\epsilon = 11405Volt$

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?b)How long are they in the air?

Answers

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\n\Rightarrow -u^2=2as-v^2\n\Rightarrow u=√(v^2-2as)\n\Rightarrow u=√(0^2-2* -9.81* 1)\n\Rightarrow u=4.45\ m/s$

a) The vertical speed when the player leaves the ground is 4.45 m/s

$v=u+at\n\Rightarrow t=(v-u)/(a)\n\Rightarrow t=(0-4.45)/(-9.81)\n\Rightarrow t=0.45\ s$

Time taken to reach the maximum height is 0.45 seconds

$s=ut+(1)/(2)at^2\n\Rightarrow 1=0t+(1)/(2)* 9.81* t^2\n\Rightarrow t=\sqrt{(1* 2)/(9.81)}\n\Rightarrow t=0.45\ s$

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.a. in what direction, relative to the shore, does dave’s boat go?
b. how long does it take dave to cross the river?
c. how far downstream is dave’s landing point?
d. how long would it take dave to cross the river if there were no current?

Answers

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

$\theta= arctan((v_y)/(v_x))=arctan((4.0 m/s)/(6.0 m/s))=arctan(0.67)=33.7^(\circ)$

relative to the direction of the river.

b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

$t=(S_y)/(v_y)=(360 m)/(4.0 m/s)=90 s$

c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

$S_x = v_x t =(6.0 m/s)(90 s)=540 m$

so, Dave's landing point is 540 m downstream.

d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).

Your classmate’s mass is 63 kg and the table weighs 500 N. Calculate the normal force on the table by the floor. Show your work!

Answers

Answer:

$F_N=1234.8N$

Explanation:

Hello.

In this case, since the normal force is opposite to the total present weight, we can compute it by considering the mass of the classmate with the gravity to compute its weight, and the weight of the table:

$F_N=63kg*9.8m/s^2+500N\n\nF_N=617.4N+500N\n\nF_N=1234.8N$

Best regards.

Two cars, one with mass mi = 1200 kg is traveling north, and the other a large car with mass m2= 3000 kg is traveling East. If they collide while each car is traveling with a speed v = 5.0 m/s, what is the car's final speed and direction (vector notation can be given as well). Assume the collision is perfectly inelastic.

Answers

Answer:

Both cars travel at < 10 , 4 > m/s

Explanation:

Conservation of Linear Momentum

The total momentum of a system of particles of masses m1 and m2 traveling at velocities v1 and v2 (vectors) is given by

$\vec p_1=m_1\vec v_1+m_2\vec v_2$

When the particles collide, their velocities change to v1' and v2' while their masses remain unaltered. The total momentum in the final condition is

$\vec p_2=m_1\vec v'_1+m_2\vec v'_2$

We know the collision is perfectly inelastic, which means both cars stick together at a common final velocity v'. Thus

$\vec p_2=m_1\vec v'+m_2\vec v'=(m_1+m_2)\vec v'$

Both total momentums are equal:

$m_1\vec v_1+m_2\vec v_2=(m_1+m_2)\vec v'$

Solving for v'

$\displaystyle v'=(m_1\vec v_1+m_2\vec v_2)/(m_1+m_2)$

The data obtained from the question is

$m_1=1200\ kg$

$m_2=3000\ kg$

The first car travels north which means its velocity has only y-component

$\vec v_1=<0,5>$

The second car travels east, only x-component of the velocity is present

$\vec v_2=<5,0>$

Plugging in the values

$\displaystyle v'=(1200<0,5>+3000<5,0>)/(1200+3000)$

$\displaystyle v'=(<0,6000>+<15000,0>)/(1500)$

$\displaystyle v'=(<15000,6000>)/(1500)=<10,4>\ m/s$

The magnitude of the velocity is

$|v'|=√(10^2+4^2)=10.77\ m/s$

And the angle

$\displaystyle tan\alpha=(4)/(10)=0.4$

$\alpha=21.8^o$

Proposed Exercise - Circular MovementConsider four pulleys connected by correals as illustrated in the figure below. One motor moves the A pulley with angular acceleration A= 20 rad/s^2/. If pulley A is initially moving with angular acceleration A= 40 rad/s^2, determine the angular speed of pulleys B and C after three seconds. Consider that the belts do not slide

Answers

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

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If the intensity of a loud car horn is 0.005 W/m^2 when you are 2 meters away from the source. Calculate the sound intensity level. A. 1.6 WB. 0.06 WC. 97 dBD. 223 dBE. 179 dB

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If the intensity of a loud car horn is 0.005 W/m^2 when you are 2 meters away from the source. Calculate the sound intensity level. A. 1.6 WB. 0.06 W
C. 97 dB
D. 223 dB
E. 179 dB