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aleksley [76]
2 years ago
5

Two sinusoidal waves travel along the same string. They have the same wavelength and frequency. Their amplitudes are ym1 = 2.5 m

m and ym2 = 4.5 mm, and their phases are π / 4 rad and π / 2 rad, respectively. What are the amplitude and phase of the resultant wave?
Physics
1 answer:
Nimfa-mama [501]2 years ago
7 0

Answer:

0.5 m

Explanation:

Givens:

ym1 = 2.5 mm

ym2 = 4.5 mm

Ф_1=π / 4

Ф_2=π / 2

We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is

Ym = (ym1 + ym2)cos(Ф_2/2)

By substitution we have  

Ym= (0.025 + 0.045)cos(π/4) = 0.496 m

The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore  

Ym^2=(ym1^2+ym2^2)

So we have Ym=√0.025^2+0.045^2

                         = 0.5 m

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NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

F_1 + F_2 = (1530\times 9.81)

F_1 + F_2 = 15009.3 N

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

F_1(1.15) = F_2(2.70 - 1.15)

F_1 = 1.35 F_2

now from above equation

F_2 + 1.35F_2 = 15009.3

now we have

F_2 = 6392.85 N

now the other force is given as

F_1 = 8616.45 N

4 0
2 years ago
D=? V=100mL M=1.5kg=___g
Nana76 [90]
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8 0
2 years ago
For a magnetic field strength of 2T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical
nydimaria [60]

Answer:

Incomplete question

This is the complete question

For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter

Explanation:

Given the magnetic field

B=2T

Lenght of rod is 1mm

L=1/1000=0.001m

Diameter of rod=1.5mm

d=1.5/1000=0.0015m

Radius is given as

r=d/2=0.0015/2

r=0.00075m

Area of the circle is πr²

A=π×0.00075²

A=1.77×10^-6m²

Given that the voltage applied is 100mV

V=0.1V

Given that resistive is 0.6 Ωm

We can calculate the resistance of the cylinder by using

R= ρl/A

R=0.6×0.001/1.77×10^-6

R=339.4Ω

Then the current can be calculated, using ohms law

V=iR

i=V/R

i=0.1/339.4

i=2.95×10^-4 A

i=29.5 mA

The force in a magnetic field of a wire is given as

B=μoI/2πR

Where

μo is a constant and its value is

μo=4π×10^-7 Tm/A

Then,

B=4π×10^-7×2.95×10^-4/(2π×0.00075)

B=8.43×10^-8 T

Then, the force is given as

F=iLB

Since B=2T

F=iL(2B)

F=2.95×10^-4×2×8.34×10^-8

F=4.97×10^-11N

7 0
2 years ago
Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.
dimulka [17.4K]

Answer:

The distance of separation is d = 0.092 \ m

Explanation:

The mass of the each ball is  m= 10 g  =  0.01 \ kg

 The negative charge on each ball is q_1 =q_2=q =  1 \mu C  =  1 *10^{-6} \ C

Now we are told that the lower ball is  restrained from moving this implies that the net force acting on it is  zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

          F =  \frac{kq_1 * q_2}{d}

=>       m* g  =  \frac{kq_1 * q_2}{d}

here k the the coulomb's  constant with a value  k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

So  

      0.01 * 9.8  =  \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}

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5 0
2 years ago
Two wires are used to suspend a sign that weighs 500 N. The two wires make an angle of 100° between each other. If each wire is
vaieri [72.5K]
1) draw a diagram.
2) label diagram. (split the 100 degrees into 50, (which is right down the middle)  to make a right angle triangle.)
3) since its a free body diagram, the forces known must be labelled. (force of gravity). this shows that the straight vertical line of the right angle triangle is Fg (force gravity). label it.
4) use trigonometry. rearrange the equation to solve for what needs to be known.
 
angles known: 50 (split 100 in half to make a right angle triangle), 90 (since its right angle), and 40 (180-90-50 = 40)

sides known: vertical lined up with the 90 degree angle. Fg. --> fg=mg=500N x 9.81m/s^2 = 4905N

use formula: sin or cos 

i used sin. sin(40) = 4905 / ?
- times '?' on both sides. :  sin(40) x '?' = 4905
-divide both sides by sin(40):  '?' = 4905/ sin(40)
--> Solve.

8 0
2 years ago
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