Ask question

Ask question

All categories

3 years ago

10

A truck is traveling 70.0 kph away from you. The driver is blowing the horn, which has a frequency of 400 Hz. The air temperatur

e is 25°C. What is the observed frequency of the horn?

1 answer:

Helga [31]3 years ago

4 0

Use this site is better https://www.cymath.com/

You might be interested in

Which of the following is the human ear able to detect?

DaniilM [7]

I believe it would be a musical note

7 0

3 years ago

Read 2 more answers

Which statement is correct? (2 points) Select one:

Olin [163]

<span>The correct answer is B - Light can travel in a vacuum, and its speed is constant if the source is moving or stationary.</span>

7 0

3 years ago

A weightlifter is attempting a biceps curl with a 200 n barbell. the moment arm of the barbell about the elbow joint is 40 cm. t

CaHeK987 [17]

Weight of the barbell W = 200 Ndistance of the joint is r = 40 cm = 0.4 mtorque created by the weight at the joint is τ = F*r = 200 N*0.4 m = 80 N.mat equilibrium condition , Στ = force*distance - 80 N.m = 0 F'*0.4 - 80 N.m = 0 F'*0.4 = 80 force F' = 200 N

4 0

3 years ago

traveling at about 30 mph, how many feet will the average driver cover from the time they see the danger until they hit the brak

barxatty [35]

Answer: 75 ft

Explanation:

Breaking distance = Speed²/ 20

= 30²/20

= 45 feet

Stopping distance = Speed + braking distance

= 30 + 45

= 75 ft

5 0

3 years ago

A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =

castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

$emf=NBA\omega sin(\omega t)$

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

$emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V$

(b) the maximum rate of change of the magnetic flux is given by:

$\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}$

(c) $emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V$

(d) The torque is given by:

$\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm$

3 0

3 years ago