A wave in the ocean has a wavelength of 2 m and a frequency of 0’5 Hz. What is the speed of this wave?

A rod of length L and electrical resistance R moves through a constant uniform magnetic field ; both the magnetic field and the

CaHeK987 [17]

Answer:

don't know what class are you you are using which mobile or laptop

6 0

2 years ago

A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.

dalvyx [7]

Answer:

a. Acceleration, a = 1.88 m/s²

b. Time, t = 7.87 seconds.

Explanation:

Given the following data;

Initial velocity, U = 14.5m/s

Final velocity, V = 29.3m/s

Distance, S = 172m

a. To find the acceleration of the speedboat;

We would use the third equation of motion;

V² = U² + 2aS

Substituting into the formula

29.3² = 14.5² + 2a*172

858.49 = 210.25 + 344a

344a = 858.49 - 210.25

344a = 648.24

a = 648.24/344

Acceleration, a = 1.88 m/s²

b. To find the time;

We would use the first equation of motion;

V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

6 0

3 years ago

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

4 0

3 years ago

Sound with frequency 1300 Hz leaves a room through a doorway with a width of 1.03 m . At what minimum angle relative to the cent

AlexFokin [52]

Answer:

about 14.7°

Explanation:

The formula for the angle of the first minimum is ...

sin(θ) = λ/a

where θ is the angle relative to the door centerline, λ is the wavelength of the sound, and "a" is the width of the door.

The wavelength of the sound is the speed of sound divided by the frequency:

λ = (340 m/s)/(1300 Hz) ≈ 0.261538 m

Then the angle of interest is ...

θ = arcsin(0.261538/1.03) ≈ 14.7°

At an angle of about 14.7°, someone outside the room will hear no sound.

6 0

2 years ago

The spacing between the plates of a 1.0 μF capacitor is 0.050 mm. a) What is the surface area of the plates? b) How much charge

nataly862011 [7]

Answer:

(a) surface area of the plate will be equal to $1.129m^2$

(b) Charge on the capacitor is equal to $1.5\times 10^{-6}C$

Explanation:

We have given spacing between the plates d = 0.05 mm = $5\times 10^{-5}m$

Value of capacitance $C=1\mu F=10^{-6}F$

(A) Capacitance of a parallel plate capacitor is equal to $C=\frac{\epsilon _0A}{d}$

So $10^{-6}=\frac{8.85\times 10^{-12}\times A}{10^{-5}}$

$A=1.129m^2$

So surface area of the plate will be equal to $1.129m^2$

(B) It is given that capacitor is charged by 1.5 volt

So voltage V = 1.5 volt

Charge on the capacitor is equal to $Q=CV$

So $Q=1.5\times 10^{-6}C$

5 0

2 years ago