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nirvana33 [79]
2 years ago
13

Which statement correctly describes the relationship between frequency and wavelength?

Physics
1 answer:
Reika [66]2 years ago
4 0

Answer:

Answer:

Option B

Explanation:

Option B talks about how the frequency of the wave increases the shorter it's wavelengths

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A woman throws a baseball straight up into the air. what is the baseball's acceleration at its highest point?
AleksandrR [38]
After the ball leaves the woman's hand, the only relevant force acting on the ball is the gravity, which produces an acceleration of a=g=9.81 m/s^2 on the baseball, directed downward. 
Gravity acts on the ball for the whole duration of the motion, so the acceleration of the ball is g=9.81 m/s^2 at every point of the motion, and therefore also when the baseball is at its highest point.
5 0
3 years ago
A ball is pushed into a spring-loaded launcher with a force of 20 N, which compresses the spring 0.08 m.
damaskus [11]

Answer:

0.8J

Explanation:

Given parameters:

Force  = 20N

Compression  = 0.08m

Unknown:

Spring constant  = ?

Elastic potential energy  = ?

Solution:

To solve this problem, we use the expression below:

           F = k e

F is the force

k is the spring constant

e is the compression

             20  = k x 0.08

              k  = 250N/m

Elastic potential energy;

       EPE  = \frac{1}{2} k e²    =  \frac{1}{2}  x 250 x 0.08²

 Elastic potential energy = 0.8J

3 0
2 years ago
In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th
lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, \theta=20.0°

Mass of package is, m=2.50\ kg

Initial speed of package is, u=2.00\ m/s

Final speed of the package at the bottom is, v=0\ m/s

Distance of travel along the incline is, d=12.0\ m

Acceleration due to gravity is, g=9.8\ m/s^2

Let the coefficient of kinetic friction be \mu.

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

a=\mu g\cos\theta-g\sin\theta ----------------- 1

Now, using equation of motion, we have:

v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)

Solving for 'a', we get:

-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2

Now, plug in the value of 'a' in equation (1). This gives,

\mu g\cos\theta-g\sin\theta=\frac{1}{6} ( Neglecting negative sign)

Plug in all the given values and solve for \mu. This gives,

9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382

Therefore, the coefficient of kinetic friction is 0.382.

5 0
4 years ago
The melting point of potassium thiocyanate determined by a student in the laboratory turned out to be 174.5 oC. The accepted val
yaroslaw [1]

Answer:

0.75%

Explanation:

Measured value of melting point of potassium thiocyanate = 174.5 °C

Actual value of melting point of potassium thiocyanate = 173.2 °C

<em>Error in the reading = |Experimental value - Theoretical value|</em>

<em>= |174.5 - 173.2|</em>

<em>= |1.3|</em>

<em>Percentage error = (Error / Theoretical value) × 100</em>

<em>= (1.3 / 173.2)×100</em>

<em>= 0.75 %</em>

∴ Percentage error in the reading is 0.75%

4 0
3 years ago
You have a string with a mass of 0.0133 kg. You stretch the string with a force of 8.89 N, giving it a length of 1.97 m. Then, y
Kazeer [188]

Answer:

(i) The wavelength is 0.985 m

(ii) The frequency of the wave is 36.84 Hz

Explanation:

Given;

mass of the string, m = 0.0133 kg

tensional force on the string, T = 8.89 N

length of the string, L = 1.97 m

Velocity of the wave is:

V = \sqrt{\frac{F_T}{M/L} } \\\\V = \sqrt{\frac{8.89}{0.0133/1.97} } \ = 36.29 \ m/s

(i) The wavelength:

Fourth harmonic of a string with two nodes, the wavelength is given as,

L = 2λ

λ = L/2

λ = 1.97 / 2

λ = 0.985 m

(ii) Frequency of the wave is:

v = fλ

f = v / λ

f = 36.29 / 0.985

f = 36.84 Hz

3 0
4 years ago
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