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eimsori [14]
3 years ago
12

What affect, if any, does increasing the speed of the plunger have on the wavelength of the waves

Physics
1 answer:
mojhsa [17]3 years ago
6 0

As the speed of the plunger increases, the wavelength of the waves decreases. The greater the frequency, the smaller the wavelength. The smaller the frequency, the greater the wavelength. When we increase the speed of the plunger, the frequency of the waves also increases, and just like with the size of the ball, it’s the speed of the plunger and the frequency of the waves are directly related.

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Neglecting air resistance, when does a struck baseball accelerate downward at a rate of 9.8 m/s2?
olga2289 [7]
A baseball is accelerated downward at 9.8 m/s^2 when the only vertical force acting upon the baseball is gravity.
3 0
4 years ago
Read 2 more answers
Keaton is asked to solve the following physics problem:
RideAnS [48]

Answer:

The answer is C "think about the problem first, systematically consider all factors, and form a hypothesis"

Explanation:

In physics there is some basic fomula that sir Isacc Newton proposed under the topic of motion. The three formulas are below;

<em>1) v=u+at</em>

<em>2)v^2=u^2+2as</em>

<em>3)s=ut+(1/2)(at^2)</em>

the variables are explained below;

u= initial velocity of the body

a=acceleration/Speed of the body

t= time taken by the body while travelling

s= displacement of the body.

Therefore to solve keatons problem, the factors(variables) in the formulas above need to be systematically considered. Since the ball was dropped from the top of the building, the initial velocity is 0 because the body was at rest. Also the acceleration will be acceleration due to gravity (9.8m/s^2)

5 0
4 years ago
A greyhound's velocity changes from rest to 19 m/s in 2 seconds. What is the greyhound's average acceleration?
Andrew [12]

It should be b)9.5m/s2

6 0
3 years ago
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 m/s when the
german

Answer: 4 s

Explanation:

Given

The ball leaves the hand of student with a speed of u=19\ m/s

When the hand is h=2.5\ m above the ground

Using the equation of motion we can write

h=ut+\dfrac{1}{2}at^2

Substitute the values

\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]

Thus, the ball will take 4 s to hit the ground.

5 0
3 years ago
A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking
Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}

The total motion has a total velocity and is

Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}

The time the worker take walking is

t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m

5 0
3 years ago
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