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2 years ago

A student says that the two terms speed and frequency of the wave refer to the same thing. What is your response?

Physics

1 answer:

rjkz [21]2 years ago

3 0

Answer:

He is wrong!

Explanation:

Frequency refers to how many wave lengths pass in a second and speed is how fast the wave is traveling for example the speed of light goes really fast but has a mid-level frequency.

Hope this helps! ;-)

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You're driving your new sports car at 85 mph over the top of a hill that has a radius of curvature of 525 m.

Bumek [7]

Explanation:

It is given that,

Speed of the sports car, v = 85 mph = 37.99 m/s

The radius of curvature, r = 525 m

Let $W_N$ is the normal weight and $W_A$ is the apparent weight of the person. Its apparent weight is given by :

$W_A=mg-\dfrac{mv^2}{r}$

So, $\dfrac{W_A}{W_N}=\dfrac{mg-\dfrac{mv^2}{r}}{mg}$

$\dfrac{W_A}{W_N}=\dfrac{g-\dfrac{v^2}{r}}{g}$

$\dfrac{W_A}{W_N}=\dfrac{9.8-\dfrac{(37.99)^2}{525}}{9.8}$

$\dfrac{W_A}{W_N}=0.719$

$\dfrac{W_A}{W_N}=71.9\%$

Hence, this is the required solution.

5 0

3 years ago

Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air

valentina_108 [34]

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: $v_f=v_0+a*t$ , where $v_f$ is the final velocity, $v_0$ is the initial velocity, $a$ the acceleration, and $t$ is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: $t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s} }{-201\frac{m}{s^2} }=1.40s$ , where $v_f=0\frac{m}{s}$ because the sled is totally stopped, $v_0=282\frac{m}{s}$ is the velocity of the sled before braking and, $a=-201\frac{m}{s^2}$ is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration: $x_f-x_0=v_0*t+\frac{1}{2}*a*t^2$ , where $x_f-x_0$ is the distance traveled, $v_0$ is the initial velocity, $t$ the time of the process and, $a$ is the acceleration of the process.

Then for this case the relationship becomes: $x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m$ .

<u>Note that the acceleration is negative because is a braking process.</u>

4 0

3 years ago

Read 2 more answers

Help!! If an atom has 7 protons and 8 neutrons, and is not in an excited state, how many electrons would Bohr say the atom has?

quester [9]

The answer I believe is 6

8 0

3 years ago

Read 2 more answers

A frog jumps for 4.0 seconds at a maximum horizontal distance of 0.8m. what is its velocity along the road?

Lerok [7]

Answer:

The frog's horizontal velocity is 0.2 m/s.

Explanation:

To solve this problem, we must first remember what velocity is and how we solve for it. Velocity can be solved for using the formula x/t, where x represents horizontal distance and t represents time (in seconds), that it takes to travel this distance. If we plug in the given numbers for these variables and solve, we get the following:

v = x/t

v = 0.8m/4s

v = 0.2 m/s

Therefore, the correct answer is 0.2 m/s. We can verify that these units are correct because the formula calls for distance divided by time, so meters per second is a sensible answer.

Hope this helps!

3 0

2 years ago

30.42

CaHeK987 [17]

Answer: Try C

Explanation:

It's the only one that makes since.

7 0

2 years ago