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Darina [25.2K]
3 years ago
10

A swimmer standing near the edge of a lake notices a cork bobbing in the water. While watching for one minute, she notices the c

ork bob (from up to down to back up) 240 times. What is the frequency in Hz of the water wave going by?
Physics
1 answer:
timurjin [86]3 years ago
5 0

Answer:

4Hz (240 cycles/60 seconds = 4 cycles/second)

Explanation:

hope this helped!

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A pronghorn, the fastest North American animal, is capable of running for at 18 m/s (40 mph) for 10 minutes. The time limit isn’
Otrada [13]

Explanation:

The increase in the body temperature of pronghorn, the fastest North American animal, results from the chemical energy of the pronghorn converting into kinetic energy with efficiency less than 100%. The remaining energy is converted into heat energy. Thus, raising the temperature of pronghorn.

Due to the chemical energy it gains both kinetic and heat energy.

3 0
3 years ago
A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend
Viefleur [7K]

a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how

long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175

s this is the time to fall from the top; it would take the same time to travel

upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175

= 0.35s

b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice

to solve this problem the time it takes to fall the final 0.13 m is: time it

takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to

fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it

takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m

is then twice this, or 0.08s

5 0
3 years ago
Read 2 more answers
15 points. give me the method.
AveGali [126]

Answer:

\boxed{{160 \:  m(s)}^{ - 1} }

Explanation:

if \:the \:  frequencies \: are \to \\   f_{1} =  640Hz  \\ and \\f_{2}   = 480Hz \:  \\ but \:  \boxed{v = f \gamma }:   f =  \frac{v}{ \gamma } \\ if \:  \gamma_{1}  -  \gamma _{2}  = 1 =  \gamma  \\ f_{1}  - f_{2}  = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 =  \boxed{{160 \:  m(s)}^{ - 1} }

5 0
2 years ago
A 0.095 kg tennis ball is traveling to the right at 40 m/s when it bounces of a wall and travels in the opposite direction it ca
Mashutka [201]

given that

mass of ball = 0.095 kg

initial velocity of ball towards the wall = 40 m/s

final velocity of the ball after it rebound = 30 m/s

now change in momentum is given as

\Delta P = m(v_f - v_i)

\Delta P = 0.095(30 - (-40))

\Delta P = 6.65 kg m/s

So change in momentum will be 6.65 kg m/s

3 0
3 years ago
If this decay has a half-life of 10.2 years, what mass of 60.8g carbon-14 will remain after 20.4 years
OLEGan [10]
20.4 years is 20.4/10.2 = 2 half-life cycles, which means a quarter of the starting mass or 15.2 g will remain after this time.
5 0
3 years ago
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