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Snowcat [4.5K]
2 years ago
15

After an initial push, a sled begins to move downhill at 1 m/s. A few seconds later its kinetic energy has increased. The sled n

ow has 64 times more kinetic energy than it did right after the push. How fast it it going?
Physics
1 answer:
Mila [183]2 years ago
6 0
1/6 is how much your neighbor will have the answer would be 8
You might be interested in
A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which
goldfiish [28.3K]

Answer:

160 Hz  ,  240 Hz  , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

f_1=\dfrac{320}{4}\ Hz

f₁=80 Hz

Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

Frequency of second harmonic = f₃

f₃= 3 f₁=3 x 80 = 240 Hz

Frequency of fifth harmonic = f₅

f₅=  5 f₁= 5 x 80 = 400 Hz

Three frequencies are as follows

160 Hz  ,  240 Hz  , 400 Hz

6 0
2 years ago
A.Whale communication. Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard n
Y_Kistochka [10]

A. 90.1 m

The wavelength of a wave is given by:

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is its frequency

For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is

\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m

B. 102 kHz

We can re-arrange the same equation used previously to solve for the frequency, f:

f=\frac{v}{\lambda}

where for the dolphin:

v = 1531 m/s is the wave speed

\lambda=1.50 cm=0.015 m is the wavelength

Substituting into the equation,

f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz

C. 13.6 m

Again, the wavelength is given by:

\lambda=\frac{v}{f}

where

v = 340 m/s is the speed of sound in air

f = 25.0 Hz is the frequency of the whistle

Substituting into the equation,

\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m

D. 4.4-8.7 m

Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:

- Wavelength corresponding to the minimum frequency (f=39.0 Hz):

\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m

- Wavelength corresponding to the maximum frequency (f=78.0 Hz):

\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m

So the range of wavelength is 4.4-8.7 m.

E. 6.2 MHz

In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so

\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m

And since the speed of the sound wave is

v = 1550 m/s

The frequency will be

f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz

3 0
2 years ago
A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average
Kaylis [27]

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

3 0
2 years ago
A spring is used to launch a coffee mug. The 20cm long spring can be compressed by a maximum of 8cm. The mug has a mass of 350 g
Zepler [3.9K]

Answer:

7503.13 N/m

Explanation:

Use principle of conservation of energy.

Here, energy stored in the spring due to compression shall be utilized in attaining the potential energy of the mug.

Given that,

Length of the spring = 20 cm = 0.20 m

Compression, x = 8 cm = 0.08 m

mass of the mug, m = 350 g = 0.35 kg

h = 7 m

use the expression for energy balance -

(1/2)*k*x^2 = m*g*h

=> k = (2*m*g*h) / x^2

input the values

k = (2*0.35*9.8*7) / 0.08^2

= 7503.13 N/m

8 0
2 years ago
Read 2 more answers
Mateo drew the field lines around the ends of two bar magnets but forgot to label the direction of the lines with arrows. At lef
Sladkaya [172]

Question:

Mateo drew the field lines around the ends of two bar magnets but forgot to label the direction of the lines with arrows. In which direction should an arrow at position 1 point?

left

right

up

down

Answer:

The correct answer is

Left

Explanation:

Magnetic circuits describe the path of a magnetic flux. In the same way electricity follows a complete closed circuit, the path of a magnetic flux is also a complete and closed circuit which leaves from the N pole, migrates through the air  and reenters the magnet through the S pole through which it passes back into the magnet to come to the N pole again.

As such the magnetic field lines emanate from the N pole which is on he right to the S pole which is on the left. Hence the arrow should point in the left direction.

3 0
2 years ago
Read 2 more answers
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