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3 years ago

How to Identify Newtons Three Laws in my Own Words?

Physics

1 answer:

ipn [44]3 years ago

5 0

I understand the struggle. Here's a way to do it.

Identify Newton's Three Laws, and let it be important that you understand what you're reading. If there's a passage, read the passage/text and look for any important detail you should add to your piece, and add any information you have learned from the passage/text and put it into your own words. You should make sure that the information you've written down is consistent with the passage/text and that your phrasing is not too close to the original.

-I anticipate that this would help you.

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1. f=

ahrayia [7]

1. F = 8 Hz
2. V = 8 m/s
3. D = 8 m
4. T = 8 sec

5 0

3 years ago

Diagnostic ultrasound of frequency 3.82 MHz is used to examine tumors in soft tissue. (a) What is the wavelength in air of such

maxonik [38]

Explanation:

It is given that,

Frequency of diagnostic ultrasound, f = 3.82 MHz = 3820 Hz

The speed of the sound in air, v = 343 m/s

(a) We need to find the wavelength in air of such a sound wave. Let it is given by λ₁

i.e. $\lambda=\dfrac{v}{\nu}$

$\lambda_1=\dfrac{343\ m/s}{3820\ Hz}$

$\lambda_1=0.089\ m$

(b) If the speed of sound in tissue is 1650 m/s .

$\lambda_2=\dfrac{v}{\nu}$

$\lambda_2=\dfrac{1650\ m/s}{3820\ Hz}$

$\lambda_2=0.43\ m$

Hence, this is the required solution.

7 0

3 years ago

Joe and Max shake hands and say goodbye. Joe walks east 0.50 km to a coffee shop, and Max flags a cab and rides north 3.45 km to

timama [110]

Answer:

3.486 km

Explanation:

Suppose Joe and Max's directions are perfectly perpendicular (east vs north). We can calculate their distance at the destinations using Pythagorean theorem:

$s = \sqrt{J^2 + M^2}$

where J = 0.5 km and M= 3.45 km are the distances between Joe and Max to their original parting point, respectively. s is the distance between them.

$s = \sqrt{0.5^2 + 3.45^2} = \sqrt{12.1525} = 3.486 km$

8 0

3 years ago

The graph at the right shows the force needed to pull a bow back as the string is pulled further and further.

Sindrei [870]

A. 9 J

In a force-distance graph, the work done is equal to the area under the curve in the graph.

In this case, we need to extrapolate the value of the force when the distance is x=30 cm. We can easily do that by noticing that there is a direct proportionality between the force and the distance:

$F=kx$

where k is the slope of the line. We can find k, for instance chosing the point at x=5 cm and F=10 N:

$k=\frac{F}{x}=\frac{10 N}{5 cm}=2 N/cm$

And now we can calculate the work by calculating the area under the curve until x=30 cm, F=60 N:

$W=\frac{1}{2} (height) (base)= \frac{1}{2}(60 N)(0.30 m)=9 J$

B. 24.5 m/s

The mass of the arrow is m=30 g=0.03 kg. The kinetic energy of the arrow when it is released is equal to the work done by pulling back the bow for 30 cm:

$W=K=\frac{1}{2}mv^2$

where m is the mass of the arrow and v is its speed. By re-arranging the formula and using W=9 J, we find the speed:

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2\cdot 9J}{0.03 kg}}=24.5 m/s$

8 0

3 years ago

A jogger runs 4.0 km [W] in 0.50 h, then turns and runs 1.0 km [E] in 0.20 h, then 1.5 km [N] in 0.25 h, then 3.0 km [E] in 0.75

GrogVix [38]

This is a sneaky trick question, to help you discover whether you know
one of the differences between velocity and speed.
=======================================
If you make a list of the distances and directions, and ignore the times,
you find these:

4 - west, (3 + 1) - east . . . . . zero in the east/west direction

1.5 - north, 1.5 - south . . . . . zero in the north/south direction

This jogger went out, had a nice jog around the neighborhood,and ended up exactly where he started.

Average velocity = (distance between start point and end point) / (time)

IF the question asked for average SPEED, then you would need the total distance, and divide it by the total time. But it asks for VELOCITY, and that only involves the straight distance between the start point and the end point, regardless of the route taken in between.

The jogger ended up exactly where he started. The distance between start and end points was zero. Average velocity is (zero) / (time) . And that fraction is going to be Zero, no matter how long or how short the trip was, and no matter how much time it took.

3 0

3 years ago

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