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jok3333 [9.3K]
3 years ago
9

A measure of how much pulling, or tension, a material can withstand before breaking is -

Physics
2 answers:
andre [41]3 years ago
6 0
It should be C......................
const2013 [10]3 years ago
5 0

Its c. I just took the test thats how i know.

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Which cell process occurs only in organisms that
kenny6666 [7]

Answer:

c meiosis

Explanation:

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3 years ago
What advice would a personal trainer give you before a workout?
Katen [24]

Answer:

Have a quick snack an hour before to get the energy, warm up

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HELP PLZZZ!!!! Hurry
Mashcka [7]

Answer:

3. When the number of turns, N is doubled, the strength of the electromagnet is also doubled

4. Doubling the voltage, doubles the strength of the electromagnet

5. The number of paper clips a 7.5 V battery would pick is approximately 28 paper clips

The number of paper clips a 7.5 V battery would pick is 59 paperclips

6. For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is approximately 7 paperclips

For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is 16 paperclips

Explanation:

3. The Magnetomotive Force, MMF = The Number of Turns on the Coil, N × The Current I Flowing in the Coil, I

∴ MMF = N × I

When the number of turns, N is doubled, the magnetomotive force, MMF is also doubled, and the strength of the electromagnet is doubled

4. Given that the voltage, V applied to the coil = The current, I flowing × The resistance, R of the coil, we have

V = I × R

Therefore, for a given constant resistance in the coil, doubling the voltage, doubles the current and therefore doubles the strength of the electromagnet

5. The average slope for the 25-coil electromagnet = (23 - 12)/(6 - 3) = 3.\bar 6

The number of paper clips a 7.5 V battery would pick = 12 + (7.5 - 3) × 11/3 = 28.5 paperclips  ≈ 28 paper clips

The average slope for the 50-coil electromagnet = (48 - 26)/(6 - 3) = 7.\bar 3

The number of paper clips a 7.5 V battery would pick = 26 + (7.5 - 3) × 22/3 = 59 paperclips

6. The slope calculated from a start point of approximately 0.4 V, is given as follows;

The slope for the 25-coil electromagnet = (12 - 6)/(3 - 0.4) = 30/13

Therefore, for the 25-coil electromagnet,  the average number of paper clips a 1 V battery would pick = 6 + (1 - 0.4) × 30/13) = 96/13 ≈ 7 paperclips

The slope for the 50-coil electromagnet = (26 - 13)/(3 - 0.4) = 5

Therefore, for the 50-coil electromagnet,  the average number of paper clips a 1 V battery would pick = 13 + (1 - 0.4) × 5 = 16 paperclips

8 0
3 years ago
Read 2 more answers
A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef
bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

3 0
3 years ago
Please answer offering a lot of points and give brainliest pleaseeeee
asambeis [7]

23 is the right answer.

3 0
3 years ago
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