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Mashutka [201]
3 years ago
11

A scientist suspends a spring from a force sensor. She then pulls on the spring causing the spring to stretch 1.0 cm. She record

s the amount of force applied to the spring. She then stretches the spring 2.0 cm and measures the new force. She repeats the process several times, each time increasing the length the spring is stretched.
Physics
1 answer:
Nadusha1986 [10]3 years ago
8 0
This situation describes the Hooke's Law which states that "When an elastic object - such as a spring - is stretched, the increased length is called its extension. The extension of an elastic object is directly proportional to the force applied to it". The formula is <span>F = k × e , F for the force, k for spring constant expressed in N/m, e for extension in m. This equation works for as long the spring is not stretch too much because once it exceeded its limit, the spring will not return to its original length the moment the load is removed.</span>
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He has 1 1/30 gallons, or 31/30 gallons, you can find this by setting all the fractions to a common denominator and adding them
7 0
3 years ago
Infer why the output force exerted by a rake must be less than input force?
adell [148]
<h3><u>Answer and Explanation</u>;</h3>
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5 0
3 years ago
The car's initial speed was 15 m / s and the distance the car travels before it comes to a complete stop after the driver applie
pentagon [3]

Initial speed of the car (u) = 15 m/s

Final speed of the car (v) = 0 m/s (Car comes to a complete stop after driver applies the brake)

Distance travelled by the car before it comes to halt (s) = 63 m

By using equation of motion, we get:

\bf \longrightarrow  {v}^{2}  =  {u}^{2}  + 2as \\  \\ \rm \longrightarrow  {0}^{2}  =  {15}^{2}  + 2 \times a \times 63 \\  \\ \rm \longrightarrow 0 = 225 + 126a \\  \\ \rm \longrightarrow 126a =  - 225 \\  \\ \rm \longrightarrow a =  -  \dfrac{225}{126}  \\  \\ \rm \longrightarrow a =  - 1.78 \: m {s}^{ - 2}

\therefore Acceleration of the car (a) = -1.78 m/s²

Magnitude of the car's acceleration (|a|) = 1.78 m/s²

5 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.23 2.23 times a second. A tack is stuck in the t
elena-14-01-66 [18.8K]

Explanation:

The given data is as follows.

       Angular velocity (\omega) = 2.23 rps

     Distance from the center (R) = 0.379 m

First, we will convert revolutions per second into radian per second as follows.

             = 2.23 revolutions per second

             = 2.23 \times 2 \times 3.14 rad/s

             = 14.01 rad/s

Now, tangential speed will be calculated as follows.

  Tangential speed, v = R \times \omega

                               = 0.379 x 14.01

                               = 5.31 m/s

Thus, we can conclude that the tack's tangential speed is 5.31 m/s.

8 0
3 years ago
(FIRST TO ANSWER CORRECTLY WILL BE THE BRAINIEST!!!)How will the temperatures of the water in the beakers compare if an equal am
bulgar [2K]

Answer:

the answer is the temperatures of both beakers' water will increase by the same amount...

Explanation:

I know this because i just did it on study island

7 0
3 years ago
Read 2 more answers
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