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4 years ago

14

How can Newton's laws be used when addressing the motion of objects?

1 answer:

romanna [79]4 years ago

6 0

They can be used to calculate state of object (moving or stationary), acceleration of the moving object, forces acting on the object & reactions of it.

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olya-2409 [2.1K]

The moon orbiting the Earth

Explanation:

The motion of the moon orbiting the earth is a circular motion. Circular motion is simply the motion of an object in circle at constant speed.

A cannonball flying from a cannon is a projectile motion and not a circular motion.
A car moving along a straight track is a linear/translational motion.
Pendulum of a grandfather clock is a simple harmonic motion.

Learn more:

Circular motion brainly.com/question/2562955

#learnwithBrainly

7 0

3 years ago

A spring with a pointer attached is hanging next to a scale marked in millimeters. Three different packages are hung from the sp

SOVA2 [1]

solution;

$the expression for force applied on the spring due to the load is\\f=k\Delta x\\here,\Delta x is the extension in the spring due to appling force\\given three case as following\\110N=k(40-x_{o})----------1\\240N=k(60-x_{o})----------2\\w=k(30-x_{o})-------------3\\$ $To calculate the accrual length of the spring,solve th equation 1 and 2\\\frac{110N}{240N}=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458(60mm-x_{o})=(40mm-x_{o})\\x_{o}(1-0.458)=(40-60(0.458))mm\\x_{o}\frac{12.52}{0.542}\\=23.1mm\\to calculate the force on the spring in case,\\solve the equation 1 and 2\\\frac{110}{w}=\frac{k(40-x_{o})}{k(60-x_{o})}\\\frac{110}{w}=\frac{(40mm-23.1mm)}{30mm-23.1mm}\\w=\frac{110}{2.45}=44.9N$

8 0

3 years ago

What is the speed of the object between 60 to 70 seconds?

lyudmila [28]

We have that the speed of the object between 60 to 70 seconds

$V=\frac{d}{65}$

From the question we are told

The speed of the object between 60 to 70 seconds

Generally the equation for Average speed is mathematically given as

$V=\frac{d}{t}$

Where

$t=\frac{60+70}{2}\\\\t=65seconds$

Therefore

$V=\frac{d}{65}$

In conclusion

The speed of the object between 60 to 70 seconds

$V=\frac{d}{65}$

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brainly.com/question/23379286?referrer=searchResults

5 0

3 years ago

A uniform steel rod of cross-sectional area A is attached to rigid supports and is unstressed at a temperature of 458F. The stee

Vedmedyk [2.9K]

Explanation:

As the given rod is attached to rigid supports as a result, the deformation occurring due to the change in temperature will cause stress in the rod.

Let us assume that P is the compressive force in the rod due to change in temperature.

So, $\Delta T = \frac{\sigma_{y}}{E_{a}}$

= $\frac{36 \times 10^{3}}{(29 \times 10^{6} \times 6.5 \times 10^{-6})}$

= $190.98^{o}F$

Now, we will calculate the actual change in temperature as follows.

$\Delta T = 320 - 45 = 275^{o}F$

This means that the actual change in temperature is more than required for yielding.

(a) Formula to calculate yielding stress is as follows.

$\sigma' = \frac{P'}{A}$

$\sigma' = -\frac{AE \alpha \Delta T}{A}$

= $-E \alpha \Delta T$

= $-29 \times 10^{6} \times 6.5 \times 10^{-6} \times 275$

= $-51.8375 \times 10^{3} psi$

Hence, stress in the bar when temperature is raised to $320^{o}F$ is $-51.8375 \times 10^{3} psi$ .

(b) Now, we will calculate the residual stress as follows.

$\sigma_{r} = -\sigma_{y} - \sigma'$

$\sigma_{r} = -36 + 51.837 ksi$

= 15.837 ksi

Therefore, stress in the bar when the temperature has returned to $45^{o}F$ is 15.837 ksi.

6 0

3 years ago

Read 2 more answers

PLEASE ITS AN Emergency IF ITS RIGHT I WILL GIVE BRAINLIEST

n200080 [17]

Answer:

all of those are pisitions

Explanation:

6 0

3 years ago