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3 years ago

Which type of acceleration would be best if you were merging onto the interstate?

Physics

1 answer:

kondaur [170]3 years ago

4 0

Answer:

If merging onto an interstate, it is best to merge at the speed of the moving traffic.

Explanation:

As a driver, you cannot always count on the other drivers to see you wanting to merge onto the interstate, and even if they see you, they might not be willing to stop just so you merge, therefore it is best advised to look for a gap in the traffic and merge at the traffic speed.

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A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

$KE = PE$

$\frac{1}{2} mv ^2 = \frac{1}{2} k A^2$

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

$A = \sqrt{\frac{mv^2}{k}}$

Replacing,

$A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}$

$A = 0.0744m$

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Replacing,

$T = 2\pi \sqrt{\frac{0.750}{13}}$

$T= 1.509s$

Now the velocity is described as,

$v = \frac{2\pi}{T} * \sqrt{A^2-x^2}$

$v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}$

We have all the values, then replacing,

$v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}$

$v = 0.2049m/s$

7 0

2 years ago

The ammonia molecule (NH3) has a dipole moment of 5.0×10?30C?m. Ammonia molecules in the gas phase are placed in a uniform elect

Neko [114]

Question (continuation)

(a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E S from parallel to perpendicular?

(b) At what absolute temperature T is the average translational kinetic energy 3/2kT of a molecule equal to the change in potential energy calculated in part (a)?

Answer:

a. 9.0 * 10^-24 Joules

b. 0.44K

Explanation:

Given

Let p = dipole moment = 5.0 * 10^-30 Cm

Let E = Magnitude = 1.8 * 10^6 N/m

The charge in electric potential = Final Charge - Initial Charge

Initial Charge = Potential Energy

Initial Energy = -pE cosФ where Ф = 0

So, initial Energy = - 5.0 * 10^-30 * 1.8 * 10^6

Initial Energy = -9 * 10^-24 Joules

Final Energy = 0

Charge = 0 - (-9.0 * 10^-24)

Charge = 9.0 * 10^-24 Joules

Absolute Temperature

Change in Kinetic Energy = Change in Potential Energy = 9.0 * 10^-24

Change in Kinetic Energy = 3/2kT where k is Steven-Boltzmann constant = 1.38 * 10^-23

So,

9.0 * 10^-24 = 3/2 * 1.38 * 10^-23 * T

T = (9.0 * 10^-24 * 2)/(3 * 1.38 * 10^-23)

T = (18 * 10^-24)/(4.14 * 10^-23)

T = 0.44K

6 0

3 years ago

If 270 watts of power is used in 42 seconds, how much work was done

navik [9.2K]

Answer: W = 11340J

Explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

Step 1: Find the Formula

P = W / T

Step 2: Make W the subject of the equation.

W = PT

Step 3: Given.

P = 270 Watts, T = 42 seconds

Step 4: Substitute these values into equation 2 .

W = 270(42)

Step 5: Simplify.

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

4 0

3 years ago

A physical count of merchandise inventory on november 30

Zigmanuir [339]

It depends on the indirect taxes and the share market values

4 0

3 years ago

A book has been lifted 1.5 meters into the air giving it 30J of potential energy. How much force was used to lift it

joja [24]

Answer:

Force on the object is 20 N

Explanation:

As we know that work done to raise the book from initial position to final position is known as potential energy stored in it

So here we know that

$U = F.s$

here we know that

U = 30 J

s = displacement = 1.5 m

so we have

$30 = F(1.5)$

$F = 20 N$

6 0

3 years ago