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3 years ago

Please need help on this

Physics

1 answer:

lara [203]3 years ago

6 0

#6 - Lower frequencies correspond to <em>longer</em> wavelengths.

#7 - To change the pitch of a sound, you have to change its <em>frequency</em>.

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I could really use some help on this question guys! Will give brainliest!

aev [14]

Answer:

I think it’s the third one

4 0

2 years ago

Read 2 more answers

A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

cupoosta [38]

Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively:

$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

$\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}$

$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0

3 years ago

What is the net force on a car moving in a straight line with a constant velocity

guapka [62]

Answer:

For example, when a car travels at a constant speed, the driving force from the engine is balanced by resistive forces such as air resistance and friction in the car's moving parts. The resultant force on the car is zero.

Explanation:

hope this helps

3 0

2 years ago

Read 2 more answers

You collect some more data on that horse at a later time interval, but now you are measuring thehorse’s velocity, not its positi

Monica [59]

Answer:

a) x(t) = 10t + (2/3)*t^3

b) x*(0.1875) = 10.18 m

Explanation:

Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.

Given:

- v(t) = 10 + 2*t^2 (radar gun)

- x*(t) = 10 + 5t^2 + 3t^3 (our coordinate)

Find:

-The position x of horse as a function of time t in radar system.

-The position of the horse at x = 2m in our coordinate system

Solution:

- The position of horse according to radar gun:

v(t) = dx / dt = 10 + 2*t^2

- Separate variables:

dx = (10 + 2*t^2).dt

- Integrate over interval x = 0 @ t= 0

x(t) = 10t + (2/3)*t^3

- time @ x = 2 :

2 = 10t + (2/3)*t^3

0 = 10t + (2/3)*t^3 + 2

- solve for t:

t = 0.1875 s

- Evaluate x* at t = 0.1875 s

x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3

x*(0.1875) = 10.18 m

3 0

3 years ago

A substance has life of 40 years. In how many years will the substance disintegrate to 1/8 of its initial size

tia_tia [17]

Answer:

5 years

Explanation:

$\frac{1}{8}(40)\\\\= 5$

8 0

1 year ago