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1 year ago

If air resistance acts on a falling object will all of its potential energy be converted into kinetic energy?

Physics

1 answer:

Mademuasel [1]1 year ago

6 0

Answer:

No, some energy will be dissipated energy due to work of air resistance.

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A wave amplitude 0.36m interferes with a second wave of amplitude 0.22m traveling in the same direction. What is the largest res

DerKrebs [107]

The largest resultant amplitude would be that created by constructive interference, basically when the two waves are of the same phase, so it would be 0.36m+0.22m= 0.58 m.

8 0

3 years ago

I need help on 6, 7, 8, and 9

Ivahew [28]

6 is b. part B on 6 is a. 7 is a. partB ON 7 b

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3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t

Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

$\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}$

Generally, the equation for Rate of flow of Liquid is mathematically given as

$\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}$

Where dP is pressure difference r is the radius

$\mu$ is the viscosity of water

L is the length of the pipe

$Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}$

$Q=5.2 \mathrm{~mL} / \mathrm{s}$

In $30s the quantity that flows out of the tube

$&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}$

In conclusion, the quantity that flows out of the tube

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

Read more about the flows rate

brainly.com/question/27880305

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5 0

1 year ago

How do you find the force of a ping-pong ball rolling down a track?

Kaylis [27]

Answer:

Weight

a) weight's vertical component = Normal upward force

b) weight's horizontal component = Friction force = (mass of ball)(acceleration)

These forces depend upon the track,

1) inclined or horizontal

2) steepness.

Explanation

The force of gravity points straight down, but a ball rolling down a ramp doesn't go straight down, it follows the ramp. Therefore, only the component of the weight which points along the direction of the ball's motion can accelerate the ball.

weight's horizontal component = Friction force = (mass of ball)(acceleration)

The other component pushes the ball into the ramp, and the ramp pushes back.

If the ramp is horizontal, then the ball does not accelerate, as gravity pushes the ball into the ramp and not along the surface of the ramp. Hope this helps. Can u give me brainliest

Explanation:

3 0

2 years ago