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2 years ago

A father pushes his child in a cart. The cart starts to move.

Physics

1 answer:

LuckyWell [14K]2 years ago

5 0

The potential energy of the car when it let go is 20,000 J.

The speed of the car at the bottom of the ramp is 20 m/s.

The given parameters;

mass of the car, m = 100 kg
height of the car, h = 20 m

The potential energy of the car is calculated as follows;

P.E = mgh

P.E = 100 x 10 x 20

P.E = 20,000 J

The speed of the car at the bottom of the ramp is calculated as follows;

$K.E = P.E\\\\\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2 \times 10 \times 20} \\\\v = 20 \ m/s$

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If a motorcycle accelerates uniformly from rest at 5 m / s2, how long does it take for it to reach a speed of 20 m / s.

irina1246 [14]

Answer:

4 seconds

Explanation:

Given:

v₀ = 0 m/s

v = 20 m/s

a = 5 m/s²

Find: t

v = at + v₀

20 m/s = (5 m/s²) t + 0 m/s

t = 4 s

6 0

2 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

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

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1 year ago

A theory is a hyothesis that has been varified by multiple investigations. True or false

irakobra [83]

A theory is a hyothesis that has been varified by multiple investigations.

true

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2 years ago

At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise betw

allsm [11]

To develop this problem it is necessary to apply the concepts related to the Dopler effect.

The equation is defined by

$f_i = f_0 \frac{c}{c+v}$

Where

$f_h$ = Approaching velocities

$f_i$ = Receding velocities

c = Speed of sound

v = Emitter speed

And

$f_h = f_0 \frac{c}{c+v}$

Therefore using the values given we can find the velocity through,

$\frac{f_h}{f_0}=\frac{c-v}{c+v}$

$v = c(\frac{f_h-f_i}{f_h+f_i})$

Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1

$v = 353(\frac{2.4-1}{2.4+1})$

$v = 145.35m/s$

Therefore the cars goes to 145.3m/s

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2 years ago

Can someone please help me with these. I don't understand any of it.

ValentinkaMS [17]

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4. the inside of the fridge and the food becomes colder

5. the coolants becomes a hot, high-pressure gas

6. as coolant transfers thermal energy to the air outside, it turns back into a liquid

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2 years ago