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2 years ago

The cheetah can run a distance of 275 m for 9 seconds. Calculate its speed.

Physics

1 answer:

AnnZ [28]2 years ago

4 0

Answer: 110000

Explanation:

26/9=30.5555555556

30.5555555556 x 60=1833.33333333

110000 x 60=110000

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You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner

omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

$Q_1 = Q_2$

$V_1A_1=V_2A_2$

Our values are given as,

$A_1=\frac{1}{2}^2*\pi=0.785 in^2$

$A_2=\frac{5}{8}^2*\pi=1.227 in^2$

Re-arrange the equation to find the first ratio of rates we have:

$\frac{V_1}{V_2}=\frac{A_2}{A_1}$

$\frac{V_1}{V_2}=\frac{1.227}{0.785}$

$\frac{V_1}{V_2}=1.56$

The second ratio of rates is

$\frac{V2}{V1}=\frac{A_1}{A2}$

$\frac{V2}{V1}=\frac{0.785}{1.227}$

$\frac{V2}{V1}=0.640$

3 0

3 years ago

Obtain the formula for the focal length of a lens in terms of object distance (u)

Alexxx [7]

Answer:

m=image distance÷object distance

6 0

1 year ago

How much kinetic energy does a system containing 3 moles of an ideal gas at 300 K possess? What is the heat capacity at constant

ElenaW [278]

Answer:

1. K.E = 11.2239 kJ ≈ 11.224 kJ

2. $C_{V} = 37.413 JK^{- 1}$

3. $Q = 10.7749 kJ$

Solution:

Now, the kinetic energy of an ideal gas per mole is given by:

K.E = $\frac{3}{2}mRT$

where

m = no. of moles = 3

R = Rydberg's constant = 8.314 J/mol.K

Temperature, T = 300 K

Therefore,

K.E = $\frac{3}{2}\times 3\times 8.314\times 300$

K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ

Now,

The heat capacity at constant volume is:

$C_{V} = \frac{3}{2}mR$

$C_{V} = \frac{3}{2}\times 3\times 8.314 = 37.413 JK^{- 1}$

Now,

Required heat transfer to raise the temperature by $15^{\circ}$ is:

$Q = C_{V}\Delta T$

$\Delta T = 15^{\circ} = 273 + 15 = 288 K$

$Q = 37.413\times 288 = 10774.9 J = 10.7749 kJ$

8 0

2 years ago

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow

Oksana_A [137]

Answer:

a) $\omega=1.36rad/s$

b) $\omega=12.99rpm$

c) $F=705.6N$

Explanation:

a) The angular velocity is related to the centripetal acceleration by the formula $a_{cp}=\omega^2r$ , which for our purposes we will write as:

$\omega=\sqrt{\frac{a_{cp}}{r}}$

Since <em>we want this acceleration to be 1.5 times that due to gravity</em>, for our values we will have:

$\omega=\sqrt{\frac{1.5g}{r}}=\sqrt{\frac{(1.5)(9.8m/s^2)}{(8m)}}=1.36rad/s$

b) 1 rpm (revolution per minute) is equivalent to an angle of $2\pi$ radians in 60 seconds:

$1\ rpm=\frac{2\pi rad}{60s} =\frac{\pi}{30}rad/s$

Which means <em>we can use the conversion factor</em>:

$\frac{1\ rpm}{\frac{\pi}{30}rad/s}=1$

So we have (multiplying by the conversion factor, which is 1, not affecting anything but transforming our units):

$\omega=1.36rad/s=1.36rad/s(\frac{1\ rpm}{\frac{\pi}{30}rad/s})=12.99rpm$

c) The centripetal force will be given by Newton's 2nd Law F=ma, so on the centripetal direction for our values we have:

$F=ma=(48kg)(1.5)(9.8m/s^2)=705.6N$

8 0

2 years ago

To test the performance of its tires, a car

yarga [219]

Answer:

0.43

Explanation:

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

There are friction forces in two directions: centripetal and tangential. The centripetal acceleration is:

ac = v² / r

ac = (35 m/s)² / 564 m

ac = 2.17 m/s²

The total acceleration is:

a = √(ac² + at²)

a = √((2.17 m/s²)² + (3.62 m/s²)²)

a = 4.22 m/s²

Sum of forces:

∑F = ma

Nμ = ma

mgμ = ma

μ = a / g

μ = 4.22 m/s² / 9.8 m/s²

μ = 0.43

5 0

3 years ago